javascript - 如何将错误消息放入 if 语句 if(...)

Question

我正在尝试从面部识别 API 获得 ajax 响应。响应是人脸的数据特征，错误信息是

{"Errors":[{"Message":"no faces found in the image","ErrCode":5002}]}

响应信息是

{"images":[{"status":"Complete","width":236,}]}]}

If/else语句应该怎么写，有face/response就提示“success”，没有face/error就提示“fail”？

$("#testDetect").click(function () {
    var file = $('#imageFile')[0].files[0]; 
    var reader  = new FileReader();
    reader.readAsDataURL(file);
    reader.onloadend = function () {
      var imageData = parseImageData(reader.result);
      var data = {};
      data.image = imageData;
      $.ajax({
      url      : "http://localhost/Karios/simple-detect/form-post.php",
      type     : "POST",
      data     :  data,
      dataType : 'text'
   }).done(function(response) {

     console.log(response);

            if (!response) { 

                alert('Hmm, unexpected response from Kairos');
            } else if (response['Errors'] && response['Errors'].size() > 0) { 
                if (response['Errors'][0]['ErrCode'] == 5002) { 

                    alert(response['Errors'][0]['Message']);

                } else {

                    alert('Some other error occurred:\n' + response['Errors']['ErrorCode'] + ': ' + response['Errors']['Message']);

                }
                
            } else { 

                alert('Face(s) detected');
      }  
    })
    }
});

如果有人脸，如果没有人脸则提示“失败”

Answer 1

实际上 jquery 为每个都给出了 .done 和 .fail，所以:

    $.ajax({
              url      : "http://localhost/Karios/simple-detect/form-post.php",
              type     : "POST",
              data     :  data,
              dataType : 'text'
           }).done(function(response) {
    try {
let jsonResponse = JSON.parse(response);
      if(jsonResponse.Errors && jsonResponse.Errors[0] && jsonResponse.Errors[0].ErrCode ===5002) {alert(jsonResponse.Errors[0].Message);}
    } catch(err) { console.log("bad response");}
                alert( "success" );
           })
          .fail(function() {
            alert( "error" );
          })
          .always(function() {
            alert( "complete" );
          });