我想在保存数据后显示sweetalert消息。但我遇到了一些问题。请纠正我的部分代码。我使用PHP作为我的后端语言，使用MYSQL作为我的数据库。这是我的USER_SAVE.php。我已经搜索了相同的场景我已经尝试了它对我不起作用的代码可能是代码已经被弃用或使用其他版本的sweetalert或者什么。
<html>
<link rel='stylesheet' href='https://cdn.rawgit.com/t4t5/sweetalert/v0.2.0/lib/sweet-alert.css'>
<?php session_start();
if(empty($_SESSION['id'])):
header('Location:../index');
endif;
include('../dist/includes/dbcon.php');
$rec= mysqli_real_escape_string($con,$_POST['rec']);
$bdo= mysqli_real_escape_string($con,$_POST['bdo']);
$can_name= mysqli_real_escape_string($con,$_POST['can_name']);
$po_ti= mysqli_real_escape_string($con,$_POST['po_ti']);
$client= mysqli_real_escape_string($con,$_POST['client']);
$rdr= mysqli_real_escape_string($con,$_POST['rdr']);
$de= mysqli_real_escape_string($con,$_POST['de']);
$remarks= mysqli_real_escape_string($con,$_POST['remarks']);
$f_back= mysqli_real_escape_string($con,$_POST['f_back']);
$datee= mysqli_real_escape_string($con,$_POST['datee']);
$status= mysqli_real_escape_string($con,$_POST['status']);
$tnum= mysqli_real_escape_string($con,$_POST['tnum']);
$query=mysqli_query($con,"SELECT * FROM accounts_at WHERE can_name='$can_name'")or die(mysqli_error());
$count=mysqli_num_rows($query);
if ($count>0)
{
echo "<script type='text/javascript'>alert('Account already exist');</script>";
echo "<script>document.location='index'</script>";
}
else{
mysqli_query($con,"INSERT INTO accounts_at(id,tnum,rec,bdo,can_name,po_ti,client,rdr,de,remarks,f_back,datee,status)
VALUES(NULL,'$tnum','$rec','$bdo','$can_name','$po_ti','$client','$rdr','$de','$remarks','$f_back','$datee','$status')")or die(mysqli_error($con));
mysqli_query($con,"INSERT INTO accounts_at_action(id,tnum,rec,bdo,can_name,po_ti,client,rdr,de,remarks,f_back,datee,status)
VALUES(NULL,'$tnum','$rec','$bdo','$can_name','$po_ti','$client','$rdr','$de','$remarks','$f_back','$datee','$status')")or die(mysqli_error($con));
/*echo "<script type='text/javascript'>
alert('Successfuly added new applicant');</script>";*/
echo "
<script type='text/javascript'>
setTimeout(function () {
swal('Successfully Added a Account!')
},1);
window.setTimeout(function(){
window.location.replace('index.php');
} ,3000);
</script>";
// echo "<script>document.location='index'</script>";
}
?>
<script src='https://cdn.rawgit.com/t4t5/sweetalert/v0.2.0/lib/sweet-alert.min.js'></script>
</html>
我认为你的mysqli_query发送了一些错误并使页面死掉。 “或死（mysqli_error（$ con））;”
所以页面不再运行到底部。我已经清除你的代码，没有任何查询只是PHP来回应脚本，它的工作正常。
<html>
<link rel='stylesheet' href='https://cdn.rawgit.com/t4t5/sweetalert/v0.2.0/lib/sweet-alert.css'>
<?php
echo "
<script type='text/javascript'>
setTimeout(function () {
swal('Successfully Added a Account!')
},1);
window.setTimeout(function(){
window.location.replace('index.php');
} ,3000);
</script>
";
?>
<script src='https://cdn.rawgit.com/t4t5/sweetalert/v0.2.0/lib/sweet-alert.min.js'></script>
</html>
您可以通过回显结果来检查您的查询
echo mysqli_query($con,"INSERT INTO accounts_at(id,tnum,rec,bdo,can_name,po_ti,client,rdr,de,remarks,f_back,datee,status)
VALUES(NULL,'$tnum','$rec','$bdo','$can_name','$po_ti','$client','$rdr','$de','$remarks','$f_back','$datee','$status')");