I am using webapp2 to make a small api.
So for example if I have:
import webapp2
class Test(webapp2.RequestHandler):
def put(self):
self.response.write("this was a test")
app = webapp2.WSGIApplication([
('/test', Test)
])
And I do a request through curl:
curl --request PUT --header "Content-Type: application/json" --data '{"content": "test"}' http://localhost:8080/test
How would I go about accessing the data '{"content": "test"}' that was passed in?
All request data would be somewhere in self.request, so in this case take a look at self.request.body to find the request's contents and check out the documentation's Common Request attributes section to see the rest of the options.
You may also want to consider looking at the entire self object in a debugger to find out about more interesting properties it has.
Related
When I make a request, I get a response in XML, but what I need is JSON. In the doc it is stated in order to get a JSON in return: Use the Accept: application/json HTTP Header.
Where do I find the HTTP Header to put Accept: application/json inside?
My guess is it is not suppose to be inside the URL-request, which looks like:
http://localhost:8080/otp/routers/default/plan?fromPlace=52.5895,13.2836&toPlace=52.5461,13.3588&date=2017/04/04&time=12:00:00
You guessed right, HTTP Headers are not part of the URL.
And when you type a URL in the browser the request will be issued with standard headers. Anyway REST Apis are not meant to be consumed by typing the endpoint in the address bar of a browser.
The most common scenario is that your server consumes a third party REST Api.
To do so your server-side code forges a proper GET (/PUT/POST/DELETE) request pointing to a given endpoint (URL) setting (when needed, like your case) some headers and finally (maybe) sending some data (as typically occurrs in a POST request for example).
The code to forge the request, send it and finally get the response back depends on your server side language.
If you want to test a REST Api you may use curl tool from the command line.
curl makes a request and outputs the response to stdout (unless otherwise instructed).
In your case the test request would be issued like this:
$curl -H "Accept: application/json" 'http://localhost:8080/otp/routers/default/plan?fromPlace=52.5895,13.2836&toPlace=52.5461,13.3588&date=2017/04/04&time=12:00:00'
The H or --header directive sets a header and its value.
Here's a handy site to test out your headers. You can see your browser headers and also use cURL to reflect back whatever headers you send.
For example, you can validate the content negotiation like this.
This Accept header prefers plain text so returns in that format:-
$ curl -H "Accept: application/json;q=0.9,text/plain" http://gethttp.info/Accept
application/json;q=0.9,text/plain
Whereas this one prefers JSON and so returns in that format:-
$ curl -H "Accept: application/json,text/*;q=0.99" http://gethttp.info/Accept
{
"Accept": "application/json,text/*;q=0.99"
}
Basically I use Fiddler or Postman for testing API's.
In fiddler, in request header you need to specify instead of xml, html you need to change it to json.
Eg: Accept: application/json. That should do the job.
Well Curl could be a better option for json representation but in that case it would be difficult to understand the structure of json because its in command line.
if you want to get your json on browser you simply remove all the XML Annotations like -
#XmlRootElement(name="person")
#XmlAccessorType(XmlAccessType.NONE)
#XmlAttribute
#XmlElement
from your model class and than run the same url, you have used for xml representation.
Make sure that you have jacson-databind dependency in your pom.xml
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.4.1</version>
</dependency>
I’m using Parse.com, and trying to retrieve one object based on conditions values using Android REST API
here is a snippet from the Parse documentation for the REST API
curl -X GET \
-H "X-Parse-Application-Id: MYAPPID" \
-H "X-Parse-REST-API-Key: MYRESTKEY" \
-G \
--data-urlencode 'where={"$relatedTo":{"object":{"__type":"Pointer","className":"Post","objectId":"8TOXdXf3tz"},"key":"likes"}}' \
https://api.parse.com/1/users
How can I achieve this in android?
It's curl based api, you have to explicitly pass all its parameter.
I recommending you to first test api using client application like Postman.
-X define term like GET,POST,PUT,DELETE
-H define header, which has key-value form data separated by ":" sign.
-G When used, all data to be used in an HTTP GET request instead of the POST request that otherwise would be used. The data will be appended to the URL with a '?' separator.
Pass all these parameters and test it, once it is working fine then implement to your application.
I have this particular cURL statement that I am trying to figure out the equivalent for it in an HTTP Request.
curl -X POST --upload-file movie-data-2013.json doc-movies-123456789012.us-east-1.cloudsearch.amazonaws.com/2013-01-01/documents/batch --header "Content-Type:application/json"
So far the HTTP Request equivalent I am able to compile is below:
Set oHTTPRequest = Server.CreateObject( "MSXML2.XMLHTTP.3.0" )
oHTTPRequest.Open "POST" _
,"doc-movies-123456789012.us-east-1.cloudsearch.amazonaws.com/2013-01-01/documents/batch" _
False
oHTTPRequest.setRequestHeader "Content-type", "application/json"
oHTTPRequest.Send
For the most part I should be set, except there's one part it's missing, and I'm unsure how it's appended to the request.
There is a part in cURL that says: --upload-file movie-data-2013.json
Not sure how it applies in an HTTP Request. Does anyone how this applies?
I'd recommend trying .Net's built-in HttpWebRequest class over MSXML. There is an example of uploading text data on MSDN using the GetRequestStream method. You should be able to easily modify the sample to instead read the data from a file, for example using File.ReadAllBytes, and then write the returned data to the request stream.
I admit I'm a greenhorn with web services.
I am attempting to call an OpenAM restful web service from a legacy unprotected tomEE+ servlet. My problem is that I don't understand what I should be creating for the second argument of:
JAXRSClientFactory.create("http://openam.mylocalAMserver.lan:8080/openam/json/authenticate", WhatClassGoesHere.class);
The OpenAM documentation provides this:
3.3.1. Authentication & Logout
$ curl --request POST --header "X-OpenAM-Username: demo" --header
"X-OpenAM-Password: changeit" --header "Content-Type:
application/json" --data "{}"
https://openam.example.com:8443/openam/json/authenticate
{ "tokenId": "AQIC5w...NTcy*", "successUrl": "/openam/console" }
Should I create a class with instance variables "tokenId", "successURL" and passing that as the second parameter to JAXRSClientFactory? Do I need to worry about all of the parameters specified? Once I figure this out, I've got to figure out how to actually pass the user name and password and invoke the service...
Thanks for your help.
This is really more of a JAXRS question, and is not specific to OpenAM. You need to write Java code to make requests and parse the JSON response. JAXRS is one way to do this- but there are others as well.
Look for a good JAXRS tutorial. The OpenAM part is very simple once you understand REST web services.
I have been given a url .. www.abc.com/details and asked to send my name and phone number on this url using POST. They have told me to set the content-type as application/json and the body as valid JSON with the following keys:
name: name of the user
phone number: phone number of the user
Now i have no clue how to send this request! Will it be something like:
http://www.abc.com/details?method=post&name=john&phonenumber=445566
or do i have to use java to send the same?
Please help
Based on what you provided, it is pretty simple for what you need to do and you even have a number of ways to go about doing it. You'll need something that'll let you post a body with your request. Almost any programming language can do this as well as command line tools like cURL.
Once you have your tool decided, you'll need to create your JSON body and submit it to the server.
An example using cURL would be (all in one line, minus the \ at the end of the first line):
curl -v -H "Content-Type: application/json" -X POST \
-d '{"name":"your name","phonenumber":"111-111"}' http://www.example.com/details
The above command will create a request that should look like the following:
POST /details HTTP/1.1
Host: www.example.com
Content-Type: application/json
Content-Length: 44
{"name":"your name","phonenumber":"111-111"}
You can post data to a url with JavaScript & Jquery something like that:
$.post("www.abc.com/details", {
json_string: JSON.stringify({name:"John", phone number:"+410000000"})
});
It is not possible to send POST parameters in the URL in a straightforward manner. POST request in itself means sending information in the body.
I found a fairly simple way to do this. Use Postman by Google, which allows you to specify the content-type (a header field) as application/json and then provide name-value pairs as parameters.
You can find clear directions at [2020-09-04: broken link - see comment] http://docs.brightcove.com/en/video-cloud/player-management/guides/postman.html
Just use your URL in the place of theirs.
You can use postman.
Where select Post as method.
and In Request Body send JSON Object.
In windows this command does not work for me..I have tried the following command and it works..using this command I created session in couchdb sync gate way for the specific user...
curl -v -H "Content-Type: application/json" -X POST -d "{ \"name\": \"abc\",\"password\": \"abc123\" }" http://localhost:4984/todo/_session
If you are sending a request through url from browser(like consuming webservice) without using html pages by default it will be GET because GET has/needs no body. if you want to make url as POST you need html/jsp pages and you have to mention in form tag as "method=post" beacause post will have body and data will be transferred in that body for security reasons. So you need a medium (like html page) to make a POST request. You cannot make an URL as POST manually unless you specify it as POST through some medium. For example in URL (http://example.com/details?name=john&phonenumber=445566)you have attached data(name, phone number) so server will identify it as a GET data because server is receiving data is through URL but not inside a request body
In Java you can use GET which shows requested data on URL.But POST method cannot , because POST has body but GET donot have body.