在以下情况下，当错误来自 PHP 作为对 ajax 请求的响应时，如何显示警报 - How to show alert when error comes from the PHP as response to ajax request in following case?

How to show alert when error comes from the PHP as response to ajax request in following case?

jQuery AJAX函数如下：

$(document).ready(function() { 
    $("#zip_code").keyup(function() {
        var el = $(this);
        var module_url = $('#module_url').val();
        if (el.val().length === 5) {
            $.ajax({
                url : module_url,
                cache: false,
                dataType: "json",
                type: "GET",
                async: false,
                data: {
                    'request_type':'ajax', 
                    'op':'get_city_state',
                    'zip_code' : el.val()
                },
                success: function(result, success) {
                    $("#city").val(result.place_name);
                    $("#state_code").val(result.state_code);
                }
            }); 
        }
    });
});

PHP代码如下：

case "get_city_state":
    // to get the city and state on zip code.
    $ret = $objUserLogin->GetCityState($request); 
    if(!$ret) { 
        $error_msg = $objUserLogin->GetAllErrors();
        $data = array();
        $data['error_message'] = $error_msg;
        $data = json_encode($data);
        echo $data;
        die;
    } else {
        $data = array();
        $data = $objUserLogin->GetResponse();
        echo json_encode($data);
        die;
    }    
    break;

现在，我可以在响应没有任何错误时打印成功，但是当发生某些错误时显示警报消息呢？如何实现？需要对上述代码进行哪些更改？请帮助我。

成功使用以下条件：

    success: function(result, success) {
      if($.inArray( "error_message", result)) {
          alert("Error message");
      } else {
          $("#city").val(result.place_name);
          $("#state_code").val(result.state_code);
      }
    }

on php file if u found data acc to ur parameter
than it ok if there is no data acc to ur parameter than this time u can call its an error so in thsi case  echo "error";
and on ajax page check for result if its value is error then do what else u want 
success: function(result, success) {
          $("#city").val(result.place_name);
          $("#state_code").val(result.state_code);
        }