So here's my full code
<!DOCTYPE html>
<html>
<body>
<h1>Encrypt</h1>
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
Enter word to encrypt<input type="text" name="in">
<input type="submit">
<hr>
</form>
<h1>Decrypt</h1>
<form>
Enter word to decrypt<input type="text" name="out">
<input type="submit">
<hr>
</form>
</body>
</html>
<?php
$encrypt = $_POST['in'];
?>
And here's the error I get
Notice: Undefined index: in in /Users/idrisk/Colourity/si/index.php on line 20
Line 20 is $encrypt = $_POST['in']; and I don't see what I'm doing wrong with it. Any ideas?
As a general practice for forms in php, always check if the submit button has been clicked.
First name your submit button:
<input type="submit" name="submit">
then further in your php:
if (isset($_POST['submit'])) {
// do your stuff, eg...
$encrypt = $_POST['in'];
}
EDIT #1: Added to that, you seem to have 2 forms and 2 submit buttons. I suggest you keep only one form, and one submit button (remove the 2nd form element and submit button).
If you really need 2 forms, name your submit buttons differently and then you can call them separately.
<input type="submit" name="submit-in">
<!-- ... -->
<input type="submit" name="submit-out">
<?php // ...
if (isset($_POST['submit-in'])) {
// do your stuff, eg...
$encrypt = $_POST['in'];
}
if (isset($_POST['submit-out'])) {
// do your stuff, eg...
$dencrypt = $_POST['out'];
}
EDIT #2: If you want to echo stuff posted in your form, make sure you do the form submission checking and variable setting before the form and then echo the variable after the form (or wherever you want).
you need to first check if the form has been sent, if it hasn't then $_POST['in'] does not yet exist thus throwing the error
May be nothing but you called a php script after closing the form /form, the body /body and then then the HTML /html
replace this code $encrypt = $_POST['in']; by this $encrypt = #$_POST['in'];
this is an error on client server when you upload this file on remote server you will not saw this. use # sign on the client server when you saw this error in future.
Related
First time i try to create a simple form using the POST method.Problem is when i click the button nothing gets echoed.
here is my insert.php file :
<?php
if(isset($_POSΤ["newitem"])){
echo $itemnew = $_POSΤ["newitem"];
}
?>
<form action="insert.php" method="POST" >
<input type="text" name="newitem">
<input type="submit" value="Save">
</form>
EDIT: I tried the GET method and it works...Any ideas why that happened? Server configurations?
NEW EDIT: So it turns out i switched method to GET and it worked.Then i switched back to POST (like the code i posted on top) and it works...I have no clue why this happened.Any suggests?
The code you have posted is perfectly valid and should work.
I'm going to guess that you do not have PHP enabled, or it is not working.
<?php ... ?> looks to the browser like a long, malformed HTML tag, and therefore ignores it, making the effect invisible.
Try right-clicking the page and selecting View Source. If you see your PHP there, then the server is indeed not processing it.
The most likely reason for this is probably the same problem I had with my very first bit of PHP code: you're trying to "run" it directly in your browser. This won't work. You need to upload it to a server (or install a server on your computer and call it from there)
Use !empty($_POST['newitem'] instead:
if(!empty($_POSΤ["newitem"])){
echo $itemnew = $_POSΤ["newitem"];
}
empty()
Try the following:
if($_POST) {
if(!empty($_POST['newitem'])) {
$itemnew = $_POSΤ['newitem'];
echo $itemnew;
// or leave it as is: echo $itemnew = $_POSΤ['newitem'];
}
}
?>
<form action="insert.php" method="POST" >
<input type="text" name="newitem">
<input type="submit" value="Save">
</form>
The if($_POST) will make sure the code is only executed on a post. The empty() function will also check if it isset() but also checks if it is empty or not.
Try this :
<?php
if(isset($_POSΤ["newitem"])){
echo $itemnew = $_POSΤ["newitem"];
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" >
<input type="text" name="newitem">
<input type="submit" value="Save">
</form>
$_SERVER['PHP_SELF']; is pre-defined variable in php.It allows the user to stay on same page after submitting the form.
I have a form on on html outside of php...
<form method="post" action="">
<input type="text" name="user"/></br>
<input type="submit" value="submit" name="login"/>
</form>
then call submit button from php and do this
if(isset($_POST["login"]))
{
print <<<this
<form method="post" action="">
<input type="submit" name="apply"/>
</form>
this;
if(isset($_POST["apply"]))
{ print "it works";}
}
Alright, so the problem is that, "it works" won't print from the second form thats inside the php. it just takes me back to where i came from. Perhaps it's a dumb question, please help though! thanks
The problem is that by the time you're checking if(isset($_POST["apply"])) the login condition becomes invalid because everything is inside the if(isset($_POST["login"])).
Try taking the if(isset($_POST["apply"])) outside the login IF.
Your "apply" code exists only INSIDE the login test code. When you submit that second form, there will be NO login form field, because you didn't include an input/textarea of that name in the second form. So the second form submits, there's no login, and the entire inner code never gets executed. You probably want:
if(isset($_POST["login"]))
{
print <<<this
<form method="post" action="" name="apply">
<input type="hidden" name="login" value="foo" /> <!-- add this line -->
etc...
I'm not sure to understand what you wanna do with this code but you obviously missed some details :
_You did not set the "action" field in your form tag, so I don't understant how you would like the PHP file to get called ?
_Your code if(isset($_POST['login'])) has no sense, you are testing the existence of a value sent by a validation button, you'd rather whrite isset($_POST['user'])
Hoping to have helped you
Your variables are declared in 2 forms, so there will be 2 calls (completely independant) to your php.
So you could have a second submit button inside your second form:
if(isset($_POST["login"]))
{
print <<<this
<form method="post" action="">
<input type="submit" name="apply" value="Second"/>
</form>
this;
}
if(isset($_POST["apply"]))
{ print "it works";}
Sorry if this is a duplicate, but I really cant find anything that could solve my problem. I can pass numbers and strings like $_SESSION['blabla']="123'; but I can't pass this $_POST value from the textfield and submit button.
Page 1 (sessions.php)
<?php session_start(); ?>
!doctype stuff here
<body>
<form id="form1" name="form1" method="post" action="sessions2.php">
<label>
<input type="text" name="damn" id="damn" />
<input type="submit" name="submit" id="submit" value="Submit" />
</label>
</form>
<?php
$omg = $_POST['damn'];
$_SESSION['damn'] = $omg;
echo $_SESSION['damn'] ;
?>
Page 2 (sessions2.php)
<?php
session_start();
$fires = $_SESSION['damn'];
echo "wth";
echo $_SESSION['damn'];
?>
PS. Sorry for the names.. I'm truly stumped.
You need to put the code that reads from $_POST in the file that you submit the form to.
Currently your process is:
Get request for sessions.php
Send form to browser
Assign $_POST['damn'] (which is undefined) to the session.
User submits form
Get request for sessions2.php
Ignore $_POST (which is now populated)
Read from the session (where the variable is still undefined).
damn is populated in the form submission request (step 4/5) not the request where you are trying to read it (step 1).
In sessions2.php
// you POST "damn" variable via form, using post method, so:
$fires = $_POST['damn'];
// and:
$_SESSION['damn'] = $fier;
// or
$_SESSION['damn'] = $_POST['damn'];
PHP code in file sessions.php doesn't work, because in form action you have session2.php.
I have a PHP form that is located on file contact.html.
The form is processed from file processForm.php.
When a user fills out the form and clicks on submit,
processForm.php sends the email and direct the user to - processForm.php
with a message on that page "Success! Your message has been sent."
I do not know much about PHP, but I know that the action that is calling for this is:
// Die with a success message
die("<span class='success'>Success! Your message has been sent.</span>");
How can I keep the message inside the form div without redirecting to the
processForm.php page?
I can post the entire processForm.php if needed, but it is long.
In order to stay on the same page on submit you can leave action empty (action="") into the form tag, or leave it out altogether.
For the message, create a variable ($message = "Success! You entered: ".$input;") and then echo the variable at the place in the page where you want the message to appear with <?php echo $message; ?>.
Like this:
<?php
$message = "";
if(isset($_POST['SubmitButton'])){ //check if form was submitted
$input = $_POST['inputText']; //get input text
$message = "Success! You entered: ".$input;
}
?>
<html>
<body>
<form action="" method="post">
<?php echo $message; ?>
<input type="text" name="inputText"/>
<input type="submit" name="SubmitButton"/>
</form>
</body>
</html>
The best way to stay on the same page is to post to the same page:
<form method="post" action="<?=$_SERVER['PHP_SELF'];?>">
There are two ways of doing it:
Submit the form to the same page: Handle the submitted form using PHP script. (This can be done by setting the form action to the current page URL.)
if(isset($_POST['submit'])) {
// Enter the code you want to execute after the form has been submitted
// Display Success or Failure message (if any)
} else {
// Display the Form and the Submit Button
}
Using AJAX Form Submission which is a little more difficult for a beginner than method #1.
You can use the # action in a form action:
<?php
if(isset($_POST['SubmitButton'])){ // Check if form was submitted
$input = $_POST['inputText']; // Get input text
$message = "Success! You entered: " . $input;
}
?>
<html>
<body>
<form action="#" method="post">
<?php echo $message; ?>
<input type="text" name="inputText"/>
<input type="submit" name="SubmitButton"/>
</form>
</body>
</html>
Friend. Use this way, There will be no "Undefined variable message" and it will work fine.
<?php
if(isset($_POST['SubmitButton'])){
$price = $_POST["price"];
$qty = $_POST["qty"];
$message = $price*$qty;
}
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form action="#" method="post">
<input type="number" name="price"> <br>
<input type="number" name="qty"><br>
<input type="submit" name="SubmitButton">
</form>
<?php echo "The Answer is" .$message; ?>
</body>
</html>
You have to use code similar to this:
echo "<div id='divwithform'>";
if(isset($_POST['submit'])) // if form was submitted (if you came here with form data)
{
echo "Success";
}
else // if form was not submitted (if you came here without form data)
{
echo "<form> ... </form>";
}
echo "</div>";
Code with if like this is typical for many pages, however this is very simplified.
Normally, you have to validate some data in first "if" (check if form fields were not empty etc).
Please visit www.thenewboston.org or phpacademy.org. There are very good PHP video tutorials, including forms.
You can see the following example for the Form action on the same page
<form action="" method="post">
<table border="1px">
<tr><td>Name: <input type="text" name="user_name" ></td></tr>
<tr><td align="right"> <input type="submit" value="submit" name="btn">
</td></tr>
</table>
</form>
<?php
if(isset($_POST['btn'])){
$name=$_POST['user_name'];
echo 'Welcome '. $name;
}
?>
simple just ignore the action attribute and use !empty (not empty) in php.
<form method="post">
<input type="name" name="name">
<input type="submit">
</form>
<?PHP
if(!empty($_POST['name']))
{
echo $_POST['name'];
}
?>
Try this... worked for me
<form action="submit.php" method="post">
<input type="text" name="input">
<input type="submit">
</form>
------ submit.php ------
<?php header("Location: ../index.php"); ?>
I know this is an old question but since it came up as the top answer on Google, it is worth an update.
You do not need to use jQuery or JavaScript to stay on the same page after form submission.
All you need to do is get PHP to return just a status code of 204 (No Content).
That tells the page to stay where it is. Of course, you will probably then want some JavaScript to empty the selected filename.
What I do is I want the page to stay after submit when there are errors...So I want the page to be reloaded :
($_SERVER["PHP_SELF"])
While I include the sript from a seperate file e.g
include_once "test.php";
I also read somewhere that
if(isset($_POST['submit']))
Is a beginners old fasion way of posting a form, and
if ($_SERVER['REQUEST_METHOD'] == 'POST')
Should be used (Not my words, read it somewhere)
HI,
I wrote this code in php.
<head>
<title>listent</title>
</head>
<body>
<form action="untitled 3.php">
<input type = "text" name = "user">
<br>
<textarea name = "address" rows = "10" cols = "40">
</textarea>
<br>
<input type = "submit" value = "heat it">
<br>
<select name="combobox" multiple[]>
<option>mehdi
<option>nine
</select>
</form>
</body>
</html>
now when i click on submit button untitled 3.php is run.
in untitled 3.php i wrote
<?php
print "welcome $user";
?>
but it has error.
Notice: Undefined variable: user in C:\xampp\htdocs\me\Untitled 3.php on line 4
welcome
what is problem?how can i solve it?
Form values don't just magically appear as variables anymore - at least not in any decently modern and properly configured PHP installation. You need to do $_GET["user"] to access the value which is sent by the form (into the URL - you might want to read about the difference between GET and POST)
And please, please use more descriptive names for your files...
PHP Globals wont survive the new page.
In you case you must use the POST variables sent by your form.
So in untitled3.php you should have
echo "welcome ".$_POST['user'];
PS : I would avoid spaces in PHP filenames.
First you should specify a Form submission method in your first page:
<form action="untitled 3.php" method="post">
Then you have access to all posted values in the $_POST array in untitled 3.php:
$user = $_POST['user'];