sweetalert2 脚本无法被 php 使用 echo 识别。我在 github 上下载了脚本文件，当我将 sweetalert 脚本放入 php 时，脚本消息不会显示在我的上。但是，当我将 sweetalert 脚本放入脚本部分时，它可以工作，但它很困难，因为当单击按钮时，会显示 sweetalert 脚本，这是我想避免的。代码如下：
更新功能：
<?php
// UPDATE FUNCTION
include '../connection.php';
if(isset($_POST['update'])) {
$assetsId = $_POST['assetsId']; // RETRIEVE THE ID OF THE assetsId TO BE UPDATED VIA HIDDEN FIELD IN HTML
$assetsName = $_POST['assetsName'];
$assetsDesc = $_POST['assetsDesc'];
$assetsQuantity = $_POST['assetsQuantity'];
$assetsLocation = $_POST['assetsLocation'];
$assetsStatus = $_POST['assetsStatus'];
$assetsUsage = $_POST['assetsUsage'];
//QUERY OF UPDATE
$update_query = "UPDATE Admin_add_assets SET
assetsName = '$assetsName',
assetsDesc = '$assetsDesc',
assetsQuantity = '$assetsQuantity',
assetsLocation = '$assetsLocation',
assetsStatus = '$assetsStatus',
assetsUsage = '$assetsUsage'
WHERE assetsId = $assetsId";
if(mysqli_query($connection, $update_query)) {
echo "
Swal.fire({
title: 'Good job!',
text: 'You clicked the button!',
icon: 'success'
});
";
} else {
echo "<script>alert('Error deleting record: " . mysqli_error($connection) . "');</script>";
}
}
?>
html 文件：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<script src="../js/sweetalert2.min.js"></script>
<link rel="stylesheet" href="style.css">
<link rel="stylesheet" href="../css/bootstrap.min.css">
<link rel="stylesheet" href="../css/sweetalert2.min.css">
<title>DASHBOARD (ADMIN)</title>
</head>
<!-- and so fort -->
如何在我的代码中显示 sweetalert2 弹出窗口
您需要告诉系统Swal.fire是一个javascript函数，因此需要用
<script>....</script>所以改变
if(mysqli_query($connection, $update_query)) {
echo "
Swal.fire({
title: 'Good job!',
text: 'You clicked the button!',
icon: 'success'
});
";
}
到
if(mysqli_query($connection, $update_query)) {
echo "
<script>
Swal.fire({
title: 'Good job!',
text: 'You clicked the button!',
icon: 'success'
});
</script>
";
}