Question

在jQuery中，当有东西加载时我可以这样做：

$.ajax({
  url: "/answer_checker.php",
  global: false, type: "POST", 
  data: ({...clipped...}), 
  cache: false,
  beforeSend: function() {
    $('#response').html("<img src='/images/loading.gif' />");
  },
  success: function(html) {
    $('#response').html(html);
  }
}

如何在普通javascript中应用beforeSend和success？我目前有这个：

function processGeneric() {
setTimeout(function () {
    if (xmlHttpObject.readyState === 0 || xmlHttpObject.readyState === 4) {
        xmlHttpObject.open("GET", requestLink, true);
        xmlHttpObject.onreadystatechange = function () {//some code to handle here};
        xmlHttpObject.send(null);
    }
}, timeout);
}

Answer 1

您创建一个函数，接受您想要的回调以及其他ajax选项，然后使用该函数：

在发送＆＃34;之前调用＆＃34;回调
拨打ajax电话，which is well-covered in this other question and its answers
通话结束后，请致电完成回叫

这些方面含糊不清：

function processGeneric(url, beforeSend, completion) {
    var xhr;

    xhr = new XMLHttpRequest();
    xhr.onreadstatechange = function() {
        if (xhr.readyState === 0 || xhr.readyState === 4) {
            if (completion) {
                completion(xhr);
            }
            done();
        }
    };
    xhr.open("GET", url, true);
    if (beforeSend) {
        if (beforeSend(xhr) === false) {
            done();
            return;
        }
    }
    xhr.send(null);
    return xhr;

    function done() {
        xhr.onreadystatechange = null;
        xhr = undefined;
    }
}

简单的javascript beforeSend和成功

1 个答案: