如何更改.done下的响应(功能(数据)?

I'm working on creating a facial detection system. So basically taking a picture using the webcam and to detect whether there's a face or not in that picture.

The response I got when there's a face is:

{"images":[{"status":"Complete","width":600,"height":450,"file":"content_5942003c94c10","faces":[{"topLeftX":259,"topLeftY":233,"height":188,"rightEyeCenterY":277,"rightEyeCenterX":314,"pitch":-8,"quality":0.20461,"confidence":0.99944,"chinTipX":350,"yaw":1,"chinTipY":430,"eyeDistance":88,"width":188,"leftEyeCenterY":280,"leftEyeCenterX":401,"attributes":{"lips":"Together","asian":0.99925,"gender":{"femaleConfidence":0.0031,"type":"M","maleConfidence":0.9969},"age":27,"hispanic":4.0e-5,"other":0.0007,"black":0,"white":1.0e-5,"glasses":"Eye"},"face_id":1,"roll":3}]}]}

The response I got when there is not a face is:

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

I want to change the response to: If there's a face, alert "success" if there is not a face, alert "fail". How should I do that? Add an if/else statement?

The Ajax:

$.ajax({
        type: 'POST',
        url: 'detect.php',
        data: data,
        dataType: 'json'
    }).done(function(data){

         console.log(data);
        $("#showCounter").html("");
        $("#detectResponse").html(data);
    });
    $(video).hide();
}

</div>

Actually you're right, you need to add if-statement inside .done()

...        
}).done(function(data){
     if (data.images.length > 0){
         alert("success");
         //Do your codes
     }else{
         alert("fail");
         //Do your codes
     }
});

Note: In your case, even no faces detected, it will still fall into .done() as it indicate the ajax return data successfully.