单击“返回”按钮后阻止重新提交表单

Question

I have 2 pages :

page1.php :
- has a form with text box and a "submit" button. Eg : <form name="frm_register" action="page1.php" method="post">

- php and mysql code to store the value of textbox to database. Javascript will redirect the page to php2.php after the value is submitted to database. Eg :

$query = "INSERT INTO traceuser (username) VALUES ('{$username}')";
$result = mysql_query($query, $connection);
echo '<script language="javascript">window.location="page2.php";</script>';

page2.php
- mysql retrieve the data from database and display on this page.

Problem : When I press "back" button, the browser will pop up a warning message saying that the form will be resubmit. How to prevent resubmit the form when click "back" button? Is it I need to clear the cache of page1.php? How to do it with php or javascript or ajax?

---

Update 1 : Thanks for the answer of replacing javascript window.location="page2.php" to php header('Location: home2.php'); . It fix 80% of problem. The rest of 20% problem show below :

  if (strtotime($_SESSION['servertime']) < time()-3){ //10800 = 3 hours 3600 = 1 hour if (($username != "") AND ($username != $_SESSION[username])){ $_SESSION['servertime'] = $servertime; $_SESSION['username'] = $username; $query = "INSERT INTO traceuser (username) VALUES ('{$username}')"; $result = mysql_query($query20, $connection); header('Location: page2.php'); exit; } else { echo "same name"; //problem here } }else{ echo "submit multiple data too fast"; //problem here too. } } 

The problem happen when do the following steps :
1) User submit data successfully, jump to page2.php view records.
2) User click "back" button, jump back to page1.php.
3) User submit data fail, stay on page1.php. (because too fast or same name)
4) User submit data successful, jump to page2.php view records.
5) User click "back" button, but browser shows warning message "form will be resubmited".

The problem is because of Step 3. Step 3 didn't run header('Location: page2.php'); , didn't jump to page2.php. So it cause Step 5 show the warning message. How to fix this problem?

---

Update 2 : I have figured out the solution to fix the 20% problem, it works perfectly. I use session['error123'] to decide whether or not want to display the error message "same name". I kill session['error123'] if success submit data to database or if success jump to page2.php. I also use header('Location: page1.php'); to redirect to own page (same page) to make the page forget about form submission previously. Example of codes :

 if ($_SESSION['error123'] == "toofast"){ echo $_SESSION['error123'] ; }elseif ($_SESSION['error123'] == "samename"){ echo $_SESSION['error123'] ; } if (strtotime($_SESSION['servertime']) < time()-3){ //10800 = 3 hours 3600 = 1 hour if (($username != "") AND ($username != $_SESSION['username'])){ $_SESSION['username'] = $username; $query = "INSERT INTO traceuser (username) VALUES ('{$username}')"; $result = mysql_query($query20, $connection); $_SESSION['error123'] = "aa"; header('Location: http://localhost/plekz/page2.php'); exit; } else { $_SESSION['error123'] = "samename"; header('Location: http://localhost/plekz/page1.php'); exit; } }else{ $_SESSION['error123'] = "toofast"; header('Location: http://localhost/plekz/page1.php'); exit; } } } 

Note : You need to buffer the output by <?php ob_start();?> because $_SESSION cannot put before header(). Buffer will stop all output including session, let header() send the output first.

Answer 1

Rather than

echo '<script language="javascript">window.location="page2.php";</script>';

you should use the header() function to redirect your user after the submission.

So in psuedo code,

click submit on page.php action page1.php page1.php submits data to database calls

header('Location: http://example.com/page2.php');

This should prevent your clicking back problem

Answer 2

You can prevent the re-submission by implementing the Post-Redirect-Get (PRG Pattern) .

Could be just a one-line if you've got the http_redirect function:

http_redirect("page2.php");

Instead of your javascript echo.

If not, that are two lines:

header("Location: http://example.com/page2.php");
exit;

Replace example.com with site's your hostname.

Related: Back button re-submit form data ($_POST) ; I am confused about PHP Post/Redirect/Get

Answer 3

Add this code in the page that is showing as offline when the user clicks the back button:

<?php
 session_start();
 header_remove("Expires");
 header_remove("Cache-Control");
 header_remove("Pragma");
 header_remove("Last-Modified");
?>

Answer 4

One way is to submit the Formdata via Ajax to a remote Script and if the Query returns success you can jump the a "Thank You" Page. So the User can hit the Back Button and the "Reload" Request doesn't pop up.

Hope the Idea helps you

Answer 5

Can you do it via an Ajax call instead? No action on the form, and the submit will call a the Ajax function. The Ajax call will execute the query, and provide a response (you can just echo a result), and you can then provide dynamic feedback based on the result. You'd never leave the page.

<form id="thisForm">
...form input fields...
<input type="button" onclick="return submitForm('thisForm')"/>
</form>

function submitForm(formId) {
    $.ajax( {
        type: "post",
        url: 'page2.php',
        data: $('#' + formId + ' input').serialize(),
        ... any other Ajax parameters...,
        success: function(data) {
        }
    });
    return false;
}

Answer 6

Create a Session like shown here

You should use session and validate the user from every page and you will amaze how SESSION works! AMAZING!