display('XXX'); 但是这样会导致X"/> display('XXX'); 但是这样会导致X">
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thinkphp框架一些小疑义

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Release: 2016-06-13 12:42:39
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thinkphp框架一些小疑问
我在用thinkphp的时候,比如验证完用户的用户名后,我想弹出一个窗口提示登录失败,然后重新display这个模版

那么我会这么写

<br />
echo "<script>alert('新产品入库成功')</script>";<br />
$this->display('XXX');<br />
Copy after login

但是这样会导致XXX的css样式加载失败,我的朋友说是因为display前面不能有输出。
请问大家碰到这样的情况怎么解决呢?
我所知道的能使用$this->success('ok'),但是除了内置的success,还有什么方法能放防止display的时候不会使样式表失效呢?或者有什么替代的方法呢?

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