I have an HTML form in a JSP file in my WebContent/jsps folder. I have a servlet class servlet.java in my default package in src folder. In my web.xml it is mapped as /servlet.
I have tried several URLs in action attribute of the HTML form:
<form action="/servlet">
<form action="/servlet.java">
<form action="/src/servlet.java">
<form action="../servlet.java">
But none of those work. They all keep returning a HTTP 404 error like below in Tomcat 6/7/8:
HTTP Status 404 — /servlet
Description: The requested resource (/servlet) is not available.
Or as below in Tomcat 8.5/9:
HTTP Status 404 — Not Found
Message: /servlet
Description: The origin server did not find a current representation for the target resource or is not willing to disclose that one exists
Or as below in Tomcat 10:
HTTP Status 404 — Not Found
Type: Status Report
Message: The requested resource (/servlet) is not available
Description: The origin server did not find a current representation for the target resource or is not willing to disclose that one exists
Why is it not working?
Introduction
This can have a lot of causes which are broken down in following sections:
Put servlet class in a package
Set servlet URL in url-pattern
#WebServlet works only on Servlet 3.0 or newer
javax.servlet.* doesn't work anymore in Servlet 5.0 or newer
Make sure compiled *.class file is present in built WAR
Test the servlet individually without any JSP/HTML page
Use domain-relative URL to reference servlet from HTML
Use straight quotes in HTML attributes
Put servlet class in a package
First of all, put the servlet class in a Java package. You should always put publicly reuseable Java classes in a package, otherwise they are invisible to classes which are in a package, such as the server itself. This way you eliminate potential environment-specific problems. Packageless servlets work only in specific Tomcat+JDK combinations and this should never be relied upon.
In case of a "plain" IDE project, the class needs to be placed in its package structure inside the "Java Sources" folder, not inside "Web Content" folder, which is for web files such as JSP. Below is an example of the folder structure of a default Eclipse Dynamic Web Project as seen in Navigator view (the "Java Sources" folder is in such project by default represented by src folder):
EclipseProjectName
|-- src
| `-- com
| `-- example
| `-- YourServlet.java
|-- WebContent
| |-- WEB-INF
| | `-- web.xml
| `-- jsps
| `-- page.jsp
:
In case of a Maven project, the class needs to be placed in its package structure inside main/java and thus not main/resources, this is for non-class files and absolutely also not main/webapp, this is for web files. Below is an example of the folder structure of a default Maven webapp project as seen in Eclipse's Navigator view:
MavenProjectName
|-- src
| `-- main
| |-- java
| | `-- com
| | `-- example
| | `-- YourServlet.java
| |-- resources
| `-- webapp
| |-- WEB-INF
| | `-- web.xml
| `-- jsps
| `-- page.jsp
:
Note that the /jsps subfolder is not strictly necessary. You can even do without it and put the JSP file directly in webcontent/webapp root, but I'm just taking over this from your question.
Set servlet URL in url-pattern
The servlet URL is specified as the "URL pattern" of the servlet mapping. It's absolutely not per definition the classname/filename of the servlet class. The URL pattern is to be specified as value of #WebServlet annotation.
package com.example; // Use a package!
import jakarta.servlet.annotation.WebServlet; // or javax.*
import jakarta.servlet.http.HttpServlet; // or javax.*
#WebServlet("/servlet") // This is the URL of the servlet.
public class YourServlet extends HttpServlet { // Must be public and extend HttpServlet.
// ...
}
In case you want to support path parameters like /servlet/foo/bar, then use an URL pattern of /servlet/* instead. See also Servlet and path parameters like /xyz/{value}/test, how to map in web.xml?
Do note that it's considered a bad practice to use a Servlet URL pattern of /* or / in an attempt to have a "front controller". So do not abuse these URL patterns in an attempt to try to catch all URLs. For an in depth explanation see also Difference between / and /* in servlet mapping url pattern.
#WebServlet works only on Servlet 3.0 or newer
In order to use #WebServlet, you only need to make sure that your web.xml file, if any (it's optional since Servlet 3.0), is declared conform Servlet 3.0+ version and thus not conform e.g. 2.5 version or lower. It should absolutely also not have any <!DOCTYPE> line. Below is a Servlet 6.0 compatible one (which matches Tomcat 10.1+, WildFly 27+ (Preview), GlassFish/Payara 7+, etc) in its entirety:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns="https://jakarta.ee/xml/ns/jakartaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="https://jakarta.ee/xml/ns/jakartaee https://jakarta.ee/xml/ns/jakartaee/web-app_6_0.xsd"
version="6.0"
>
<!-- Config here. -->
</web-app>
And below is a Servlet 5.0 compatible one (which matches Tomcat 10.0.x, WildFly 22+ (Preview), GlassFish/Payara 6+, etc).
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns="https://jakarta.ee/xml/ns/jakartaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="https://jakarta.ee/xml/ns/jakartaee https://jakarta.ee/xml/ns/jakartaee/web-app_5_0.xsd"
version="5.0"
>
<!-- Config here. -->
</web-app>
And below is a Servlet 4.0 compatible one (which matches Tomcat 9+, WildFly 11+, GlassFish/Payara 5+, etc).
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
version="4.0"
>
<!-- Config here. -->
</web-app>
Or, in case you're not on Servlet 3.0+ yet (e.g. Tomcat 6 or older), then remove the #WebServlet annotation.
package com.example;
import javax.servlet.http.HttpServlet;
public class YourServlet extends HttpServlet {
// ...
}
And register the servlet instead in web.xml like this:
<servlet>
<servlet-name>yourServlet</servlet-name>
<servlet-class>com.example.YourServlet</servlet-class> <!-- Including the package thus -->
</servlet>
<servlet-mapping>
<servlet-name>yourServlet</servlet-name>
<url-pattern>/servlet</url-pattern> <!-- This is the URL of the servlet. -->
</servlet-mapping>
Note thus that you should not use both ways. Use either annotation based configuarion or XML based configuration. When you have both, then XML based configuration will override annotation based configuration.
javax.servlet.* doesn't work anymore in Servlet 5.0 or newer
Since Jakarta EE 9 / Servlet 5.0 (Tomcat 10, TomEE 9, WildFly 22 Preview, GlassFish 6, Payara 6, Liberty 22, etc), the javax.* package has been renamed to jakarta.* package.
In other words, please make absolutely sure that you don't randomly put JAR files of a different server in your WAR project such as tomcat-servlet-api-9.x.x.jar merely in order to get the javax.* package to compile. This will only cause trouble. Remove it altogether and edit the imports of your servlet class from
import javax.servlet.*;
import javax.servlet.annotation.*;
import javax.servlet.http.*;
to
import jakarta.servlet.*;
import jakarta.servlet.annotation.*;
import jakarta.servlet.http.*;
In case you're using Maven, you can find examples of proper pom.xml declarations for Tomcat 10+, Tomcat 9-, JEE 9+ and JEE 8- in this answer: How to properly configure Jakarta EE libraries in Maven pom.xml for Tomcat? The alternative is to downgrade the server to an older version, e.g. from Tomcat 10 back to Tomcat 9 or older, but this is clearly not the recommended way to go.
Make sure compiled *.class file is present in built WAR
In case you're using a build tool such as Eclipse and/or Maven, then you need to make absolutely sure that the compiled servlet class file resides in its package structure in /WEB-INF/classes folder of the produced WAR file. In case of package com.example; public class YourServlet, it must be located in /WEB-INF/classes/com/example/YourServlet.class. Otherwise you will face in case of #WebServlet also a 404 error, or in case of <servlet> a HTTP 500 error like below:
HTTP Status 500
Error instantiating servlet class com.example.YourServlet
And find in the server log a java.lang.ClassNotFoundException: com.example.YourServlet, followed by a java.lang.NoClassDefFoundError: com.example.YourServlet, in turn followed by jakarta.servlet.ServletException: Error instantiating servlet class com.example.YourServlet.
An easy way to verify if the servlet is correctly compiled and placed in classpath is to let the build tool produce a WAR file (e.g. rightclick project, Export > WAR file in Eclipse) and then inspect its contents with a ZIP tool. If the servlet class is missing in /WEB-INF/classes, or if the export causes an error, then the project is badly configured or some IDE/project configuration defaults have been mistakenly reverted (e.g. Project > Build Automatically has been disabled in Eclipse).
You also need to make sure that the project icon has no red cross indicating a build error. You can find the exact error in Problems view (Window > Show View > Other...). Usually the error message is fine Googlable. In case you have no clue, best is to restart from scratch and do not touch any IDE/project configuration defaults. In case you're using Eclipse, you can find instructions in How do I import the javax.servlet / jakarta.servlet API in my Eclipse project?
Test the servlet individually without any JSP/HTML page
Provided that the server runs on localhost:8080, and that the WAR is successfully deployed on a context path of /contextname (which defaults to the IDE project name, case sensitive!), and the servlet hasn't failed its initialization (read server logs for any deploy/servlet success/fail messages and the actual context path and servlet mapping), then a servlet with URL pattern of /servlet is available at http://localhost:8080/contextname/servlet.
You can just enter it straight in browser's address bar to test it invidivually. If its doGet() is properly overriden and implemented, then you will see its output in browser. Or if you don't have any doGet() or if it incorrectly calls super.doGet(), then a "HTTP 405: HTTP method GET is not supported by this URL" error will be shown (which is still better than a 404 as a 405 is evidence that the servlet itself is actually found).
Overriding service() is a bad practice, unless you're reinventing a MVC framework — which is very unlikely if you're just starting out with servlets and are clueless as to the problem described in the current question ;) See also Design Patterns web based applications.
Regardless, if the servlet already returns 404 when tested invidivually, then it's entirely pointless to try with a HTML form instead. Logically, it's therefore also entirely pointless to include any HTML form in questions about 404 errors from a servlet.
Use domain-relative URL to reference servlet from HTML
Once you've verified that the servlet works fine when invoked individually, then you can advance to HTML. As to your concrete problem with the HTML form, the <form action> value needs to be a valid URL. The same applies to <a href>, <img src>, <script src>, etc. You need to understand how absolute/relative URLs work. You know, an URL is a web address as you can enter/see in the webbrowser's address bar. If you're specifying a relative URL as form action, i.e. without the http:// scheme, then it becomes relative to the current URL as you see in your webbrowser's address bar. It's thus absolutely not relative to the JSP/HTML file location in server's WAR folder structure as many starters seem to think.
So, assuming that the JSP page with the HTML form is opened by http://localhost:8080/contextname/jsps/page.jsp (and thus not by file://...), and you need to submit to a servlet located in http://localhost:8080/contextname/servlet, here are several cases (note that you can here safely substitute <form action> with <a href>, <img src>, <script src>, etc):
Form action submits to an URL with a leading slash.
<form action="/servlet">
The leading slash / makes the URL relative to the domain, thus the form will submit to
http://localhost:8080/servlet
But this will likely result in a 404 as it's in the wrong context.
Form action submits to an URL without a leading slash.
<form action="servlet">
This makes the URL relative to the current folder of the current URL, thus the form will submit to
http://localhost:8080/contextname/jsps/servlet
But this will likely result in a 404 as it's in the wrong folder.
Form action submits to an URL which goes one folder up.
<form action="../servlet">
This will go one folder up (exactly like as in local disk file system paths!), thus the form will submit to
http://localhost:8080/contextname/servlet
This one must work!
The canonical approach, however, is to make the URL domain-relative so that you don't need to fix the URLs once again when you happen to move the JSP files around into another folder.
<form action="${pageContext.request.contextPath}/servlet">
This will generate
<form action="/contextname/servlet">
Which will thus always submit to the right URL.
Use straight quotes in HTML attributes
You need to make absolutely sure you're using straight quotes in HTML attributes like action="..." or action='...' and thus not curly quotes like action=”...” or action=’...’. Curly quotes are not supported in HTML and they will simply become part of the value. Watch out when copy-pasting code snippets from blogs! Some blog engines, notably Wordpress, are known to by default use so-called "smart quotes" which thus also corrupts the quotes in code snippets this way. On the other hand, instead of copy-pasting code, try simply typing over the code yourself. Additional advantage of actually getting the code through your brain and fingers is that it will make you to remember and understand the code much better in long term and also make you a better developer.
See also:
Our servlets wiki page - Contains some hello world examples
How to call servlet class from HTML form
doGet and doPost in Servlets
How do I pass current item to Java method by clicking a hyperlink or button in JSP page?
Other cases of HTTP Status 404 error:
HTTP Status 404 - Servlet [ServletName] is not available
HTTP Status 404 - The requested resource (/ProjectName/) is not available
HTTP Status 404 - The requested resource (/) is not available
JSP in /WEB-INF returns "HTTP Status 404 The requested resource is not available"
Referencing a resource placed in WEB-INF folder in JSP file returns HTTP 404 on resource
Browser can't access/find relative resources like CSS, images and links when calling a Servlet which forwards to a JSP
Scenario #1: You accidentially re-deployed from the command line while tomcat was already running.
Short Answer: Stop Tomcat, delete target folder, mvn package, then re-deploy
Scenario #2: request.getRequestDispatcher("MIS_SPELLED_FILE_NAME.jsp")
Short Answer: Check file name spelling, make sure case is correct.
Scenario #3: Class Not Found Exceptions
(Answer put here because: Question# 17982240 )
(java.lang.ClassNotFoundException for servlet in tomcat with eclipse )
(was marked as duplicate and directed me here )
Short Answer #3.1: web.xml has wrong package path in servlet-class tag.
Short Answer #3.2: java file has wrong import statement.
Below is further details for Scenario #1:
1: Stop Tomcat
Option 1: Via CTRL+C in terminal.
Option 2: (terminal closed while tomcat still running)
------------ 2.1: press:Windows+R --> type:"services.msc"
------------ 2.2: Find "Apache Tomcat #.# Tomcat#" in Name column of list.
------------ 2.3: Right Click --> "stop"
2: Delete the "target" folder.
(mvn clean will not help you here)
3: mvn package
4: YOUR_DEPLOYMENT_COMMAND_HERE
(Mine: java -jar target/dependency/webapp-runner.jar --port 5190 target/*.war )
Full Back Story:
Accidentially opened a new git-bash window and
tried to deploy a .war file for my heroku project via:
java -jar target/dependency/webapp-runner.jar --port 5190 target/*.war
After a failure to deploy, I realized I had two git-bash windows open,
and had not used CTLR+C to stop the previous deployment.
I was met with:
HTTP Status 404 – Not Found Type Status Report
Message /if-student-test.jsp
Description The origin server did not find a current representation
for the target resource or is not willing to disclose that one
exists.
Apache Tomcat/8.5.31
Below is further details for Scenario #3:
SCENARIO 3.1:
The servlet-class package path is wrong
in your web.xml file.
It should MATCH the package statement at top
of your java servlet class.
File: my_stuff/MyClass.java:
package my_stuff;
File: PRJ_ROOT/src/main/webapp/WEB-INF/web.xml
<servlet-class>
my_stuff.MyClass
</servlet-class>
SCENARIO 3.2:
You put the wrong "package" statement
at top of your myClass.java file.
For example:
File is in: "/my_stuff" folder
You mistakenly write:
package com.my_stuff
This is tricky because:
1: The maven build (mvn package) will not report any errors here.
2: servlet-class line in web.xml can have CORRECT package path. E.g:
<servlet-class>
my_stuff.MyClass
</servlet-class>
Stack Used:
Notepad++ + GitBash + Maven + Heroku Web App Runner + Tomcat9 + Windows10:
Check if you have entered the correct URL Mapping as specified in the Web.xml
For example:
In the web.xml, your servlet declaration maybe:
<servlet>
<servlet-name>ControllerA</servlet-name>
<servlet-class>PackageName.ControllerA</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ControllerA</servlet-name>
<url-pattern>/theController</url-pattern>
</servlet-mapping>
What this snippet does is <url-pattern>/theController</url-pattern>will set the name that will be used to call the servlet from the front end (eg: form) through the URL. Therefore when you reference the servlet in the front end, in order to ensure that the request goes to the servlet "ControllerA", it should refer the specified URL Pattern "theController" from the form.
eg:
<form action="theController" method="POST">
</form>
If you're using IntelliJ, this is what fixed it for me:
Go to the Tomcat configuration:
Configuration > Deployment Tab
Scroll down and add / to the Application Context dropdown
Solution for HTTP Status 404 in NetBeans IDE:
Right click on your project and go to your project properties, then click on run, then input your project relative URL like index.jsp.
Project->Properties
Click on Run
Relative URL:/index.jsp (Select your project root URL)
My issue was that my method was missing the #RequestBody annotation. After adding the annotation I no longer received the 404 exception.
Do the following two steps. I hope, it will solve the "404 not found" issue in tomcat server during the development of java servlet application.
Step 1: Right click on the server(in the server explorer tab)->Properties->Switch Location from workspace metadata to tomcat server
Step 2: Double Click on the server(in the server explorer tab)->Select Use tomcat installation option inside server location menu
I removed the old web library such that are spring framework libraries. And build a new path of the libraries. Then it works.
An old thread, but since I didn't find it elsewhere, here is one more possibility:
If you're using servlet-api 3.0+, then your web.xml must NOT include metadata-complete="true" attribute
This tells tomcat to map the servlets using data given in web.xml instead of using the #WebServlet annotation.
First of all, run your IDE as Admin. After that, right click the project folder -> Project Facets and make sure that the Java Version is set correct. On my PC. (For Example 1.8) Now it should work.
Don't just start your server, for example Wildfly, using the cmd. It has to be launched within the IDE and now visit your localhost URL. Example: http://localhost:8080/HelloWorldServlet/HelloWorld
The fix that worked for me is(if you are using Maven): Rightclick your project, Maven -> Update project. This might give you some other error with the JDK and other Libraries(in my case, MySQL connector), but once you fix them, your original problem should be fixed!
If you would like to open a servlet with javascript without using 'form' and 'submit' button, here is the following code:
var button = document.getElementById("<<button-id>>");
button.addEventListener("click", function() {
window.location.href= "<<full-servlet-path>>" (eg. http://localhost:8086/xyz/servlet)
});
Key:
1) button-id : The 'id' tag you give to your button in your html/jsp file.
2) full-servlet-path: The path that shows in the browser when you run the servlet alone
Mapping in web.xml is what i have done :-
If there's another package made for new program then we must mention :-
packagename.filename between opening and closing of servlet-class tag in xml file.
If you are mapping your files in xml and they are not working or showing errors , then comment on the annotation line of code in the respective files.
Both methods dont work with one another , so either i use annotation method of files mentioned when we create servlet or the way of mapping , then i delete or comment the annotation line. Eg:
<servlet>
<servlet-name>s1</servlet-name>
<servlet-class>performance.FirstServ</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>s1</servlet-name>
<url-pattern>/FirstServ</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>s2</servlet-name>
<servlet-class>performance.SecondServ</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>s2</servlet-name>
<url-pattern>/SecondServ</url-pattern>
</servlet-mapping>
Commenting the annotation line of code in the respective file,if mapping in xml is done.
//#WebServlet("/FirstServ")
//#WebServlet("/SecondServ")
If someone is here who is using MySQL and felt that the code was working the previous day and now it doesn't, then I guess you must open MySQL CLI or MySQL Workbench and just make the connection to the database once. Once it gets connected, then the database also gets connected to the Java Application. I used to get the Hibernate Dialect error stating something wrong with com.mysql.jdbc.Driver. I think MySQL in some computers has a startup problem. This solved for me.
If you are a student and new to Java there might be some issue going on with your web.xml file.
Try removing the web.xml file.
Secondly check that your path variables are properly set or not.
Restart tomcat server Or your PC.
Your problem will be surely solved.
I was facing this issue too, I was receiving a 404 when accessing a URL pattern that I knew was linked to a Servlet. The reason is because I had 2 Servlets with their #WebServlet name parameter set as the same string.
#WebServlet(name = "ServletName", urlPatterns = {"/path"})
public class ServletName extends HttpServlet {}
#WebServlet(name = "ServletName", urlPatterns = {"/other-path"})
public class OtherServletName extends HttpServlet {}
Both of the name parameters are the same. If you're using the name parameter, make sure they are unique compared to all other Servlets on your application.
I had the same issue. Tried all of this but didn't help. I managed to solve this issue by adding element tags to beginning and end of the xml file. ill leave my xml file below for reference.
<?xml version="1.0" encoding="UTF-8"?>
<element>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
version="3.1">
<servlet>
<servlet-name>InsertServlet</servlet-name>
<servlet-class>com.worklog.InsertServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>InsertServlet</servlet-name>
<url-pattern>/insert</url-pattern>
</servlet-mapping>
</web-app>
</element>
I was having the same issue. I was developing a mvc based REST API where there was no explicit html configuration or files. The API was using Swagger to generate a user interface. The problem started when I introduced Swagger version "3.0.0". I reverted back to Swagger "2.9.2" This solved my problem.
<!-- Swagger -->
<dependency>
<groupId>io.springfox</groupId>
<artifactId>springfox-swagger-ui</artifactId>
<version>2.9.2</version>
</dependency>
<dependency>
<groupId>io.springfox</groupId>
<artifactId>springfox-swagger2</artifactId>
<version>2.9.2</version>
</dependency>
Please check context root cannot be empty.
If you're using eclipse:
right click, select properties, then web project settings. Check the context root cannot be empty
When I open the services.xml in PhpStorm with Symfony Plugin enabled, it's able to resolve all the services, I can Ctrl+Click and go the Service Definition , except doctrine.orm.entity_manager.
It says unable to resolve symbol 'doctrine.orm.entity_manager'
Here is the services.xml file
<?xml version="1.0" ?>
<container xmlns="http://symfony.com/schema/dic/services"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://symfony.com/schema/dic/services http://symfony.com/schema/dic/services/services-1.0.xsd">
<services>
<service id="example_manager" class="Vendor\XysBundle\Manager\ExampleManager">
<argument type="service" id="doctrine.orm.entity_manager" />
</service>
</services>
</container>
PhpStorm Details:
PhpStorm 2016.3.1
Build #PS-163.9735.1, built on December 6, 2016
JRE: 1.8.0_112-release-408-b2 amd64
JVM: OpenJDK 64-Bit Server VM by JetBrains s.r.o
appDevDebugProjectContainer.xml file exists in the app/cache directory .
How can I get get this working ?
I have the same issue. If you open up the vendor\doctrine\doctrine-bundle\Resources\config\orm.xml file where doctrine services are defined, you can see, there is no service with the id "doctrine.orm.entity_manager" but only "doctrine.orm.entity_manager.abstract"
The definition looks like:
<service id="doctrine.orm.entity_manager.abstract" class="%doctrine.orm.entity_manager.class%" abstract="true" />
As you can see the abstract attribute is set to true. This means this service can serve as a parent of other services, and when you define child services with this abstract parent you don't have to define the method calls or the parameters injected into the constructor for example, instead these definitions will be inherited from the parent.
If you investigate a bit deeper you will find that %doctrine.orm.entity_manager.class% is defined in the same file as a parameter that actually references to the Doctrine Entity Manager:
<parameter key="doctrine.orm.entity_manager.class">Doctrine\ORM\EntityManager</parameter>
So I guess we should use doctrine.orm.entity_manager.abstract, however doctrine.orm.entity_manager is recognised as well, although I don't know how.
For further explanation of abstract service definitions have a look at this Symfony doc page: https://symfony.com/doc/current/service_container/parent_services.html
I have a Asp.net C# MVC 3 application implementing the Sharp Architecture. I have been trying to get Quartz.net to setup and work nicely with Castle Windsor for a few days without any luck. Based on what I know, I have setup everything correctly, but continue to have issues.
In my Global.cs file, creating my Container and trying to register quartz jobs:
var container = new WindsorContainer(new XmlInterpreter("quartz_jobs.xml"));
container.AddFacility("quartznet", new QuartzFacility());
In my quartz_jobs.xml file I have the following contents:
<?xml version="1.0" encoding="utf-8" ?>
<quartz xmlns="http://quartznet.sourceforge.net/JobSchedulingData"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
version="1.0"
overwrite-existing-jobs="true">
<job>
<job-detail>
<name>DeleteLoansWithoutClientsJob</name>
<job-type>EasyOptions.Web.Mvc.Code.Jobs.DeleteLoansWithoutClientsJob, EasyOptions.Web.Mvc</job-type>
<durable>true</durable>
</job-detail>
<trigger>
<cron>
<name>DeleteLoansWithoutClientsJobTrigger</name>
<group>MyJobs</group>
<description>A description</description>
<job-name>DeleteLoansWithoutClientsJob</job-name>
<job-group>MyJobs</job-group>
<cron-expression>0 0/1 * * * ?</cron-expression>
</cron>
</trigger>
</job>
Problem is, you're pointing Windsor to the Quartz.NET config file.
There are two separate configurations: Windsor's and Quartz.NET's. Windsor is usually configured with code nowadays (i.e. fluent config), though it still supports XML configuration. However the Quartz.NET facility doesn't currently support code config, you have to use Windsor's XML config (at least for this, other components/facilities may still be configured via code). Then there's Quartz.NET, usually configured via an external quartz_jobs.xml file.
I recommend using the Quartz.NET facility sample app as reference. In particular, here's the sample Windsor config and the sample Quartz.NET config.
EDIT: if Quartz.NET says it can't find quartz_jobs.xml in a web application you need to include the web root in the configuration path: "~/quartz_jobs.xml" (instead of plain "quartz_jobs.xml")
I've written a blog post on how to integrate Quartz.NET with an IoC container. My example code uses Castle Windsor.
The blog post can be found here: http://thecodesaysitall.blogspot.com/2012/02/integrate-quartznet-with-your-favourite.html
I have the following three components defined in the Caste-Windsor XML configuration for my application:
<component id="StringFactory"
service="IStringFactory, MyApp"
type="DefaultStringFactory, MyApp"
lifestyle="singleton"
/>
<component id="TheString"
type="System.String"
factoryId="StringFactory"
factoryCreate="CreateString"
>
<parameters>
<name>SomeString</name>
</parameters>
</component>
<component id="TheTarget"
service="ITarget, MyApp"
type="TheTarget, MyApp"
lifestyle="transient"
>
<parameters>
<aString>${TheString}</aString>
</parameters>
</component>
And the following facility defined:
<facility id="factory.support"
type="Castle.Facilities.FactorySupport.FactorySupportFacility, Castle.MicroKernel"
/>
When I run the application and set a breakpoint in the constructor of the TheObject class, the value passed in as the aString parameter is "${TheString}" when I expect it to resolve to the value of the component with that name.
Also, I have a breakpoint in the StringFactory constructor and CreateString method, neither of which are hit. I know the configuration is being used as other components are resolving correctly.
What am I missing or doing wrong here?
UPDATE
In light of the huge tangient this topic has taken, I've refactored the code above to remove anything to do with connection strings. The original intent of this post was about injecting a property with the value returned from a method on another object. Somehow that point was lost in a discussion about why I'm using XML versus code-based configuration and if this is a good way to inject a connection string.
The above approach is far from an original idea and it was pulled from several other discussions on this topic and our requirements are what they are. I'd like help understanding why the configuration as it is in place (whether the right approach or not) isn't working as expected.
I did verify that the first two components are being instantiated correctly. When I call Container.Resolve("TheString"), I get the correct value back. For whatever reason, The parameter syntax is not working correctly.
Any ideas?
While not a definitive solution to what I need to do in my application, I believe I've figured out what is wrong with the code. Or at least I've found a way to make it work which hints at the original problem.
I replaced the String type for TheString with a custom class. That's it. Once I did that, everything worked fine.
My guess is that it has something to do with the fact that I was trying to use a ValueType (primitive) as a component. I guess Castle doesn't support it.
So, knowing that's the case, I can now move on to figuring out if this approach is really going to work or if we need to change direction.
UPDATE
For the sake of completeness, I thought I'd go ahead and explain what I did to solve my problem AND satisfy my requirements.
As before, I have access to my configuration settings through an IConfigurationService defined as:
<component id="ConfigurationService"
service="MyApp.IConfigurationService, MyApp"
type="MyApp.RuntimeConfigurationService, MyApp"
lifestyle="singleton"
/>
This is automatically injected into my (new) IConnectionFactory which is responsible for generating IDbConnection objects based on the connection strings defined in the application's configuration file. The factory is declared as:
<component id="ConnectionFactory"
service="MyApp.Factories.IConnectionFactory, MyApp"
type="MyApp.Factories.DefaultConnectionFactory, MyApp"
lifestyle="singleton"
/>
In order to resolve what connection is used by my repository, I declare each connection as a component using the ConnectionFactory to create each instance:
<component id="MyDbConnection"
type="System.Data.IDbConnection,
System.Data, Version=2.0.0.0, Culture=neutral,
PublicKeyToken=b77a5c561934e089"
factoryId="ConnectionFactory"
factoryCreate="CreateConnection"
lifestyle="transient"
>
<parameters>
<connectionStringName>MyDB</connectionStringName>
</parameters>
</component>
Notice the fully described reference to System.Data. I found this is necessary whenever referencing assemblies in the GAC.
Finally, my repository is defined as:
<component id="MyRepository"
service="MyApp.Repositories.IMyRepository, MyApp"
type="MyApp.Sql.SqlMyRepository, MyApp.Sql"
lifestyle="transient"
>
<parameters>
<connection>${MyDbConnection}</connection>
</parameters>
</component>
Now everything resolves correctly and I don't have ANY hard-coded strings compiled into my code. No connection string names, app setting keys or whatever. The app is completely reconfigurable from the XML files which is a requirement I must satisfy. Plus, other devs that will be working with the solution can manage the actual connection strings in the way they are used to. Win-win.
Hope this helps anyone else that runs into a similar scenario.
You don't really need XML registrations here, since you probably don't need to swap components or change the method used without recompiling. Writing a configurable app does not imply having to use XML registrations.
The problem with this particular XML registration you posted is that the connection string is a parameter, but it's treated like a service.
Doing this with code registrations is much easier, e.g.:
var container = new WindsorContainer();
container.Register(Component.For<IConfigurationService>().ImplementedBy<RuntimeConfigurationService>());
container.Register(Component.For<ITheRepository>().ImplementedBy<TheRepository>()
.LifeStyle.Transient
.DynamicParameters((k, d) => {
var cfg = k.Resolve<IConfigurationService>();
d["connectionString"] = cfg.GetConnectionString();
k.ReleaseComponent(cfg);
}));
Or if you don't want to depend on IConfigurationService, you could do something like:
container.Register(Component.For<ITheRepository>().ImplementedBy<TheRepository>()
.LifeStyle.Transient
.DependsOn(Property.ForKey("connectionString")
.Is(ConfigurationManager.ConnectionStrings[ConfigurationManager.AppSettings["connName"]].ConnectionString))