将两个查询变量传递给 jquery 返回数据

Question

我想发送两个查询变量并将这些值发送到 jquery ajax 数据值。现在我在 ajax 页面中什么都没有。如果我使用一个查询语句并传递一个值，我会得到一些输出

$.ajax({
  url: 'get-details',
  type: 'GET',
  data: 'id='+user_id,
  dataType: 'JSON',
  success: function(data,data1,textStatus, jqXHR){ //want to add data1
    var id = data.id;
    var alt=data1.column; 
    alert(data1.column);
  },
  error: function(jqXHR, textStatus, errorThrown){

  },
});  

public function getDetails(Request $request)
{
  $request_data = $request->all();
$user_id = $request_data['id'];
$user_data =DB::table('travel_request')
        ->join('department', 'travel_request.department_id', '=', 'department.id')
         ->select('travel_request.*')->where('travel_request.id',$user_id)->first();  
$req=DB::table('traveling_details')->select('traveling_details.*')->get(); //has multiple value
return response()->json($user_data,$req);
}  

Answer 1

如果可行就试试这个

$data['user_data'] = DB::table('travel_request')
        ->join('department', 'travel_request.department_id', '=', 'department.id')
         ->select('travel_request.*')->where('travel_request.id',$user_id)->first();  

$data['req'] = DB::table('traveling_details')->select('traveling_details.*')->get(); //has multiple value

return response()->json($data);

你在视图文件中的ajax函数应该是这样的

$.ajax({
  url: 'get-details',
  type: 'GET',
  data: 'id='+user_id,
  dataType: 'JSON',
  success: function(data){ //want to add data1

      var user_data = data.user_data;
      var req = data.req;

      console.log(user_data);
      console.log(req);

      // to access the column of the user_data
      var YourVariableName = user_data.ColumnNameHere;

  },
  error: function(jqXHR, textStatus, errorThrown){

  },
});