javascript - 当某些事情为真/成功时，如何使 PHP 脚本同时重定向并显示弹出框？

Question

在我的 PHP 代码中，我试图让它在用户成功发送电子邮件后重定向回联系人页面，然后显示一个弹出框，通知用户电子邮件已成功发送:但是，该程序只运行顶行，不运行弹出框代码。我该如何解决这个问题？

<? php
$name = $_POST['name'];
$email = $_POST['email'];
$message = $_POST['message'];
$from = 'From: #';
$to = '#';
$subject = 'Customer Message - Online Form';
$body = "From: ".$name.
"\r\n E-Mail: ".$email.
"\r\n Message: \r\n".$message;

if (isset($_POST['submit'])) {
  if (mail($to, $subject, $body, $from)) {
    header('Location: http://www.google.com.au');
    echo "<script>alert('Message successfully sent.');</script>";
  } else {
    echo '<p>Something went wrong, go back and try again!</p>';
  }
}

?>

Answer 1

在打印警报之前，您正在通过设置标题进行重定向。尝试在 JS 中重定向。

<? php
$name = $_POST['name'];
$email = $_POST['email'];
$message = $_POST['message'];
$from = 'From: #';
$to = '#';
$subject = 'Customer Message - Online Form';
$body = "From: ".$name.
"\r\n E-Mail: ".$email.
"\r\n Message: \r\n".$message;

if (isset($_POST['submit'])) {
  if (mail($to, $subject, $body, $from)) {
    echo "<script>alert('Message successfully sent.'); window.location='http://www.google.com.au'</script>";
    //header('Location: http://www.google.com.au');
  } else {
    echo '<p>Something went wrong, go back and try again!</p>';
  }
}

?>

注意

这会在切换页面之前引起 JS 警告。如果你想改变你登陆的页面，你需要发送一个 $_POST 或 $_GET 变量并在你登陆的页面上监听它