php 使用html表单将记录添加到mysql数据库后显示弹出消息
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Display popup message after record has been added to mysql database using html form
提问by dames
I have the following html form:
我有以下 html 表单：
<html>
<body>
<form action="insert.php" method="post">
Firstname: <input type="text" name="firstname" />
Lastname: <input type="text" name="lastname" />
Age: <input type="text" name="age" />
<input type="submit" />
</form>
</body>
</html>
And this is the .php file:
这是 .php 文件：
<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_db", $con);
$sql="INSERT INTO Persons (FirstName, LastName, Age)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
mysql_close($con);
?>
How do I get a popup message that displays for 2 seconds after a record has been submitted that says for example "Record added"
如何在提交记录后显示显示 2 秒的弹出消息，例如“已添加记录”
回答by NDM
What you are looking for has nothing to do with PHP, as it is client side...
您正在寻找的与 PHP 无关，因为它是客户端...
There are 2 ways to do this:
有两种方法可以做到这一点：
- after you submit render the popup and remove it after x time with javascript...
- submit the form with ajax (javascript) and parse the result, and add popup in javascript.
- 在您提交渲染弹出窗口并在 x 时间后使用 javascript 将其删除后...
- 使用ajax（javascript）提交表单并解析结果，并在javascript中添加弹出窗口。
回答by codingbiz
Try the following. You can also enhance it.
请尝试以下操作。你也可以增强它。
On your form page add the css for popup
在您的表单页面上添加用于弹出窗口的 css
//adduser.php
#popup {
visibility: hidden;
background-color: red;
position: absolute;
top: 10px;
z-index: 100;
height: 100px;
width: 300px
}
and your Popup div
和你的弹出 div
//still adduser.php
<div id="popup">
Record added successfully
</div>
Output this with PHP after success
成功后用PHP输出
//still adduser.php - probably at the bottom of the page
<?php
$recordAdded = false;
if(isset($_GET['status'] && $_GET['status'] == 1)
$recordAdded = true;
if($recordAdded)
{
echo '
<script type="text/javascript">
function hideMsg()
{
document.getElementById("popup").style.visibility = "hidden";
}
document.getElementById("popup").style.visibility = "visible";
window.setTimeout("hideMsg()", 2000);
</script>';
}
?>
The script unhide the popup to show the message and then hides it after 2 seconds
脚本取消隐藏弹出窗口以显示消息，然后在 2 秒后隐藏它
You can enhance the animation with jQuery and the div with CSS
您可以使用 jQuery 增强动画，使用 CSS 增强 div
UPDATE:
更新：
This assumes you will change your .htmlthat contains the form file to .php
这假设您将.html包含表单文件的内容更改为.php
//insert.php
if (!mysql_query($sql,$con))
{
//die('Error: ' . mysql_error());
}
else
{
// 1 record added
//if number of rows affected > 0
header("Location: backtoform.php?status=1"); //redirects back to form with status=1
}
Also try out the suggested jQuery/Ajax method mentioned in the comments
还可以尝试评论中提到的建议的 jQuery/Ajax 方法
