我的javascript / ajax代码第一次运行,但之后没有。再次调用该函数时,GetAttribute元素为null。我尝试使用createElement和AppendChild,但它做同样的事情。如果我不需要getAttribute它会工作正常,但我无法使用getAttribute方法的函数。任何帮助将不胜感激。
function ajaxFunction(Picked) {
var getdate = new Date();
if(xmlhttp) {
var Pic1 = document.getElementById("Pic1").getAttribute("name");
var Pic2 = document.getElementById("Pic2").getAttribute("name");
if (Picked === Pic1 ){
var Chosen = Pic1;
var NotChosen = Pic2;
}
else {
var Chosen = Pic2;
var NotChosen = Pic1;
}
xmlhttp.open("POST","choice.php",true);
xmlhttp.onreadystatechange = handleServerResponse;
xmlhttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
xmlhttp.send("Chosen=" + Chosen + "&NotChosen=" + NotChosen );
}
}
function handleServerResponse() {
if (xmlhttp.readyState == 4) {
if(xmlhttp.status == 200) {
var response = xmlhttp.responseText;
response = response.split("|");
document.getElementById('Pic1').innerHTML = response[0];//New Pic
document.getElementById('Pic2').innerHTML = response[1];
}
else {
alert("Error. Please try again");
}
}
}
答案 0 :(得分:1)
将if
更改为
if (Picked === Pic1)
通过撰写if (Picked = Pic1 )
,您将 Picked
分配给Pic1
。