全部，
我正在使用 Head First Ajax 一书学习 Ajax。在第一章中，他们给出了一些代码示例，我对其进行了一些简化。我添加了一堆 alert 来了解发生了什么。这是代码:
HTML + Ajax (index.php):
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Rob's Rock 'n' Roll Memorabilia</title>
<link rel="stylesheet" href="css/default.css" />
<script>
function createRequest() {
try {
request = new XMLHttpRequest();
} catch (tryMS) {
try {
request = new ActiveXObject("Msxml2.XMLHTTP");
} catch (otherMS) {
try {
request = new ActiveXObject("Microsoft.XMLHTTP");
} catch (failed) {
request = null;
}
}
}
return request;
}
function getDetails(img){
var title = img.title;
alert("getDetails1");
request = createRequest();
alert("getDetails2");
if (request == null) {
alert("Unable to create request");
return;
}
var url= "getDetails.php?ImageID=" + escape(title);
alert("getDetails3");
request.open("GET", url, true);
alert("getDetails4");
request.onreadystatechange = displayDetails;
alert("getDetails5");
request.send(null);
alert("getDetails6");
}
function displayDetails() {
alert("displayDetails1");
if (request.readyState == 4) {
alert("Request.readyState is 4");
if (request.status == 200) {
alert("Request.status is 200");
detailDiv = document.getElementById("description");
alert("displayDetails2");
detailDiv.innerHTML = request.responseText;
alert("displayDetails3");
}else{
alert("Request.status not 200");
return;
}
}else{
alert("Request.readystate not 4");
return;
}
}
</script>
</head>
<body>
<div id="wrapper">
<div id="thumbnailPane">
<img src="images/SISL_Avatar2.JPG"
title="SISL" id="SISL" onclick="getNextImage()" />
<img src="images/itemGuitar.jpg" width="301" height="105" alt="guitar"
title="itemGuitar" id="itemGuitar" onclick="getDetails(this)"/>
<img src="images/itemShades.jpg" alt="sunglasses" width="301" height="88"
title="itemShades" id="itemShades" onclick="getDetails(this)" />
<img src="images/itemCowbell.jpg" alt="cowbell" width="301" height="126"
title="itemCowbell" id="itemCowbell" onclick="getDetails(this)" />
<img src="images/itemHat.jpg" alt="hat" width="300" height="152"
title="itemHat" id="itemHat" onclick="getDetails(this)" />
</div>
<div id="detailsPane">
<img src="images/blank-detail.jpg" width="346" height="153" id="itemDetail" />
<div id="description"></div>
</div>
</div>
</body>
</html>
<?php
$details = array (
'itemGuitar' => "<p>Pete Townshend once played this guitar while his own axe was in the shop having bits of drumkit removed from it.</p>",
'itemShades' => "<p>Yoko Ono's sunglasses. While perhaps not valued much by Beatles fans, this pair is rumored to have been licked by John Lennon.</p>",
'itemCowbell' => "<p>Remember the famous \"more cowbell\" skit from Saturday Night Live? Well, this is the actual cowbell.</p>",
'itemHat' => "<p>Michael Jackson's hat, as worn in the \"Billie Jean\" video. Not really rock memorabilia, but it smells better than Slash's tophat.</p>"
);
if (isset($_REQUEST['ImageID'])){echo $details[$_REQUEST['ImageID']];}
?>
这是 Ajax (getDetails.php) 调用的 URL:
<?php
$details = array (
'itemGuitar' => "<p>Pete Townshend once played this guitar while his own axe was in the shop having bits of drumkit removed from it.</p>",
'itemShades' => "<p>Yoko Ono's sunglasses. While perhaps not valued much by Beatles fans, this pair is rumored to have been licked by John Lennon.</p>",
'itemCowbell' => "<p>Remember the famous \"more cowbell\" skit from Saturday Night Live? Well, this is the actual cowbell.</p>",
'itemHat' => "<p>Michael Jackson's hat, as worn in the \"Billie Jean\" video. Not really rock memorabilia, but it smells better than Slash's tophat.</p>"
);
echo $details[$_REQUEST['ImageID']];
?>
问题:为什么函数 displayDetails 在就绪状态 4 运行两次？
当我运行上面的代码时，代码似乎运行了两次 displayDetails() 函数。我首先收到 displayDetails1 警报信号，表明我已进入该功能。然后我收到与 readyState 不是 4 相关的警报，然后又不是 4，然后是 4(Request.readyState 是 4)。然后状态警报告诉我状态是 200。到目前为止，没有意外。
在那之后我得到了一些奇怪的东西。我得到了 displayDetails2 警报，然后根据功能修改了页面，我得到了 displayDetails3 警报。然后我希望退出该功能。相反，我再次得到 displayDetails1，Request.readyState 是 4(第二次!)，Request.status 是 200、displayDetails2 和 displayDetails3 警报，就好像整个函数已经运行了第二次。这是为什么？
附言:
1) 在第二轮之后，我得到了我期望的 getDetails6 警报。
2) 页面正常运行——从用户的 Angular 来看，如果警报被禁用，没有什么异常。
3) 我在 WampServer 本地开发，在 WinXP 机器上(我知道......)。
4)如果我添加:
function alert(msg) {
console.log(msg);
}
我的脚本日志只注册了一个readyState is 4...
决议
我做了3个测试:
1 - 只有警报，我得到两个 readyState is 4 警报。
2 - 如果我记录警报，日志只显示一个 readyState is 4 警报。
3 - 如果我同时记录并显示警报弹出窗口(使用 this function )，我会收到两个 readyState is 4 警报(日志显示)。
我对此的看法是，警报本身会导致脚本执行延迟并导致函数有效运行两次。
最佳答案
javascript alert 阻塞了您的 UI 线程，可能足够长，您的浏览器可以完成 AJAX 请求的加载。由于您直到警报之后才检查 request.readyState，因此它可以在您检查之前由浏览器更新。
尝试修改您的事件处理程序:
function displayDetails() {
var rs = request.readyState;
alert("displayDetails1");
if (rs == 4) {
alert("Request.readyState is 4");
//rest of code...
您只会看到一个“Request.readyState is 4”的警报
关于php - 当 readyState 为 4 时，为什么我在脚本中调用的 Ajax 函数会连续运行两次？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/14714908/