关于变量警告的问题。
2.php的代码是
$xx=123;
3.php的代码是
require '2.php';
echo $xx;
可以输出变量$xx;
但我把3.php的代码改成
require dirname(__FILE__).'/2.php';
echo $xx;
一样可以输出123,但变量$xx显示出警告undefined variable '$xx'
这是什么原因了,是不是因为第二次引用文件时绝对路径的原因,如何去除警告。
------解决方案--------------------
- PHP code
echo dirname(_FILE_).'/2.php'; //看看输出的路径是否正确。 <br><font color="#e78608">------解决方案--------------------</font><br>这就是全部代码? 感觉没有错误呀。 <br><font color="#e78608">------解决方案--------------------</font><br>一定漏了什么<br><br>没定义怎么输出123? <br><font color="#e78608">------解决方案--------------------</font><br>根据你的思路我测试了 <br>require dirname(__FILE__).'/2.php'; <br>我的是有问题的<br>Warning: require(/usr/web/seaboat/XXX.php) [function.require]: failed to open stream: No such file or directory in /usr/web/seaboat/htdocs/XXX.php on line 19<br><br>Fatal error: require() [function.require]: Failed opening required '/usr/web/seaboat/XXX.php' (include_path='.:/usr/share/pear:/usr/local/lib/Smarty/libs:/usr/web/seaboat/htdocs') in /usr/web/seaboat/htdocs/XX.php on line 19 <div class="clear"> </div>

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