当我按下发送时,服务器收到此错误 Incorrect Server Response
这是我的控制器
public function mediauploadpost(Request $request){
$CKEditor = $request->input('CKEditor');
$funcNum = $request->input('CKEditorFuncNum');
$message = $url = '';
if (Input::hasFile('upload')) {
$file = Input::file('upload');
if ($file->isValid()) {
$filename =rand(1000,9999).$file->getClientOriginalName();
$file->move(public_path().'/wysiwyg/', $filename);
$url = url('wysiwyg/' . $filename);
} else {
$message = 'An error occurred while uploading the file.';
}
} else {
$message = 'No file uploaded.';
}
return '<script>window.parent.CKEDITOR.tools.callFunction('.$funcNum.', "'.$url.'", "'.$message.'")</script>';
}
Run Code Online (Sandbox Code Playgroud) 在我的视图页面中,我有这条路线:
{!! Form::open(['url' => 'forumcomment/' . $forum->slug, 'files'=>false, 'id' => 'qw-commentform' ,'class' => 'qt-clearfix']) !!}
<hr class="qt-spacer-s"><div class="input-field">
{!! Form::textarea('comment', null, ['class'=>'materialize-textarea', 'id'=>'my-editor', 'required'=>'required','aria-required'=>true]) !!}
<label for="comment" class="">Comment*</label></div>
<hr class="qt-spacer-s">
{!! Form::submit('Post Comment', array( 'class'=>'qt-btn qt-btn-primary qt-btn-xl' )) !!}
{!! Form::close() !!}
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混合内容错误如何获得安全路线?
我正在使用 Laravel 5.5 我添加了这个依赖项来制作简单的搜索引擎
https://github.com/nicolaslopezj/searchable
但是当我尝试搜索任何东西时遇到问题
SQLSTATE[42000]: Syntax error or access violation: 1055 'myreview.movies.name' isn't in GROUP BY (SQL: select count() as aggregate from (select movies., max((case when LOWER(movies.name) LIKE car then 150 else 0 end) + (case when LOWER(movies.name) LIKE car% then 50 else 0 end) + (case when LOWER(movies.name) LIKE %car% then 10 else 0 end) + (case when LOWER(movies.description) LIKE car then 150 else 0 end) + (case when LOWER(movies.description) LIKE car% then 50 else 0 end) …
我正在使用 spatie/laravel-permission 包来获取角色和权限
但这里的种子永久角色和权限出现问题是数据库列表链接
我制作这些数据播种机用于播种
$this->call(UsersTableSeeder::class);
$this->call(PermissionsTableSeeder::class);
$this->call(RolesTableSeeder::class);
$this->call(RolehaspermissionTableSeeder::class);
$this->call(ModelhasrolesTableSeeder::class);
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许可表播种者
DB::table('roles')->insert([
'name' => 'Administrator',
'guard_name' => 'web',
]);
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角色表播种者
DB::table('roles')->insert([
'name' => 'Admin',
'guard_name' => 'web',
]);
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RolehaspermissionTableSeeder
DB::table('role_has_permissions')->insert([
'permission_id' => '1',
'role_id' => '1',
]);
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模型hasrolesTableSeeder
DB::table('model_has_roles')->insert([
'role_id' => '1',
'model_id' => '1',
'model_type' => 'App\User',
]);
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这是错误的屏幕截图
帮我找种子谢谢
我正在使用 laravel 5.5 试图限制 ckeditor 字符串,但它的输出是这样的
我的数据库
<p> something something something something something </p>
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我正在尝试这样
{!! substr($content,0, 20) !!}
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输出
<p> something ...
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而不是关闭段落,这就是为什么我的页面崩溃并且无法正常工作
我如何将 ckeditor 字符串转换为普通字符串并限制它
谢谢
laravel ×4
php ×4
ckeditor ×2
laravel-5 ×2
laravel-5.5 ×2
javascript ×1
laravel-5.6 ×1
mysql ×1
ubuntu-16.04 ×1