Laravel5.x如何让404返回一个json？

Question

比如正常的一个请求，返回方式如下：

return response($result, $code);  // 返回json

异常的请求，比如该路由没有被定义，该请求的方法没有被定义。如何也返回一个json对象呢，写在404.blade.php里肯定不合适，因为写进去的话会返回一个string

Answer 1

在Middleware或者App\Exceptions\Handler里捕获
Symfony\Component\HttpKernel\Exception\NotFoundHttpException

如果是Middleware

use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;

public function handle($request, Closure $next) {
    try {
        return $next($request);
    } catch (NotFoundHttpException $e) {
        return response()->json(['msg'=>'NotFound']);
    }
}

---

2017.04.04 更新

已知laravel5的默认Exceptions\Handler会优先匹配404异常，所以建议在Handler进行处理。

修改app/Exceptions/Handler.php的render方法如下

/**
 * Render an exception into an HTTP response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Exception  $exception
 * @return \Illuminate\Http\Response
 */
public function render($request, Exception $exception)
{
    if (is_a($exception, \Symfony\Component\HttpKernel\Exception\NotFoundHttpException::class) && $request->expectsJson()) {
        return response()->json(['msg'=>'NotFound']);
    } else {
        return parent::render($request, $exception);
    }
}