javascript - 如何处理ajax 201

Question

当进行 ajax 调用时，请参见下面的示例 success 确实会返回 201 状态。你如何更好地处理这些，即 success 函数中的 200、201？

$.ajax({
    type: "POST",
    dataType: "json",
    url: "http://api.domain.com/sms",
    data: {
      // Send value in mobile input field.
      mobile: $("#mobile").val(),
    },
    // On successful AJAX call do the following.
    success: function(data) {
      $('#messageText').text('SMS successfully sent');
    },
    error: function(jqXhr) {
      data = JSON.parse(jqXhr.responseText);
    }
});

Answer 1

使用statusCode对象:

var handle200 = function(data, textStatus, jqXHR) {
    alert('200'); // success codes have the success signature
};

var handle201 = function(data, textStatus, jqXHR) {
    alert('201'); // success codes have the success signature
    // test it if you are in doubt:
    console.log(data);
    console.log(textStatus);
    console.log(jqXHR);
};

var handle404 = function(jqXHR, textStatus, errorThrown) {
    alert('404'); // failing codes have the error signature
});

var request = $.ajax({
    type: 'POST',
    url: '/myresource/posttarget',
    data: { name: 'john' },
    statusCode: {
        200: handle200,
        201: handle201,
        404: handle404
    }
});