在PHP中保存数据后发出甜警报_Php_Sweetalert

在PHP中保存数据后发出甜警报

php

在PHP中保存数据后发出甜警报,php,sweetalert,Php,Sweetalert,我想在保存数据后显示sweetalert消息。但我遇到了一些问题。请更正我的部分代码。我使用PHP作为后端语言，使用MYSQL作为数据库。这是我的用户_SAVE.php。我已经搜索了相同的场景我已经尝试了代码它对我不起作用可能是代码已经被弃用或使用了其他版本的sweetalert或其他什么 <html> <link rel='stylesheet' href='https://cdn.rawgit.com/t4t5/sweetalert/v0.2.0/lib/sweet-

我想在保存数据后显示sweetalert消息。但我遇到了一些问题。请更正我的部分代码。我使用PHP作为后端语言，使用MYSQL作为数据库。这是我的用户_SAVE.php。我已经搜索了相同的场景我已经尝试了代码它对我不起作用可能是代码已经被弃用或使用了其他版本的sweetalert或其他什么

 <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>


我认为您的mysqli_查询是发送了一些错误并导致页面死亡。“或死亡（mysqli_错误（$con））；”
因此页面不再运行到底部。
我已经清除了你的代码，没有任何查询，只是用PHP来回显脚本，效果很好
<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>

但我遇到了一些问题
哪些是？请提供具体不起作用的内容与您的预期内容的详细信息。您遇到了什么错误？请查看您的查询中一定存在问题，请尝试打印您得到的行数，我尝试了此操作，结果成功$a=1；如果（$a>0）{echo”警报（'Hello this大于0'）；“；}或者{echo”警报（'Hello this小于0'）；“；}您不需要在swal（）周围设置setTimeout（）。
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')");