api返回实现

$result = User::find($id);
if(empty($result)){
  throw new ApiException('获取失败');
}
else{
  return json_decode($result);
}

api返回信息

{
  "msg": "",
  "data": "获取失败",
  "status": 0
}

1，添加异常类

namespace App\Exceptions;


class ApiException extends \Exception
{

  function _construct($msg='')
  {
    parent::_construct($msg);
  }

}

2，修改laravel异常类u。。。

namespace App\Exceptions;


public function render($request, Exception $e)
{
  if ($e instanceof ApiException){
    $result = [
      "msg" => "",
      "data"=>$e->getMessage(),
      "status"=>0
    ];
    return response()->json($result);
  }
  return parent::render($request, $e);

考虑开发配置时

public function render($request, Exception $e)
{
 
  if(config('app.debug')){
    return parent::render($request,$e);
  }
  return $this->handle($request,$e);
}

public function handle($request,Exception $e){
  if ($e instanceof ApiException){
    $result = [
      "msg" => "",
      "data"=>$e->getMessage(),
      "status"=>0
    ];
    return response()->json($result);
  }

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