Fun*_*end 5 php laravel laravel-5
我有一个index.blade.php页面,我有一个显示警报成功或错误的代码:
\n\n <!-- ALERTS -->\n @if (isset($msg))\n @if ($msg_type == \'error\')\n <div id="alert" class="alert alert-danger" role="alert">{{ $msg }}</div>\n @else\n <div id="alert" class="alert alert-success" role="alert">{{ $msg }}</div>\n @endif\n <script>setTimeout("document.getElementById(\'mensajecarrito\').style.display=\'none\'",6000);</script>\n @endif\n <!-- //ALERTS -->\nRun Code Online (Sandbox Code Playgroud)\n\n这个索引有一个表单,这个表单的url POST到post_designs url,这是他的控制器方法:
\n\npublic function postDesign() {\n\n if (Session::has(\'FUNCTION\'))\n $function = Session::get(\'FUNCTION\');\n\n if (Input::has(\'FUNCTION\'))\n $function = Input::get(\'FUNCTION\'); \n\n if (isset($function))\n { \n switch($function)\n {\n case \'createDesign\': \n\n try\n { \n //do something good\n $msg_type = \'success\'; \n $msg = \'Design created successfully\'; \n }\n catch(FotiApiException $e)\n {\n $msg_type = \'error\'; \n $msg = \'Unexpected error\'; \n }\n\n break;\n } \n }\n\n\n\n return back()->with(array(\'msg\' => $msg, \'msg_type\' => $msg_type));\n }\nRun Code Online (Sandbox Code Playgroud)\n\n问题是,当返回到我的index.blade.php 时,$msg 和$msg_type 变量为空...
\n\n我不\xc2\xb4t明白,我的项目是Laravel 5.2.3,我有正确的路线中间件:\n\n\n
Route::group([\'middleware\' =>[ \'web\']], function () {\n\n # Index\n Route::get(\'/\',[\'as\'=> \'index\',\'uses\' => \'WebController@getIndex\']);\n\n // # POSTS\n Route::post(\'/post-design\', [\'as\' => \'post_design\', \'uses\' => \'WebController@postDesign\']);\n\n});\nRun Code Online (Sandbox Code Playgroud)\n
小智 5
你可以使用 withInput()
return back()->withInput(array('msg' => $msg, 'msg_type' => $msg_type));
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如果您想使用刀片中的数据,则保存在会话中的数据。尝试使用 old('msg') 或 old('msg_type')
@if(old('msg'))
<p>msg:{{ old('msg') }}<p>
@endif
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php
$msg = $request->old('msg');
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我是 laravel 5.2 的新手,关于路线没有必要,但路线
['middleware' =>[ 'web']]
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场景 laravel 5.2 中的所有路由都在其中,所以尝试将其删除
并返回具有 2 个值的视图,尝试使其如下:
return back()->with('msg', $msg)->with('msg_type', $msg_type);
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我希望能解决这个问题:)
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