Question

js传值到php，不能接收到值

function load_point(q) {
alert(q);
// Change this depending on the name of your PHP file
downloadUrl("phpsqlajax_genxml2.php?q="+q, function(data) {
......
});
}

function downloadUrl(url, callback) {
var request = window.ActiveXObject ?
new ActiveXObject('Microsoft.XMLHTTP') :
new XMLHttpRequest;

request.onreadystatechange = function() {
if (request.readyState == 4) {
request.onreadystatechange = doNothing;
callback(request, request.status);
}
};

request.open('GET', url, true);
request.send(null);
}

下面是phpsqlajax_genxml2.php文件
<?php
$q=$_GET["q"];
echo $q;
?>
php文件当中不能接收参数q 不知道什么原因
js中alert(q);能打印q值
http://bbs.csdn.net/topics/390937188?page=1#post-398561277

Answer 1

用jquery吧
function downloadUrl(url, callback) {

$.ajax({
url:url,
dataType:'html',
type:'get',
data:{'q':q}

success:function(data,status){
callback(data,status);
},
error:function(){
alert("请求出错");

}

});

}
希望能帮到你