I need to take a data of username, then check in db, which aunthefecator user has got and depends on aunthefecator send him the window where he can enter it and login in his account!
I have tryed to make it like a 1 form and the method was sent to another form but it doesn not work.
<form name="logon" action="check_login.php" method="POST" id="login">
<input type="text" name="username">
<input type="password" name="password">
<button type="submit">
<span>Log IN</span>
</button>
</form>
I just redirecting on this page without getting data.
You must put this code in top of your php page, or if you want to check login in another page , you should put this code in destination page that you want to check login authentication. Remember to put your code about checking user and pass from database.
<?php
if(isset($_POST[“username”] && $_POST[“password”]){
$user = $_POST[“username”];
$pass = $_POST[“password”];
//Check your db here and then replace your url that you want to reload client if user and pass was correct.
header(‘Location: http://yourURL.com’);
}else{
echo (“username or password was not correct”);
}
?>
I want to create a fully dynamic chat UI for my website,
But it reloads the whole page if a person submits the button this will directly show the div
<div class="messeging" id="msg">
<?php print $message->getName() ." : " . $chat->message . ""; ?>
</div>
Without reload msgs are save in some xml file path like example (../user/xml)
HTML
<form action="action.php" method="post">
<input type="text" id="input" value="php echo">
<input type="submit" value="send" onclick="showDiv()">
</form>
<div class="messeging" id="msg">
<?php print $message->getName() ." : " . $chat->message . ""; ?>
</div>
i don't know javascript/ajax well how to solve this
Try jquery ajax, on submit button send last msg for save into db and fetch all record and show them into chat page. so each time you send msg it will fetch all msg. if you don't know about how ajax work then let me know.
Begin by looking at the event on forms called 'onSubmit'. https://developer.mozilla.org/en-US/docs/Web/API/GlobalEventHandlers/onsubmit
Once you understand that, you can 'event.preventDefault()' in your 'onSubmit' method handler. Or in English, keep a form's default behavior to submit to the action from occurring.
After preventing the default action, you will need to do something with the form, and the easiest in latest browsers is to use the fetch API. https://developer.mozilla.org/en-US/docs/Web/API/Fetch_API
You should be able to submit the data in the form to some endpoint that returns the HTML you wish to inject.
Once the HTML has been successfully returned, simple document.createElement methods with append (https://developer.mozilla.org/en-US/docs/Web/API/ParentNode/append) will need to be used.
Good luck learning the web!
I think that this problem occurs often on a web application development. But I'll try to explain in details my problem.
I'd like to know how to correct this behavior, for example, when I have a block of code like this :
<?
if (isset($_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";
}
In my project, I want the signed-in user to insert data and wait for approval but the problem is where to save that unapproved data like if 400 signed-in users have inserted data then how can I manage that data until it is approved. Do I have to make it's another database table to save that unapproved data ?? or use the same database table ? if yes then how ?? Or did I can do this with javascript or jquery??. My database is in MySQL and using PHP. My form looks like this
<form action="#" method="post" id="FormVerification">
<input type="text" name="Input1" required value="<?php echo $Variable; ?>" placeholder="Enter Some Value">
<input type="Submit" name="Submit" value="Submit">
</form>
<?php
if(isset($_POST['Submit']))
{
//do something for varification
}
?>
Any Suggestions ?? ?
If it's only activation/approval you need then the easiest approach would be to insert a column in the actual table and call it "is_approved" or something similar.
Have that default to 0, and when an admin approves it set it to 1.
You can use that column to retrieve data not yet approved (where is_approved=0), and also to display the approved a data (is_approved=1).
Other than that you'd have the option of copying the table and name it something like pending_submission. Once the submission is approve you'd move the data to the approved table.
Hello and thank you for viewing my question. I am a complete beginner and am looking for simple ways to do the following...
What I have in seperate linked documents:
HTML, CSS, Javascript, PHP
What I am having trouble with:
I need to use something like JSON (although I would also accept XML requests or Ajax at this point if they work) to transfer variables from Javascript to PHP. I need the variables to search in a database, so they need to be literally available within PHP (not only seen on a pop-up message or something).
I have seen a LOT of different ways to do this, I have even watched tutorials on YouTube, but nothing has worked for me yet. The things I am having the biggest problem with is that when I add a submit button to my form it doesn't submit my form and I don't know why.
Form code snippet:
<form id="form" name="input" method="post" action="javascript:proofLength();">
<input id="userinput" type="text" autofocus />
<input id="submit" type="button" value="submit" onsubmit="post();">
</form>
The second to last line there doesn't work. Do I need javascript to submit the form? Because I really thought that in this case it was part of the functionality of the form just like method="post"...
The other thing is that for JSON, I have no idea what to do because my variables are determined by user input. Therefore, I cannot define them myself. They are only defined by document.getElement... and that doesn't fit the syntax of JSON.
Those are really my main problems at the moment. So if anyone could show me a simple way to get this variable transfer done, that would be amazing.
After this I will need to search/compare in my database with some php/sql (it's already connecting fine), and I need to be able to return information back to a in HTML based on what I find to be true. I saw one example, but I am not sure that was very applicable to what I am doing, so if you are able to explain how to do that, that would be great also.
Thank you very, very much.
April
You don't need ajax to submit this form. You don't even need javscript. Just do this:
<form id="form" name="input" method="post" action="mytarget.php">
<input id="userinput" name="userinput" type="text" autofocus />
<input id="submit" type="submit" value="submit" />
</form>
This will send the form data to mytarget.php (can be changed of course)
See that i have added the name attribute to your text-field in the form and i changed the type of the button to submit.
Now you can work the Data in mytarget.php like this:
<?
$username = $_POST['userinput'];
echo "Your name is: ".$username;
?>
You wanted to have a check for length in the submit. There are two ways to this:
Before the input is send (the server is not bothered)
Let the server Check the input
for 1 you will have to append a event listener, like this:
var form = document.getElementById("form");
form.addEventListener("submit", function(event){
console.log("test");
var name = form.elements['userinput'].value;
if(name.length < 3){
alert("boy your name is short!");
event.preventDefault();
}
});
Enter a name with less then 3 characters and the form will not be submitted. test here: http://jsfiddle.net/NicoO/c47cr/
Test it Serverside
In your mytarget.php:
<?
$username = $_POST['userinput'];
if(strlen($username) > 3)
echo "Your name is: ".$username;
else
echo "your name was too short!";
?>
You may also do all this with ajax. You will find a lot of good content here. But I'd recommend a framework like jQuery to do so.
The problem is in this line
<form id="form" name="input" method="post" action="javascript:proofLength();">
The action should be a PHP page (or any other type of server script) that will process the form.
Or the proofLength function must call submit() on the form
In the php page you can obtain variable values using $_GET["name"] or $_POST["name"]
To summarize; your code should look like this
<form id="form" name="input" method="post" action="yourpage.php">
<input id="userinput" type="text" autofocus />
<input id="submit" type="button" value="submit">
</form>
and for your php page:
<?php
$userinput = $_POST["userinput"];
//Do what ever you need here
?>
If you want to do something in your javascript before submitting the form, refer to this answer