Question

我试图进行简单的回调，所以我可以习惯它，但是当页面加载时它不起作用。

$(document).ready(getVideoId(function (response) {
        alert(response);
    }));

function getVideoId(callback){
var http;
var url = base_url + "main/firstVideo";

if (window.XMLHttpRequest){
    // code for IE7+, Firefox, Chrome, Opera, Safari
    http=new XMLHttpRequest();
}
else if (window.ActiveXObject){
    // code for IE6, IE5
    http=new ActiveXObject("Microsoft.XMLHTTP");
}
else{
    alert("Your browser does not support XMLHTTP!");
}

http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", 0);
http.setRequestHeader("Connection", "close");
http.onreadystatechange = function() {

    if(http.readyState == 4 && http.status == 200) {
        callback(http.responseText);
    }
http.open("POST", url, true);
http.send(null);
};
}

想法是我会从数据库中取回video_id并提醒它。 Base_url定义为变量。

Answer 1

你只是从它自己的回调中打开AJAX请求，它永远不会被激活。

http.onreadystatechange = function() {

    if(http.readyState == 4 && http.status == 200) {
        callback(http.responseText);
    }
}; // << close the function here
http.open("POST", url, true);
http.send(null);

AJAX回调不起作用。

1 个答案: