Question

可以使用try catch块捕获TokenMismatchException吗？而不是在VerifyCsrfToken.php第46行...＆＃34;中显示显示＆＃34; TokenMismatchException的调试页面，我希望它显示实际页面并只显示错误消息。

我对CSRF没有任何问题，我只是希望它仍然显示页面而不是调试页面。

复制（使用firefox）：步骤进行：

打开页面（http://example.com/login）
清除Cookie（域，路径，会话）。我在这里使用web开发人员工具栏插件。
提交表格。

实际结果：＆＃34;糟糕，看起来出了问题＆＃34;页面显示。预期结果：仍显示登录页面，然后传递错误“＃34;令牌不匹配＆＃34;什么的。

请注意，当我清除Cookie时，我没有刷新页面，以便令牌生成新密钥并强制它出错。

更新（已添加表格）：

        <form class="form-horizontal" action="<?php echo route($formActionStoreUrl); ?>" method="post">
        <input type="hidden" name="_token" value="<?php echo csrf_token(); ?>" />
        <div class="form-group">
            <label for="txtCode" class="col-sm-1 control-label">Code</label>
            <div class="col-sm-11">
                <input type="text" name="txtCode" id="txtCode" class="form-control" placeholder="Code" />
            </div>
        </div>
        <div class="form-group">
            <label for="txtDesc" class="col-sm-1 control-label">Description</label>
            <div class="col-sm-11">
                <input type="text" name="txtDesc" id="txtDesc" class="form-control" placeholder="Description" />
            </div>
        </div>
        <div class="form-group">
            <label for="cbxInactive" class="col-sm-1 control-label">Inactive</label>
            <div class="col-sm-11">
                <div class="checkbox">
                    <label>
                        <input type="checkbox" name="cbxInactive" id="cbxInactive" value="inactive" />&nbsp;
                        <span class="check"></span>
                    </label>
                </div>
            </div>
        </div>
        <div class="form-group">
            <div class="col-sm-12">
                <button type="submit" class="btn btn-primary pull-right"><i class="fa fa-save fa-lg"></i> Save</button>
            </div>
        </div>
    </form>

这里没什么好看的。只是一个普通的形式。就像我所说的那样，表格工作得非常好。就在我陈述上述步骤时，由于TOKEN过期而导致错误。我的问题是，表格是否应该采用这种方式？我的意思是，当我清除cookie和会话时，我还需要重新加载页面吗？是CSRF在这里工作的吗？

Answer 1

您可以在 App \ Exceptions \ Handler.php

中处理TokenMismatchException异常

<?php namespace App\Exceptions;
use Exception;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
use Illuminate\Session\TokenMismatchException;


class Handler extends ExceptionHandler {


    /**
     * A list of the exception types that should not be reported.
     *
     * @var array
     */
    protected $dontReport = [
        'Symfony\Component\HttpKernel\Exception\HttpException'
    ];
    /**
     * Report or log an exception.
     *
     * This is a great spot to send exceptions to Sentry, Bugsnag, etc.
     *
     * @param  \Exception  $e
     * @return void
     */
    public function report(Exception $e)
    {
        return parent::report($e);
    }
    /**
     * Render an exception into an HTTP response.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Exception  $e
     * @return \Illuminate\Http\Response
     */
    public function render($request, Exception $e)
    {
        if ($e instanceof TokenMismatchException){
            // Redirect to a form. Here is an example of how I handle mine
            return redirect($request->fullUrl())->with('csrf_error',"Oops! Seems you couldn't submit form for a long time. Please try again.");
        }

        return parent::render($request, $e);
    }
}

Answer 2

更好的 Laravel 5解决方案

App \ Exceptions \ Handler.php
使用新的有效CSRF令牌将用户返回到表单，这样他们就可以重新提交表单而无需再次填写表单。

public function render($request, Exception $e)
    {
         if($e instanceof \Illuminate\Session\TokenMismatchException){
              return redirect()
                  ->back()
                  ->withInput($request->except('_token'))
                  ->withMessage('Your explanation message depending on how much you want to dumb it down, lol!');
        }
        return parent::render($request, $e);
    }

我也非常喜欢这个想法：

https://github.com/GeneaLabs/laravel-caffeine

Answer 3

而不是尝试捕获异常只是将用户重定向回同一页面并让他/她重复操作。

在App \ Http \ Middleware \ VerifyCsrfToken.php中使用此代码

<?php
namespace App\Http\Middleware;
use Closure;
use Redirect;
use Illuminate\Foundation\Http\Middleware\VerifyCsrfToken as BaseVerifier;
class VerifyCsrfToken extends BaseVerifier
{
    /**
     * The URIs that should be excluded from CSRF verification.
     *
     * @var array
     */
    protected $except = [
        //
    ];

    public function handle( $request, Closure $next )
    {
        if (
            $this->isReading($request) ||
            $this->runningUnitTests() ||
            $this->shouldPassThrough($request) ||
            $this->tokensMatch($request)
        ) {
            return $this->addCookieToResponse($request, $next($request));
        }

        // redirect the user back to the last page and show error
        return Redirect::back()->withError('Sorry, we could not verify your request. Please try again.');
    }
}

Answer 4

Laravel 5.2：修改 App \ Exceptions \ Handler.php ，如下所示：

<?php

namespace App\Exceptions;

use Exception;
use Illuminate\Validation\ValidationException;
use Illuminate\Auth\Access\AuthorizationException;
use Illuminate\Database\Eloquent\ModelNotFoundException;
use Symfony\Component\HttpKernel\Exception\HttpException;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;

use Illuminate\Session\TokenMismatchException;

class Handler extends ExceptionHandler
{
    /**
     * A list of the exception types that should not be reported.
     *
     * @var array
     */
    protected $dontReport = [
        AuthorizationException::class,
        HttpException::class,
        ModelNotFoundException::class,
        ValidationException::class,
    ];

    /**
     * Report or log an exception.
     *
     * This is a great spot to send exceptions to Sentry, Bugsnag, etc.
     *
     * @param  \Exception  $e
     * @return void
     */
    public function report(Exception $e)
    {
        parent::report($e);
    }

    /**
     * Render an exception into an HTTP response.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Exception  $e
     * @return \Illuminate\Http\Response
     */
    public function render($request, Exception $e)
    {
        if ($e instanceof TokenMismatchException) {
            abort(400); /* bad request */
        }
        return parent::render($request, $e);
    }
}

在AJAX请求中，您可以使用abort（）函数响应客户端，然后使用AJAX jqXHR.status在客户端处理响应非常容易，例如通过显示消息并刷新页面。别忘了在jQuery ajaxComplete事件中捕获HTML状态代码：

$(document).ajaxComplete(function(event, xhr, settings) {
  switch (xhr.status) {
    case 400:
      status_write('Bad Response!!!', 'error');
      location.reload();
  }
}

Answer 5

Laravel 8似乎在处理异常方面略有不同，并且上述解决方案在我全新安装的Laravel中均无效。因此，我要发布最终得到的工作，并希望对其他人有所帮助。参见Laravel Docs here。

这是我的App \ Exceptions \ Handler.php文件：

<?php

namespace App\Exceptions;

use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;

class Handler extends ExceptionHandler
{
    /**
     * A list of the exception types that are not reported.
     *
     * @var array
     */
    protected $dontReport = [
        //
    ];

    /**
     * A list of the inputs that are never flashed for validation exceptions.
     *
     * @var array
     */
    protected $dontFlash = [
        'password',
        'password_confirmation',
    ];

    /**
     * Register the exception handling callbacks for the application.
     *
     * @return void
     */
    public function register()
    {
        $this->renderable(function (\Symfony\Component\HttpKernel\Exception\HttpException $e, $request) {
            if ($e->getStatusCode() == 419) {
                // Do whatever you need to do here.
            }
        });
    }

}

Answer 6

很好。 Laravel 8肯定以不同的方式做到了。下面的代码块不适用于laravel 8。

  if ($exception instanceof \Illuminate\Session\TokenMismatchException) {
    return redirect()->route('login');
  }

但是这个确实做到了：

  $this->renderable(function (\Symfony\Component\HttpKernel\Exception\HttpException $e, $request) {
    if ($e->getStatusCode() == 419) {
      return redirect('/login')->with('error','Your session expired due to inactivity. Please login again.');
    }
  });

Laravel捕获TokenMismatchException

6 个答案: