Question

由于某种原因，回调不会触发。我删去了所有多余的代码，这就是结构。也许我看不到这里有什么问题？

(function() {
  registerSecendStep = new Page('registerSecendStep', function(name) { //CLIKED THE BUTTON
    $('.register-second-step-button').click(function() {
      function doAjax(testReferral) {
        alert(testReferral); //THIS NEVER GETS PRINTED
      }

      showHint(function() { doAjax(testReferral); }); //START THE CALL
    });
  });

  function showHint(callback) { //CALLBACK
    xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function() {
      if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
        alert('called'); //THIS GETS ALERTED
        callback(true);
      }
    }
    xmlhttp.open("GET", helper.app.HomeSite + "/ajaxresponder?w=" + str, true);
    xmlhttp.send();
  }
})();

任何帮助表示赞赏

Answer 1

testReferral从未在回调中定义。试试：

showHint(function(testReferral) {doAjax(testReferral);});

JavaScript回调函数无法启动

1 个答案: