使用ajax传递多个值并在另一页中回显 [英] Pass multiple values using ajax and echo in another page

问题描述

function getData(des, poid, divid){
            $.ajax({
                type: 'GET',
                url: 'loaddata.php?des=' + des + '&poid='+poid, //call storeemdata.php to store form data

                success: function(data) {
                    var ajaxDisplay = document.getElementById(divid);
                    ajaxDisplay.innerHTML = data;
                }
            });
        }

//Function call
<select name="descr" id="descr" onchange="getData(this.value,this.value,   'displaydata')">

这是代码，我在url中传递了多个值.当我尝试在下一页中回显这些值时，它仅回显一个值(des)两次.任何帮助将不胜感激.

This is the code and I passed multiple values in url. when I try to echo those values in next page it echo only one value (des) two times.Any help would be appreciated.

$slno = $_GET['des'];
$poid = $_GET['poid'];
echo $slno;
echo $poid;

推荐答案

我找到了一种方法.感谢所有研究它的人来帮助我.这是代码，它就像一个魅力.

I found a way to do it. Thanks to everyone who looked into it to help me. Here is the code and it works like a charm.

function getData(des, divid){
    var poid = $("select[name='poid']").val();//this line pass second value "poid" in my case
            $.ajax({
                type: 'GET',
                url: 'loaddata.php?',
                data: {des : des, poid : poid},
                success: function(data) {
                    var ajaxDisplay = document.getElementById(divid);
                    ajaxDisplay.innerHTML = data;
                }
            });
        }

函数调用

<select name="des" id="des" onchange="getData(this.value, 'displaydata')">

ajax.php文件

ajax.php file

$slno = $_GET['des'];
$pid1 = $_GET['poid'];
echo $slno;
echo $pid1;

这篇关于使用ajax传递多个值并在另一页中回显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！