javascript - var ajaxreq = false,ajaxCallback;

问题描述

问 题

/**
 * ajax库
 */
// global variables to keep track of the request and the function to call when done
var ajaxreq = false,ajaxCallback;
// ajaxRequest: Sets up a request
function ajaxRequest(filename) {
    try {
        // Firefox /IE7 /Others
        ajaxreq = new XMLHttpRequest();
    } catch (error) {
        try {
            // IE5 / IE6
            ajaxreq = new ActiveXOject("Microsoft.XMLHTTP");
        } catch (error) {
            return false;
        }
    }
    ajaxreq.open("GET",filename);
    ajaxreq.onreadystatechange = ajaxResponse;
    ajaxreq.send(null);
}
// ajaxResponse: Waits for response and calls a function
function ajaxResponse(){
    if(ajaxreq.readyState != 4) return;
    if(ajaxreq.status == 200) {
        // if the request succeed...
        if(ajaxCallback) ajaxCallback();
    }else alert("Request failed: " + ajaxreq.statusText);
    return true;
}

开头的var ajaxreq = false,ajaxCallback;是声明ajaxreq对象吗？ajaxCallback在这个地方是什么意思？初学JavaScript，请前辈指教哈

function ajaxResponse(){

if(ajaxreq.readyState != 4) return;
if(ajaxreq.status == 200) {
    // if the request succeed...
    if(ajaxCallback) ajaxCallback();
}else alert("Request failed: " + ajaxreq.statusText);
return true;

}
请大神帮忙解释下这个4 和 200吧？还有别的值吗？

解决方案

var ajaxreq = false,ajaxCallback;
只是声明了 两个变量， 形同c语言的 int a=b,c;
ajaxCallback 这里只是声明，实际在用时是个回调，ajaxCallback = function(){//……}

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