Javascript 如何将XMLHttpRequest中的查询字符串参数发送到php页面
我有以下php变量的名称Javascript 如何将XMLHttpRequest中的查询字符串参数发送到php页面,javascript,php,ajax,Javascript,Php,Ajax,我有以下php变量的名称 $name=$_POST["name"]; <script> pge = '<?php echo $name ;?>'; var url = "searchCustomerByNameApi2.php" //some code xmlhttp.open("GET", url, true); xmlhttp.send(); $name=$\u POST[“name”]; pge
$name=$_POST["name"];
<script>
pge = '<?php echo $name ;?>';
var url = "searchCustomerByNameApi2.php"
//some code
xmlhttp.open("GET", url, true);
xmlhttp.send();
$name=$\u POST[“name”];
pge='';
var url=“searchCustomerByNameApi2.php”
//一些代码
open(“GET”,url,true);
xmlhttp.send();
如何将pge变量与url一起作为一个查询字符串发送,并在php页面上访问它???您可以使用函数从对象创建查询字符串: 并将该结果附加到您的url:
$name=$_POST["name"];
<script>
function encodeQueryData(data) {
let ret = [];
for (let d in data)
ret.push(encodeURIComponent(d) + '=' + encodeURIComponent(data[d]));
return ret.join('&');
}
pge = '<?php echo $name ;?>';
var data = {pge: pge};
var query =encodeQueryData(data);
var url = "searchCustomerByNameApi2.php";
url += "?" +query;
xmlhttp.open("GET", url, true);
xmlhttp.send();
$name=$\u POST[“name”];
函数encodeQueryData(数据){
设ret=[];
for(让d输入数据)
ret.push(encodeURIComponent(d)+'='+encodeURIComponent(data[d]));
返回ret.join('&');
}
pge='';
var data={pge:pge};
var query=encodeQueryData(数据);
var url=“searchCustomerByNameApi2.php”;
url+=“?”+查询;
open(“GET”,url,true);
xmlhttp.send();
简单您的javascript代码如下所示
<script>
var name = '';
name = '<?php print $_POST["name"]?>';
var url = "searchCustomerByNameApi2.php?name"+name;
xhttp.open("GET", url, true);
xhttp.send();
</script>
你试过下面的吗
var url = "searchCustomerByNameApi2.php";
var test = "name=<?php echo $_POST["name"]; ?>";
xmlhttp.open("GET", url, true);
xmlhttp.send(test);
var url=“searchCustomerByNameApi2.php”;
var test=“name=”;
open(“GET”,url,true);
发送(test);
希望这能解决您的问题。试试
“searchCustomerByNameApi2.php?pge=“+pge
”,然后像$\u REQUEST['pge']
一样在php页面上获取它。非常感谢它添加了这个答案,请接受并投票,如果有用的话
var url = "searchCustomerByNameApi2.php";
var test = "name=<?php echo $_POST["name"]; ?>";
xmlhttp.open("GET", url, true);
xmlhttp.send(test);