Zend Framework: How to create controller in WINXP on command line

问题

I am using zend framework with windows XP and WAMP server.

I have created zend project using following DOS commands:

SET PATH=%PATH%;C:\wamp\bin\php\php5.3.0    
C:\>wamp\www> C:\wamp\www\ZendFramework\bin\zf.bat create project test

and my project is created at:

C:\wamp\www\test

Now I have two questions. Q.1: In which folder I will run my create controller zend command using DOS?

C:\wamp\www\test>
or
C:\wamp\www\test\application\controller>

Q:2 I have tried both above folder locations to run create controller command. I have following error. how to solve this?

COMMAND
C:\>wamp\www\test> C:\wamp\www\ZendFramework\bin\zf.bat create controller test

ERROR:
    Fatal Error: Cannot Redeclare class Zend_Loader in C:\wamp\www\test\library\Zend\Loader.php on line 31

Thanks

回答1:

See this related post. If you have copied the ZF framework to the C:\wammp\www\test\library folder then you will need to comment out this line in your test\application\configs\application.ini file so that the ZEND_LOADER class isn't included twice:

;includePaths.library = APPLICATION_PATH "/../library"

回答2:

Using xampp the procedure looks like this:

Create Project
working directory: htdocs

zf.bat create project zftest
cd zftest

Create module
working directory: htdocs/zftest

zf.bat create module testmodule

Create controller
working directory: htdocs/zftest

zf.bat create controller testcontroller index-action testmodule

Notice the working directory of each command

Xampp is an application stack for several different platforms, it's use is mainly to be able to setup a working development platform in a fast and effecient way.

The indexaction parameter is there to create a default index action in the created controller.

来源：https://stackoverflow.com/questions/1734166/zend-framework-how-to-create-controller-in-winxp-on-command-line

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