小编liz*_*liz的帖子

如何在插件中获取视图对象?

我有一个控制器动作助手,我将用户对象保存到视图中(在init函数中),如下所示:

Zend_Layout::getMvcInstance()->getView()->user = $user;
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我想在控制器插件的preDispatch方法中访问该对象,因此我不必在数据库中查找用户.我尝试这样做:

$user = Zend_Layout::getMvcInstance()->getView()->user;
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但它返回一个空对象.我希望它,因为我做错了,而不是因为我在我的登录逻辑中编写了一个catch 22.有没有其他方法来访问该对象?

plugins zend-framework

4
推荐指数
1
解决办法
4905
查看次数

需要帮助升级PEAR才能安装PHPUnit

我正在尝试用PEAR安装PHPUnit.这就是我正在运行的......

PEAR Version: 1.7.2
PHP Version: 5.3.8-ZS5.5.0
Zend Engine Version: 2.3.0
Running on: Windows NT L-PC 6.1 build 7600 (Windows 7 Home Premium Edition) i586
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当我尝试安装时,我收到一条错误消息,说我需要升级PEAR ...

C:\Program Files (x86)\Zend\ZendServer\bin>pear install phpunit/PHPUnit
Did not download optional dependencies: phpunit/PHP_Invoker, use --alldeps to download automatically
phpunit/PHPUnit requires PEAR Installer (version >= 1.9.4), installed version is 1.7.2
phpunit/PHPUnit requires package "channel://pear.symfony-project.com/YAML" (version >= 1.0.2)
phpunit/PHPUnit can optionally use package "phpunit/PHP_Invoker" (version >= 1.1.0)
phpunit/File_Iterator requires PEAR Installer (version >= 1.9.2), installed version is …
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php pear phpunit zend-server

3
推荐指数
1
解决办法
3180
查看次数

WAMP虚拟主机AllowOverrides所有抛出500错误

我有一个虚拟主机在WAMP上完美设置,就像这样......

#
# Use name-based virtual hosting.
#
NameVirtualHost *:80

#
# VirtualHost example:
# Almost any Apache directive may go into a VirtualHost container.
# The first VirtualHost section is used for all requests that do not
# match a ServerName or ServerAlias in any <VirtualHost> block.
#


<VirtualHost *:80>
    DocumentRoot "C:/wamp/www"
    ServerName localhost
    ServerAlias localhost

</VirtualHost>

<VirtualHost *:80>
    DocumentRoot "D:\Work\BOT\public"
    ServerAlias bot.dev
    ServerName bot.dev
    <Directory "D:\Work\BOT\public">
        Options Indexes FollowSymLinks
        AllowOverride None
        Order allow,deny
        Allow from all
    </Directory> 
</VirtualHost> …
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apache wamp virtualhost

3
推荐指数
1
解决办法
6432
查看次数

PHP POST无法正常工作

<?php echo $_POST['ss'];?>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
<input name="ss" type="text" />
<input type="submit" name="submit">
</form>
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单击提交时,此代码应打印文本框name ="ss"中输入的内容.
但它不打印.使用method ="get"但不使用post,问题是什么.

html php forms post echo

3
推荐指数
3
解决办法
3万
查看次数

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php ×2

apache ×1

echo ×1

forms ×1

html ×1

pear ×1

phpunit ×1

plugins ×1

post ×1

virtualhost ×1

wamp ×1

zend-framework ×1

zend-server ×1