我已经更新了ajax,但仍然没有通过,我快疯了,请有人帮我。它必须工作:(
function ajaxFunction(gotid){
alert (gotid)
$.ajax({
type: "POST",
url: "profile/php/getorder.php",
data: {
dataString: gotid
},
success: function(msg){
alert( "Data Saved: " + msg );
}
});
}
和 PHP
<?php
$ordid = $_POST["dataString"];
$ordid = mysql_real_escape_string($ordid);
echo $ordid;
?>
为什么看不到警报(“数据已保存:” + msg);? 帮助 :(