'"php.exe"'无法识别为内部或外部命令

Question

我正在尝试使用wamp创建一个新的zend项目.

Microsoft Windows XP [Version 5.1.2600]
(C) Copyright 1985-2001 Microsoft Corp.

C:\> cd wamp\www\zendtest*

C:\wamp\www\zendtest>C:\wamp\www\zend\bin\zf.bat create project quickstart*


'"php.exe"' is not recognized as an internal or external command,
operable program or batch file.


C:\wamp\www\zendtest>

我的环境变量路径是

%SystemRoot%\ system32;%SystemRoot%;%SystemRoot%\ System32\Wbem; C:\ PROGRA~1\IBM\SQLLIB\BIN; C:\ PROGRA~1\IBM\SQLLIB\FUNCTION; C:\ Program Files \的Zend\MySQL51\BIN; C:\瓦帕\ WWW\Zend的\ BIN;

哪个部分我认为是错误的.

Answer 1

在zf.bat文件中替换此行

SET PHP_BIN=C:\wamp\bin\php\php5.3.4\php.exe