我是ajax的新手,我正在尝试将html页面数据发送到php而不重新加载当前页面。当我点击提交按钮时,它只是闪烁“处理......”
html文件
<!DOCTYPE html>
<head>
<script type="text/javascript">
function ajax_post(){
var hr = new XMLHttpRequest();
var url = "simple.php";
var fn = document.getElementById("first_name").value;
var ln = document.getElementById("last_name").value;
var vars = "firstname="+fn+"&lastname="+ln;
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("status").innerHTML = return_data;
}
}
hr.send(vars);
document.getElementById("status").innerHTML = "processing...";
}
</script>
</head>
<body>
First Name: <input id="first_name" name="first_name" type="text"> <br><br>
Last Name: <input id="last_name" name="last_name" type="text"> <br><br>
<input name="myBtn" type="submit" value="Submit Data" onclick="ajax_post();"> <br><br>
<div id="status"></div>
</body>
</html>`
Php档案
<?php
echo 'Thank you '. $_POST['firstname'] . ' ' . $_POST['lastname'] . ', says the PHP file';
?>
修改1
我在onreadystatechange中添加了警告消息。它分别显示三条警报消息2,3,4。
alert(hr.readyState);
答案 0 :(得分:0)
我可以看到很多事情:
</script>
关闭脚本中没有任何开头标记的脚本。vars
没有创建查询字符串,因为它缺少?
。因此,完整的更改将是:
<head>
<script type="text/javascript">
function ajax_post(){
var hr = new XMLHttpRequest();
var url = "simple.php";
var fn = document.getElementById("first_name").value;
var ln = document.getElementById("last_name").value;
var vars = "?firstname="+fn+"&lastname="+ln;
//----------^---put this at start.
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
document.getElementById("status").innerHTML = return_data;
}
}
hr.send(vars);
document.getElementById("status").innerHTML = "processing...";
}
</script>
</head>
或者您可以将其发送为:
hr.send({ firstname:fn, lastname:ln });