【发布时间】:2026-01-29 10:40:02
【问题描述】:
这是我的代码
function myfoo() {
$.ajax({
type: "POST",
url: "update.php",
dataType: 'json',
async: false,
data: {
Eventid: 1,
Seats: 'Seats'
},
success: function(r) {}
});
}
$(window).on('beforeunload', function() {
return 'Are you sure you want to leave?';
});
$(window).on('unload', function() {
console.log('calling ajax');
myfoo();
});
update.php 代码
<?php
session_start();
include 'conn.php';
if(isset($_SESSION['Seats'])) {
$seatS=$_SESSION['Seats'];
$Eventid=$_SESSION['Eventid'];
$cats = explode(" ", $seatS);
$cats = preg_split('/,/', $seatS, -1, PREG_SPLIT_NO_EMPTY);
foreach($cats as $key => $cat ) {
$cat = mysql_real_escape_string($cats[$key]);
$cat = trim($cat);
if($cat !=NULL) {
$stmt = $con->prepare('UPDATE fistevent SET `Status`=" " where `Event_Id`=? AND `seats`="'.$cat.'" AND `Status`="Hold" ');
$stmt->bind_param("s", $_SESSION['Eventid']);
$stmt->execute();
session_destroy();
}
}
}
?>
【问题讨论】:
-
你有什么问题?
-
Javascript 控制台有错误吗?
-
@Barmar sir iam 试图通过窗口卸载更新记录。但卸载后记录未更新
-
把它放在问题中,而不是评论。
-
当您尝试离开页面时是否会收到提示?
标签: php ajax onbeforeunload onunload