为什么当我通过 Ajax 将下拉列表中选定的选项传递给 PHP 时，它总是给我第一个选项？
这是我的代码：
<html>
<head>
<script>
comenzar = function(){
document.getElementById('gene').onclick = xmlhttpPost("generador.php");
}
xmlhttpPost = function(strURL){
xmlHttpReq = false;
self = this;
self.xmlHttpReq = new XMLHttpRequest();
self.xmlHttpReq.open('POST', strURL, true);
self.xmlHttpReq.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
self.xmlHttpReq.onreadystatechange = function(){
if(self.xmlHttpReq.readyState == 4){
update(self.xmlHttpReq.responseText);
}
}
self.xmlHttpReq.send(getquerystring());
}
getquerystring = function(){
form = document.getElementById('formu');
combi = document.getElementById('sele').options[document.getElementById('sele').selectedIndex].value;
qstr = "tres=" + escape(combi);
return qstr;
}
update = function(resu){
document.getElementById('tag').innerHTML = resu;
}
window.onload = comenzar;
</script>
</head>
<body>
<form id=formu>
<select id=sele>
<option>C</option>
<option>v</option>
</select>
<button id=gene><p>generate</p></button>
</form>
<p id=tag></p>
</body>
</html>
//generador.php
<?php
echo( "tu combinacion" . $_POST['tres'] );
?>
This is my code:
<html>
<head>
<script>
comenzar = function(){
document.getElementById('gene').onclick = xmlhttpPost("generador.php");
}
xmlhttpPost = function(strURL){
xmlHttpReq = false;
self = this;
self.xmlHttpReq = new XMLHttpRequest();
self.xmlHttpReq.open('POST', strURL, true);
self.xmlHttpReq.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
self.xmlHttpReq.onreadystatechange = function(){
if(self.xmlHttpReq.readyState == 4){
update(self.xmlHttpReq.responseText);
}
}
self.xmlHttpReq.send(getquerystring());
}
getquerystring = function(){
form = document.getElementById('formu');
combi = document.getElementById('sele').options[document.getElementById('sele').selectedIndex].value;
qstr = "tres=" + escape(combi);
return qstr;
}
update = function(resu){
document.getElementById('tag').innerHTML = resu;
}
window.onload = comenzar;
</script>
</head>
<body>
<form id=formu>
<select id=sele>
<option>C</option>
<option>v</option>
</select>
<button id=gene><p>generate</p></button>
</form>
<p id=tag></p>
</body>
</html>
//generador.php
<?php
echo( "tu combinacion" . $_POST['tres'] );
?>
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评论（1）
问题不在于价值的获取。挂钩是这两件事：
...您可能认为您设置了要执行到按钮的 onclick-Event 的某些内容，但您没有。这个函数会立即执行，而不是onclick。
你应该这样写：
另一个问题：
The problem is not the getting of the value. The hook are these two things:
...you may think that you set something to be executed to the onclick-Event of the button, but you don’t. This function will be executed immediately, not onclick.
You should write it that way:
The other problem:
The default type of a
<button>is submit, so if you click on the button, the form will be sent.You can either set the type of the button to "button" or cancel the submitting of the form via JavaScript, otherwise your Ajax request is useless, because the form will be sent (and the page reloaded).