I'm trying to generate a namespace through PhpStorm's templating system (File and Code Templates), I can't see any helpers / means of getting this through PhpStorm's documentation.
I was wondering if there was a way to automatically define the namespace based on the directory:
<?php
// I'm wanting to apply to $namespace
namespace $namespace;
class ${NAME} implements
\Zend\Mail\Mailable
{
/** \Zend\Mail\Factory */
private $factory;
...
}#set( $factory = "factory" ) #set( $this- = "this-" ) #set( $directory = ?)
If my directory/filename is zend/mail/model/notice/send.php I'd like this to generate a namespace of Zend\Mail\Model\Notice; I can get the class name through the ${NAME} variable.
I'm aware I can have the template request the namespace through a dialog when the file is being created, however I was wondering if it's possible to automate this?
if you want , you can use groovyScript.
for example, you can create a variable $FOLDER_OF_FILE$ and set this groovyScript to take the folder of file.
groovyScript("String[] p=_editor.getVirtualFile().getPath().split('/');String prevLast = p[p.length-2];")
you can change and adapt my example for your needs
There are no predefined variables for file path, please vote for IDEA-136387 to be notified on any progress with this feature.
Related
I have a root folder named "root".
In this folder I have 2 more directories, each with one folder, each with a script:
/root/script01/client_script01/main.lua
In this script I have this:
local function OpenWindow()
stuff
end
And
/root/script02/client_script02/main.lua
I want to use OpenWindow() function in second script!
You are correct that you can call OpenWindowfrom client_script02/main.lua if you do not use the local keyword in the definition.
This however is not best practice. I am unsure of the specifics of your environment or intent, but in most cases it is better to create a lua module and use the require function to load it.
This is better because it shows the relationship between the files, showing that client_script02/main.lua requires client_script01/main.lua to be loaded to operate properly.
Your module could look something like this:
local client_script01 = {}
client_script01.OpenWindow = function()
--stuff
end
return client_script01
The other script something like this:
local cs01 = require('client_script01')
do
cs01.OpenWindow()
--stuff
end
You would also need to adjust your file structures to better suit this convention, based on how the require function preforms searching: lua-users - Package Path
I have changed Yii2 advanced directory structure like following(it's working well):
app-folder
-admin
-assets
-.htaccess
-index.php
-assets
-protected
-backend
...
-common
...
-frontend
...
...
-uploads
...
Now, I am trying to add a namespace as namespace protected\base; into protected/base/AnyFile.php file and use it in a controller as use protected\base\AnyFile;. But, my project is giving error:
syntax error, unexpected 'protected' (T_PROTECTED), expecting identifier (T_STRING) or function (T_FUNCTION) or const (T_CONST) or \\ (T_NS_SEPARATOR)
I saw this issue on the website: Yii2 Custom / Shorter Namespace. However, It didn't work on my condition.
First of all protected is reserved keyword (token T_PROTECTED). You can keep direcory name but you need to change namespace root alias.
In your alias config file protected/common/bootstrap.php write:
Yii::setAlias('app', dirname(dirname(__DIR__))); // set path to protected directory
And then use namespace app\base; and use app\base\AnyFile;.
See Class Autoloading section of the guide https://www.yiiframework.com/doc/guide/2.0/en/concept-autoloading
I made a bunch of classes and put them inside the directory common/a/b/c.
So, inside file
Class1.class.php
I have:
<?php
namespace x\y\b\c;
class Class1
...
Namespace is different from the directory structure organization, because I wanted that way. (x\y should map to common/a directory)
On my common/config/bootstrap.php I tried this:
Yii::setAlias('common', dirname(__DIR__));
Yii::setAlias('x/y', '#common/a');
And tried importing this class in another file using
use x\y\b\c\Class1;
With no success. But if I use:
Yii::$classMap['x\y\b\c\Class1'] = __DIR__ . '/../../common/a/b/c/Class1.class.php';
instead of setAlias, it works.
I wonder if it's possible to have namespace different from the directory structure without using composer and how can I do this instead of mapping every class inside common/a/b/c
In bootstrap.php.
Yii::setAlias('common', dirname(__DIR__));
Yii::setAlias('x/y', dirname(__DIR__).'/models');
Inside my "models" folder, there is folder "b" and folder "c" is inside folder "b".
models > b > c
I have a model file named "LoginForm.php" and resides in folder "c".
At the top of this file are these few lines.
namespace x\y\b\c;
use Yii;
use yii\base\Model;
class LoginForm extends Model
Inside my SiteController I have this.
use x\y\b\c\LoginForm;
In one of the action function, i can successfully call this model.
$model = new LoginForm();
I had an old ipy_user_conf.py in which I included a simple function into the user namespace like this:
import IPython.ipapi
ip = IPython.ipapi.get()
def myfunc():
...
ip.user_ns['myfunc'] = myfunc
Then, I could use myfunc in the prompt.
However, I updated to IPython 0.12.1 and now the ip_user_conf.py does not work. I haven't seen how to translate such a custom function for prompts to the new configuration model.
Which is the way to do this?
Best regards,
Manuel.
UPDATE: Changed the subject to question
After reading a bit of the documentation (and peeking at the source code for leads) I found the solution for this problem.
Simply now you should move all your custom functions to a module inside your .ipython directory. Since what I was doing was a simple function that returns the git branch and status for the current directory, I created a file called gitprompt.py and then I included the filename in the exec_file configuration option:
c.InteractiveShellApp.exec_files = [b'gitprompt.py']
All definitions in such files are placed into the user namespace. So now I can use it inside my prompt:
# Input prompt. '\#' will be transformed to the prompt number
c.PromptManager.in_template = br'{color.Green}\# {color.LightBlue}~\u{color.Green}:\w{color.LightBlue} {git_branch_and_st} \$\n>>> '
# Continuation prompt.
c.PromptManager.in2_template = br'... '
Notice that in order for the function to behave as such (i.e called each time the prompt is printed) you need to use the IPython.core.prompts.LazyEvaluation class. You may use it as a decorator for your function. The gitprompt.py has being placed in the public domain as the gist: https://gist.github.com/2719419
I am using flash.utils.getDefinitionByName in an attempt to grab an art asset. I use this function quite a bit and haven't had trouble until now. Check it:
assetName = Assets.MegaBerry; // works
assetName = getDefinitionByName("Assets.MegaBerry") as Class; // doesn't work
What the heck?? Error response for the second line is "Variable not found."
If it matters: Assets is a file in my root source directory (it has no package; Assets is the fully qualified name) and I've tried putting:
import Assets;
at the top with no luck.
For reference, in Assets.as I have:
[Embed(source = "../art/Inventory/MegaBerry.png")]
public static var MegaBerry:Class;
Your problem is that embedding the resource into the Assets class will create a static variable of type Class that belongs to that class - which is what you are referencing when you use Assets.MegaBerry: A variable(!) of type Class.
It does not, however, register the MegaBerry class to a fully qualified class name. To do this, you have to use - who would have guessed it - registerClassAlias at some point in your application:
registerClassAlias("Assets.MegaBerry", Assets.MegaBerry);
After that, it will be available everywhere else when calling getDefinitionByName.
** EDIT **
Well that's some unexpected behavior... It turns out, the class that was embedded is in fact automatically registered, but under {className}_{variableName}, instead of the notation you would expect. So using:
getDefinitionByName("Assets_MegaBerry") as Class;
should to the trick.
registerClassAlias also works, but then you need to call getClassByAliasinstead of getDefinitionByName. Sorry for the mix-up.
** END EDIT **
You can also use the Embed tag to inject the resource into a separate class file, which you can then reference as expected by using getDefinitionByName, or simply using an import:
package assets {
[Embed(source="../art/Inventory/MegaBerry.png"]
public class MegaBerry extends BitmapData {
}
}
Instead of calling
assetName = getDefinitionByName("Assets.MegaBerry") as Class;
, instead just use:
assetName = Assets["MegaBerry"];
try:
[Embed(source = "../art/Inventory/MegaBerry.png" , symbol="MegaBerry")]
public static var MegaBerry:Class;
In actionscript, objects actually have a name property that is different from the actual variable name as it shows in code.
For example, if you create a variable as follows,
var myBerry = new MegaBerry();
Then getDefinitionByName("myBerry") will return null.
Only when you set the name of the variable by writing myBerry.name = "myBerry", will getDefinitionByName("myBerry") return what you want it to. The name of the object doesn't necessarily have to be equal to the variable name in code.
In your specific case, I don't think you need to use any of that anyways. Have you tried assetName = new MegaBerry() ?
If you want to find out what the fully qualified name of you class really is, you may do the following:
trace(getQualifiedClassName(Assets.MegaBerry));
You may do that from inside Assets.as, for instance.
You can feed that string back to getDefinitionByName() and get a reference to the class.
trace(getDefinitionByName(getQualifiedClassName(SomeClass)));
// output [class SomeClass]
And remember, getDefinitionByName() only gets you references for classes that are in the same scope as the getDefinitionByName call itself. So, if you are loading external SWFs, getting class references will depend on the application domain you are using and the place, where this code executes.