javascript - AJAX 请求 - 获取成功请求的 HTTP 代码/响应 header

Question

我正在尝试从我的 AJAX 请求中获取 HTTP 响应代码/响应 header 。这是我的原始脚本:

$("#callContact1").click(function() {
  $.ajax({
      url: "https://www.server.com?type=makecall",
      data: {},
      type: "GET"
    })
    .then(function(data) {
      $('#ajaxResponse').html(data).show();
    })
    .fail(function(xhr) {
      var httpStatus = (xhr.status);
      var ajaxError = 'There was an requesting the call back. HTTP Status: ' + httpStatus;
      console.log('ajaxError: ' + ajaxError);
      //make alert visible 
      $('#ajaxResponse').html(ajaxError).show();
    })
})

工作正常。我已经更新了它以尝试获取 HTTP 响应代码/ header 并在 console.log 中查看它，但我在那里没有看到任何东西。这是我更新的脚本:

$("#callContact1").click(function() {
  console.log('starting call back request');
  $.ajax({
      url: "https://www.server.com?type=makecall",
      data: {},
      type: "GET"
    })
    .then(function(data) {
      $('#ajaxResponse').html(data).show();
      var httpStatus = (data.status);
      var httpResponseCode = (data.getAllResponseHeaders);
      console.log('httpStatus: ' + httpStatus);
      console.log('httpResponseCode: ' + httpResponseCode);
    })
    .fail(function(xhr) {
      var httpStatus = (xhr.status);
      var ajaxError = 'There was an requesting the call back. HTTP Status: ' + httpStatus;
      console.log('ajaxError: ' + ajaxError);
      //make alert visible 
      $('#ajaxResponse').html(ajaxError).show();
    })
})

但我在控制台中没有得到任何信息(虽然请求已成功执行)。我还注意到更新脚本第二行的输出也没有出现在控制台中。

Answer 1

修改以上代码为

.then(function(data,status,xhr) {
      $('#ajaxResponse').html(data).show();
      var httpStatus = status;
      var httpResponseCode = (xhr.status);
      console.log('httpStatus: ' + httpStatus);
      console.log('httpResponseCode: ' + httpResponseCode);
    })