Form problem

seaking197

There are two pages a.php and b.php
 a.php filling form data
 b.php is used to extract the form content and add records!
 But clicking back on the menu will add an extra record. This will judge a few more.
 Excuse me: Is there any good way, or is there a way to delete the content from the post after the submission is successful!

knicks

I usually do this

<?php
if(isset($_POST['submit'])&&$_POST['submit'] == "OK"&&$_POST['lastAction']>$_SESSION['userLastAction'])
{
$_SESSION['userLastAction']++;
...
}
?>

<form name="form1" method=post action="<?=$_SERVER['PHP_SELF'] ?>">
...

<input type=Hidden name=lastAction value=<?=$_SESSION['userLastAction']+1 ?>>
<input type=reset name=reset value="reset">&nbsp;
<input type=submit name=submit value="OK"
</form>

seaking197

Unique idea!
I solved it in this way, but I just can't return, can you return like that>?
<form action="a.php?act=regiest">
</form>
After adding
Refresh this page
location.href='b.php'

seaking197

But when act is passed with hidden, an error will occur!

qijun2005

Cookie+timestamp to prevent

knicks

Can return
Through judgment, the content in the IF will not be executed

yezi4321

The problem description is unclear!

On which page is "return"? When is "return"? After successfully inserted into the database? Which page is returned?

The data actually exists in the browser's buffer. So after refreshing the page, a piece of data will be inserted. The solution is to use header("location:".$_SERVER['PHP_SELF']);