[英]ajax request $_get not working
我在用jquery在php中發送get請求時遇到問題。 它只是無法檢測到獲取值,這是我的代碼:
function getopmerking($id) {
document.getElementById("popup_box").style.display = "block";
$httpreq = new XMLHttpRequest();
$httpreq.onreadystatechange = function () {
document.getElementById("txtouders").innerText = $httpreq.responseText;
document.getElementById("requesteddata").innerText = $id;
}
$httpreq.open("GET", "files/request.php?q=1",true);
$httpreq.send();
}
和我的PHP代碼:
$id = $_GET["q"];
fetchData();
function fetchData()
{
$drow = mysql_fetch_array(mysql_query("SELECT * FROM `tblreservering` where fldllnid=$id;"));
if(!empty($drow))
{
$drow['fldopmerking'];
}else{
echo "id is: ".$id."geen gegevens gevonden!";
// i only get "id is: geen gegevens gevonden!" as output since $id is nothing.
}
}
如果您想在JQuery中執行此操作,則將如下所示:
jQuery的:
function getopmerking($id) {
$("#popup_box").show();
$.get("files/request.php?q=1", function(resp){
$("#txttouders").text(resp);
$("#requesteddata").text($id);
});
}
PHP:
$id = $_GET["q"];
fetchData($id);
function fetchData($getId){
$query = sprintf("SELECT * FROM `tblreservering` where fldllnid=%d;", mysql_real_escape_string($getId));
$drow = mysql_fetch_array(mysql_query($query));
if(!empty($drow))
{
$drow['fldopmerking'];
}else{
echo "id is: ".$getId."geen gegevens gevonden!";
// i only get "id is: geen gegevens gevonden!" as output since $id is nothing.
}
}
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