I think that this problem occurs often on a web application development. But I'll try to explain in details my problem.
I'd like to know how to correct this behavior, for example, when I have a block of code like this :
<?
if (isset($_POST['name'])) {
... operation on database, like to insert $_POST['name'] in a table ...
echo "Operation Done";
die();
}
?>
<form action='page.php' method='post' name="myForm">
<input type="text" maxlength="50" name="name" class="input400" />
<input type="submit" name="Submit" />
</form>
When the form gets submitted, the data get inserted into the database, and the message Operation Done is produced. Then, if I refreshed the page, the data would get inserted into the database again.
How this problem can be avoided? Any suggestion will be appreciated :)
Don't show the response after your create action; redirect to another page after the action completes instead. If someone refreshes, they're refreshing the GET requested page you redirected to.
// submit
// set success flash message (you are using a framework, right?)
header('Location: /path/to/record');
exit;
Set a random number in a session when the form is displayed, and also put that number in a hidden field. If the posted number and the session number match, delete the session, run the query; if they don't, redisplay the form, and generate a new session number. This is the basic idea of XSRF tokens, you can read more about them, and their uses for security here: http://en.wikipedia.org/wiki/Cross-site_request_forgery
Here is an example:
<?php
session_start();
if (isset($_POST['formid']) && isset($_SESSION['formid']) && $_POST["formid"] == $_SESSION["formid"])
{
$_SESSION["formid"] = '';
echo 'Process form';
}
else
{
$_SESSION["formid"] = md5(rand(0,10000000));
?>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<input type="hidden" name="formid" value="<?php echo htmlspecialchars($_SESSION["formid"]); ?>" />
<input type="submit" name="submit" />
</form>
<?php } ?>
I ran into a similar problem. I need to show the user the result of the POST. I don't want to use sessions and I don't want to redirect with the result in the URL (it's kinda secure, I don't want it accidentally bookmarked). I found a pretty simple solution that should work for the cases mentioned in other answers.
On successfully submitting the form, include this bit of Javascript on the page:
<script>history.pushState({}, "", "")</script>
It pushes the current URL onto the history stack. Since this is a new item in history, refreshing won't re-POST.
UPDATE: This doesn't work in Safari. It's a known bug. But since it was originally reported in 2017, it may not be fixed soon. I've tried a few things (replaceState, etc), but haven't found a workaround in Safari. Here are some pertinent links regarding the issue:
Safari send POST request when refresh after pushState/replaceState
https://bugs.webkit.org/show_bug.cgi?id=202963
https://github.com/aurelia/history-browser/issues/34
Like this:
<?php
if(isset($_POST['uniqid']) AND $_POST['uniqid'] == $_SESSION['uniqid']){
// can't submit again
}
else{
// submit!
$_SESSION['uniqid'] = $_POST['uniqid'];
}
?>
<form action="page.php" method="post" name="myForm">
<input type="hidden" name="uniqid" value="<?php echo uniqid();?>" />
<!-- the rest of the fields here -->
</form>
I think it is simpler,
page.php
<?php
session_start();
if (isset($_POST['name'])) {
... operation on database, like to insert $_POST['name'] in a table ...
$_SESSION["message"]="Operation Done";
header("Location:page.php");
exit;
}
?>
<html>
<body>
<div style='some styles'>
<?php
//message here
echo $_SESSION["message"];
?>
</div>
<form action='page.php' method='post'>
<!--elements-->
</form>
</body>
</html>
So, for what I needed this is what works.
Based on all of the above solutions this allows me to go from a form to another form, and to the n^ form , all the while preventing the same exact data from being "saved" over and over when a page is refreshed (and the post data from before lingers onto the new page).
Thanks to those who posted their solution which quickly led me to my own.
<?php
//Check if there was a post
if ($_POST) {
//Assuming there was a post, was it identical as the last time?
if (isset($_SESSION['pastData']) AND $_SESSION['pastData'] != $_POST) {
//No, Save
} else {
//Yes, Don't save
}
} else {
//Save
}
//Set the session to the most current post.
$_session['pastData'] = $_POST;
?>
We work on web apps where we design number of php forms. It is heck to write another page to get the data and submit it for each and every form. To avoid re-submission, in every table we created a 'random_check' field which is marked as 'Unique'.
On page loading generate a random value and store it in a text field (which is obviously hidden).
On SUBMIT save this random text value in 'random_check' field in your table. In case of re-submission query will through error because it can't insert the duplicate value.
After that you can display the error like
if ( !$result ) {
die( '<script>alertify.alert("Error while saving data OR you are resubmitting the form.");</script>' );
}
No need to redirect...
replace die(); with
isset(! $_POST['name']);
, setting the isset to isset not equal to $_POST['name'], so when you refresh it, it would not add anymore to your database, unless you click the submit button again.
<?
if (isset($_POST['name'])) {
... operation on database, like to insert $_POST['name'] in a table ...
echo "Operation Done";
isset(! $_POST['name']);
}
?>
<form action='page.php' method='post' name="myForm">
<input type="text" maxlength="50" name="name" class="input400" />
<input type="submit" name="Submit" />
</form>
This happen because of simply on refresh it will submit your request again.
So the idea to solve this issue by cure its root of cause.
I mean we can set up one session variable inside the form and check it when update.
if($_SESSION["csrf_token"] == $_POST['csrf_token'] )
{
// submit data
}
//inside from
$_SESSION["csrf_token"] = md5(rand(0,10000000)).time();
<input type="hidden" name="csrf_token" value="
htmlspecialchars($_SESSION["csrf_token"]);">
I think following is the better way to avoid resubmit or refresh the page.
$sample = $_POST['submit'];
if ($sample == "true")
{
//do it your code here
$sample = "false";
}
Related
I currently make my first Wordpress website using Java script snippets for a countdown that refreshes itself constantly and allows me to click a button once every 6 hours. So I managed that the time refreshes itself but I need one permanent variable that tells me if I already clicked the button or not (since it should work wether I refresh the page, go to another side or even log in with another user).
Basically I just want one variable that changes between 0 and 1.
I implemented a hidden input field and it works just fine as long as I stay on the side (refreshing the side works as well) but as soon as I change the page it sets the variable back. I tried to implement a global variable in the function.php of my theme (and a function as well) but it still doesn't work.
I tried it like this:
global $x;
echo $x;
And this:
function displayX() {
global $x;
$x = "0";
echo $x;
}
The thing is I don't want to set a value because the variable needs to be changeable any time.
That's my current html:
<input type="text" id="id1" value="<?php echo $x; ?>" <="" input="">
But it just doesn't work.
My second approach was to make the input field permanent (but updateable) - again this approach works as long as I don't change the side.
I tried it like this:
<span id="00">
<input type="text" id="id1">
</span>
Can anybody please help me? Also please specifiy where I have to set the global variable since there are so many function.php files.
THANK YOU!
Easiest way to do that is using of update_option and get_option.
update_option is for save data to database that will be permanent.
get_option is for fetching data from database.
<form method="post">
<input type="text" name="permanent" id="permanent" value="<?php echo get_option('permanent_data'); ?>"/>
<input type="submit" name="save" value="submit"/>
</form>
You can catch form data in backend using an action like this:
in functions.php
add_action('init','save_permanent');
function save_permanent(){
if(isset($_POST['save'])){
update_option('permanent_data', $_POST['permanent']);
}
}
Below code checks that if form is submitted:
if(isset($_POST['save'])){
update_option('permanent_data', $_POST['permanent']);
}
If form submitted it gets value of input text that named permanent
Mentioned code permanent the data in database as name of permanent_data
If you want to fetch stored data you just call get_option(option name) like this:
<?php echo get_option('permanent_data'); ?>
For first question you can do it in such way:
in functions.php
<?php
if(if(isset($_POST['save']))){
update_option('already_clicked', '1');
}
And for fetch stored data you can use:
<?php
if(get_option('already_clicked') == '1'){
//do somthing
}
?>
I am new to AJAX and am in the process of converting some regular HTML forms to AJAX.
My existing implementation is as follows - form (on page1.php) posts to page2.php which does some validation on post data and redirects to an error page if something is missing. If the input is fine, it includes page3.php which processes the request and redirects back to page1.php.
php/page1.php
<form method="post" action="/php/page2.php" >
<input type="text" name="input1" placeholder="Howdy..." />
<input type="text" name="input2" placeholder="Howdy..." />
<input type="submit" value="Submit" />
</form>
php/page2.php
<?php
// perform some validation on inputs
if (empty($_POST['input1']))
{
$location ='Location: /php/error.php';
header($location);
exit;
}
// Inputs are fine
include('/php/page3.php');
?>
page3.php
<?php
// do some form processing
// redirect back to page1.php
$location = 'Location: /php/page1.php";
header($location);
exit;
?>
To convert to AJAX, I am using #SSL's solution on this SO link How to show loading gif when request goes Ajax
http://jsfiddle.net/clickthelink/Uwcuz/1/
The error from validation and success page are both displayed back on page1.php via the callback function.
php/page2.php
<?php
// perform some validation on inputs
if (empty($_POST['input1']))
{
// Echo erorr code isntead of redirect
echo "Please enter input1";
return;
//$location ='Location: /php/error.php';
//header($location);
//exit;
}
// Inputs are fine
include('/php/page3.php');
?>
page3.php
<?php
// do some form processing
// Echo success instead of redirect
echo "SUCCESS";
// redirect back to page1.php
//$location = 'Location: /php/page1.php";
//header($location);
//exit;
?>
This part is working fine.
My question (finally) is how do I handle users who have javascript disabled? I know the form will get submitted appropriately but I wont get the redirect back in case of the error or success. I would like to retain header() redirect type of functionality in this case also. Is this possible? I would appreciate the help.
You want to detect if this is an xhr request, and default to the non-ajax behavior if it is not.
I would look at $_SERVER['HTTP_X_REQUESTED_WITH']
Keep your current form setup as-is, if it is working for you without javascript.
For javascript enabled browsers you can hijack the 'submit' event on the form. Capture the event and post the form, via ajax, to scripts/pages that handle and return the data in a javascript-friendly format for final consumption.
For example, using jquery:
<form method="post" action="/php/page2.php" id="js-form" >
<input type="text" name="input1" placeholder="Howdy..." />
<input type="text" name="input2" placeholder="Howdy..." />
<input type="submit" value="Submit" />
</form>
<script>
$(document).ready(function(){
$('#js-form').on('submit',function(e){
// logic to submit ajax form and handle response
// return false to cancel native browser form submission.
return false;
});
});
</script>
Another idea is to keep the pages you already have, but send a flag with the ajax request to disable the browser redirect headers. For example, add 'src=ajax' when submitting the form via ajax. Then in the script use logic to say:
<?php
if( !empty($_REQUEST['src'] && $_REQUEST['src'] == 'ajax' ) {
// add redirect logic here.
}
?>
I have created a comment-reply system for my blog in php. Comments are stored in a table called comments(comments_id, comment, comment_date, user, flag). I use a script which displays comments and near to each comment there is a link called "delete" in order for the user to delete its own comment in case want to do so.
my php script for displaying comments is :
<?php
// ... above code
$comments .= $row['comment']; // comment printed succesfully here
if($comment_user_id == $session_user_id){
$comments .="<table border='1' style='display:inline-table;'><td><h2><font size='2'>
<form action='deletepost.php' method='post'>
<input type='hidden' name='var' value='$comment_id;'>
<input type='submit' value='delete'>
</form>
</h2></font></td></table>";
}
?>
in order to delete a comment, I update the table comments and set flag=1, so my script will not display comments having their flag=1 in table. In order to do this I use script deletepost.php
<?php
$comment = mysql_real_escape_string($_POST['var']);
if(isset($comment) && !empty($comment)){
mysql_query("UPDATE `comments` SET `flag`=1 WHERE (`user`='$session_user_id' AND `comments_id`='$comment')");
header('Location: wall.php');
}
?>
My script until now works perfect without problem and the user that posts a comment can delete its own comment without any error. The problem started when I decided to insert a lightbox in javascript, so that the user will be asked before deleting a comment. So I have changed my first script to the following:
<?php
// ... above code
$comments .= $row['comment']; // comment printed succesfully here
if($comment_user_id == $session_user_id){
$comments .="<table border='1' style='display:inline-table;'><td><h2><font size='2'>
// at this point problem occurs when inserting javascript lightbox
<a href = javascript:void(0) onclick = document.getElementById('light').style.display='block';document.getElementById('fade').style.display='block'><h2><font color=green size=3>Delete All</font></h2></a>
<div id=light class=white_content>DELETE THIS COMMENT?
<form action='deletepost.php' method='post'>
<input type='hidden' name='var' value='$comment_id;'>
<input type='submit' value='delete'>
</form>
<a href = javascript:void(0) onclick = document.getElementById('light').style.display='none';document.getElementById('fade').style.display='none'><div id=pading><button>Cancel</button></div></a></div>
<div id=fade class=black_overlay></div>
</h2></font></td></table>";
}
?>
By using the javascript lightbox as shown above, when the user will press delete, a lightbox starts and asks user if wants to delete the comment. The problem is that now when the user press delete button, it is not deleting the certain comment but the last comment that finds in comments table. Probably there is something else I need to write in my javascript to correct this, in order to know which comment to delete (set its flag to 1). Any idea how to fix it?
Make sure the 'deletepost.php' is getting the correct comment id($comment) that you wanted to delete.
May be that can be the cause.
I have a simple counter for the form button clicks which works well but when the user refreshes the page it adds onto the counter again. I would be grateful if someone would help with the coding I have wrote below.
PHP CODE
<?php
if( isset($_POST['clicks']) ) {
incrementClickCount();
}
function getClickCount()
{
return (int)file_get_contents("clickcount.txt");
}
function incrementClickCount()
{
$count = getClickCount() + 1;
file_put_contents("clickcount.txt", $count);
}
?>
HTML and PHP FORM CODE
<form action="<?php $_SERVER['PHP_SELF']; ?>" method="post">
<input type="submit" value="click me!" name="clicks">
</form> <div>Click Count: <?php echo getClickCount(); ?></div>
To prevent this you need to implement the POST/REDIRECT/GET pattern.
Afroze has a good point. Using a header refresh will reset all form data, but it forces the page to refresh twice (when the form is first submitted, and second time when the header forces another refresh). There is another method of preventing a "refresh submission" without forcing a second refresh.
Try this (I'll explain the code below):
<?php
//begin a session
session_start();
//generate a MD5 hash from concatenated form $_POST values
$formDataHash = md5($_POST['clicks'] . $_POST['otherFormContent']);
//if form has been submitted before, and the data passed from the form
//is the same as the previous submission, return false (don't do anything)
if(isset($_SESSION['formData']) && $_SESSION['formData'] == $formDataHash)
{
return false;
}
else
{
//increase click count
incrementClickCount();
//store the submission values
$_SESSION['formData'] = $formDataHash;
}
The code above will detect a each form submission, but won't process duplicates. It also won't refresh the page twice (which isn't ideal for users with slow connections).
Just redirect the request to the same page after processing the form post. This can be done as follows in PHP:
<?php header('Location: http://www.example.com/'); ?>
If the user refreshes (after this redirect) nothing would be resubmitted.
I a have PHP form where I collect a bunch of values from text inputs, but for one input I have the input filled in via javascript (user selects a date from a calendar, that date then populates a text input). I've setup a simplified version of this:
<?php
$displayForm = true;
if ($_POST['submitFlag'] == 1) {
// Form was submitted. Check for errors and submit.
$displayForm = false;
$installationTime = $_POST['installation-time'];
// send e-mail notification
$recipients = "test#test.com";
$subject = "Test Email - Test Form Submission";
$message = wordwrap('Someone has filled out the secure form on test.com. Here\'s what they had to say:
Installation Time: ' . $installationTime .'
');
$headers = "From: test#test.com";
mail($recipients, $subject, $message, $headers);
// Output thank you message
?>
<h2>Thank You!</h2>
<?php if($installationTime == NULL){echo 'test failed: value submitted was null.';}else{echo 'test passed: value submitted was not null.';} ?>
<p>Your form has been submitted. Thank you for your interest in test.com.</p>
<?php
}
if ($displayForm) {
// If form was not submitted or errors detected, display form.
?>
<div class="note"><span class="required">*</span> Click me to set value of input.</div>
<form name="contactForm" id="contactForm" method="post" enctype="multipart/form-data" action="<?php echo $_SERVER['PHP_SELF']; ?>?state=submit">
<label for="installation-time" class="service-time">The time you have selected for installation is: <span class="required">*</span></label>
<input type="text" name="installation-time" id="installation-time" disabled value="<?php echo $_POST['installation-time']; ?>" />
<input type="hidden" name="submitFlag" id="submitFlag" value="1" />
<input type="submit" name="submit" id="submit" value="Sign-Up" />
</form>
<?php
} // End of block displaying form if needed.
?>
And then in jQuery I do one of these:
$('.note').click(function(){
$('#installation-time').val('test string');
});
When I submit the form, the PHP variable that's supposed to collect that value is null. Every other input in the form works, and if I remove the javascript and manually enter the exact same text that I had set with JavaScript into the input it works as well.
The question really is why populating a field with javascript as opposed to manually typing the exact same string into a text input would break things. Again there are no errors and the input is populated correctly on the front end. Somehow posting the form just doesn't pick up on the value when it's set by javascript vs. typed manually. There has to be something really fundamental I'm missing here.
Any idea what's going on here? I've spent hours puzzling over this to no avail.
Update:
Code updated, test page:
http://dev.rocdesign.info/test/
Solution: can't post a disabled input. I actually tested that back in the beginning and must have missed that removing the "disabled" on the input made it work, so I mistakenly ruled it out and moved on.
Thanks for the responses everyone. And for anyone else with this problem: use a hidden input to post the value.