更改下拉列表上的ajax函数值 [英] ajax function on change dropdown Value
本文介绍了更改下拉列表上的ajax函数值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
大家好！在PHP page1中，我的代码在这里.
Hie Everyone! In PHP page1 my code is here..
<html>
.
...
<select id="customer">...</select>
..
....
<div id="show"></div>
//and Java script function (ajax call)
<script>
$('#customer').change(function(){
var Id = $(this).val();
$.ajax({
type: "GET",
url: "page2.php",
data: "ID="+id,
success: function( data ) {
document.getElementById("show").innerHTML = data;
}
});
});
</script>
</html>
在php page2中作为代码.
In php page2 as code..
<?php
$ID=$_GET['ID'];
...
//db connection code
..
$sql="select * from Table1 where id='$ID'";
//result code..
//while loop..
//echo something..
// all working without error..
?>
因此，当我尝试执行此操作时.它不显示成功数据，或者可能是Ajax函数不起作用.但不会发出任何警报.请帮忙.
So, when I was trying to do this.It does not showing the success data or may be Ajax function not work.I had check with alert(data); but does not Alert anything. please help.
推荐答案
您将在$ get_id变量的前面给出echo.但是，请确保page2.php页面中只有一个回显.
You will give echo infront of the $get_id variable. But you will make sure only one echo in the page2.php page.
<?php
echo $get_id=$_GET['pass_id'];
...
//db connection code
..
$sql="select * from Table1 where id='$get_id'";
//result code..
//while loop..
//echo something..
// all working without error..
?>
然后在page1.php中检查您的ajax响应.使用警报功能.
Then in page1.php check your ajax response. using alert function.
<script>
$('#customer').change(function(){
var id = $(this).val();
$.ajax({
type: "GET",
url: "page2.php",
data: "pass_id="+id,
success: function( data ) {
alert(data);
document.getElementById("show").innerHTML = data;
}
});
});
</script>
这篇关于更改下拉列表上的ajax函数值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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