PHP表单提交后如何显示成功消息？

Question

这里是代码。我想要这样--

表单提交 --> page2.php --> 重定向 --> page1.php （这是消息。弹出窗口或其他）

page1.php

<form action="page2.php" method="post" enctype="multipart/form-data" class="form-inline subscribe-form">

                    <input type="name" name="name" placeholder="Jack">
                </div>
                <button type="submit" name="sub" value="sub" >Submit</button>

            </form>

page2.php

    <?php
//include necessary

if(isset($_POST['sub'])) {

    $nameget = mysqli_real_escape_string($dbconnect, $_POST['name']);
    $sqlentry = .....bla bla......//insert into DB
}

$getsql = mysqli_query($dbconnect, $);


if($getsql){
    mysql_close($dbconnect);
    header('location:page1.php');

}



?>

Answer 1

你在哪里：

header('location:page1.php');

在位置上附加一个变量，例如：

header('location:page1.php?status=success');

在 page1.php 上，执行如下操作：

if( $_GET['status'] == 'success'):
    echo 'feedback message goes here';
endif;

Answer 2

这样你的flash消息就不会在刷新后一次又一次地出现。

<?php session_start();
if(isset($_SESSION['msg']) && $_SESSION['msg'] != ''){
    echo $_SESSION['msg'];
    unset($_SESSION['msg']);
}
?>
    <form action="page2.php" method="post" 
enctype="multipart/form-data" class="form-inline subscribe-form">

                    <input type="name" name="name" placeholder="Jack">
                </div>
                <button type="submit" name="sub" value="sub" >Submit</button>    


           </form>

还有

<?php
session_start();
//include necessary

if(isset($_POST['sub'])) {

    $nameget = mysqli_real_escape_string($dbconnect, $_POST['name']);
    $sqlentry = .....bla bla......//insert into DB
}

$getsql = mysqli_query($dbconnect, $);
  if($getsql){
        mysql_close($dbconnect);
        $_SESSIOM['msg'] = 'Value Inserted or whatever';
        header('location:page1.php');        
    } 
?>