Refresh page after form submitting - php

I have a little problem. I want to reload my page after submitting a form.
<form method="post" action="">
<textarea cols="30" rows="4" name="update" id="update" maxlength="200" ></textarea>
<br />
<input type="submit" value=" Update " id="update_button" class="update_button"/>
</form>

only use
echo "<meta http-equiv='refresh' content='0'>";
right after insert query before }
example
if(isset($_POST['submit']))
{
SQL QUERY----
echo "<meta http-equiv='refresh' content='0'>";
}

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <!-- notice the updated action -->
<textarea cols="30" rows="4" name="update" id="update" maxlength="200" ></textarea>
<br />
<input name="submit_button" type="submit" value=" Update " id="update_button" class="update_button"/> <!-- notice added name="" -->
</form>
on your full page, you could have this
<?php
// check if the form was submitted
if ($_POST['submit_button']) {
// this means the submit button was clicked, and the form has refreshed the page
// to access the content in text area, you would do this
$a = $_POST['update'];
// now $a contains the data from the textarea, so you can do whatever with it
// this will echo the data on the page
echo $a;
}
else {
// form not submitted, so show the form
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <!-- notice the updated action -->
<textarea cols="30" rows="4" name="update" id="update" maxlength="200" ></textarea>
<br />
<input name="submit_button" type="submit" value=" Update " id="update_button" class="update_button"/> <!-- notice added name="" -->
</form>
<?php
} // end "else" loop
?>

If you want the form to be submitted on the same page then remove the action from the form attributes.
<form method="POST" name="myform">
<!-- Your HTML code Here -->
</form>
However, If you want to reload the page or redirect the page after submitting the form from another file then you call this function in php and it will redirect the page in 0 seconds. Also, You can use the header if you want to, just make sure you don't have any content before using the header
function page_redirect($location)
{
echo '<META HTTP-EQUIV="Refresh" Content="0; URL='.$location.'">';
exit;
}
// I want the page to go to google.
// page_redirect("http://www.google.com")

LOL, I'm just wondering why no one had idea about the PHP header function:
header("Refresh: 0"); // here 0 is in seconds
I use this, so user is not prompt to resubmit data if he refresh the page.
See Refresh a page using PHP for more details

You can maybe use :
<form method="post" action=" " onSubmit="window.location.reload()">

<form method="post" action="">
<table>
<tr><td><input name="Submit" type="submit" value="refresh"></td></tr>
</table>
</form>
<?php
if(isset($_POST['Submit']))
{
header("Location: http://yourpagehere.com");
}
?>

action attribute in <form method="post" action="action="""> should be just action=""

You want a form that self submits? Then you just leave the "action" parameter blank.
like:
<form method="post" action="" />
If you want to process the form with this page, then make sure that you have some mechanism in the form or session data to test whether it was properly submitted and to ensure you're not trying to process the empty form.
You might want another mechanism to decide if the form was filled out and submitted but is invalid. I usually use a hidden input field that matches a session variable to decide whether the user has clicked submit or just loaded the page for the first time. By giving a unique value each time and setting the session data to the same value, you can also avoid duplicate submissions if the user clicks submit twice.

//insert this php code, at the end after your closing html tag.
<?php
//setting connection to database
$con = mysqli_connect("localhost","your-username","your-
passowrd","your-dbname");
if(isset($_POST['submit_button'])){
$txt_area = $_POST['update'];
$Our_query= "INSERT INTO your-table-name (field1name, field2name)
VALUES ('abc','def')"; // values should match data
// type to field names
$insert_query = mysqli_query($con, $Our_query);
if($insert_query){
echo "<script>window.open('form.php','_self') </script>";
// supposing form.php is where you have created this form
}
} //if statement close
?>
Hope this helps.

Related

Simple submit form to go to a page

I have some numbered pages:
1.php
2.php
3.php
etc.
I want to create a textbox that the user enter any number: 2 for example, and hit enter or Go button, and they will go to the page 2.php depending on the number entered.
I know how to link to a specific page as in form action="....", but I am not sure how to echo the user input and translate it as link (whether using html or php).
Ex:
<form method="POST">
<input type="number" value="" />
<input type="submit" value="Go" />
</form>
You need to add an action attribute to your form and a name attribute to your number input. The file from your action attribute will "catch" the POST variables and do the logic needed to redirect your user. Change your form tag to:
<form method="POST" action="redirect.php">
<input type="number" value="" name="redirect" />
<input type="submit" value="Go" />
</form>
Then create the redirect.php file that gets the POST variables and does the redirection:
<?php
$redirectPage = (int) $_POST['redirect'];
$redirectUrl = "http://www.example.com/{$redirectPage}.php";
header("Location: $redirectUrl");
printf('moved.', $redirectUrl);
Beware that there's no input validation nor error handling included.
I think, the best available option in your case would be the one using client-side javascript to dynamically change the form's action attribute base on the number entered in the input box.
A quick and dirty solution to fulfil such a task might look like this
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">
function submitAction(formElement) {
var // get the input box element
el = document.getElementById('form-action-number'),
// get a number specified by user in the input box
num = parseInt(el.value),
// validate that it's really a number and is greater than zero
// (you don't want someone to request -666.php right? :)
// build a page url using the correct number
page = !isNaN(num) && num > 0 ? num.toFixed(0).toString() + '.php' : undefined;
if (page) { // the page url is valid
// set form's action attribute to an url specified by page variable
formElement.setAttribute('action', page);
// returning true will allow the form to be submitted
return true;
}
// you might think of a better way to notify user that the input number is invalid :)
console.error('INVALID NUMBER SPECIFIED!');
// returning false will prevent form submission
return false;
}
</script>
</head>
<body>
<!-- When user clicks Go, the return value of submitAction function will be used to decide if the form should be submitted or not -->
<form method="POST" onsubmit="return submitAction(this)">
<input id="form-action-number" type="number" value="" />
<input type="submit" value="Go" />
</form>
</body>
</html>
With PHP you can do something like this:
<?php
// Check if the POST value has been set
if(isset($_POST['my_number'])) {
// Redirect to the corresponding page
header('Location: ' . $_POST['my_number'] . '.php');
}
?>
<form method="POST">
<input name="my_number" type="number" value="" />
<input type="submit" value="Go" />
</form>
This is like DaMeGeX's answer but uses javascript to go to the new page.
<?php
// Check if the POST value has been set
if(isset($_POST['my_number'])) {
// Redirect to the corresponding page
echo "<script> window.location.href = '".$_POST['number'].".php' </script>";
}
?>
<form method="POST">
<input name="my_number" type="number" value="" />
<input type="submit" value="Go" />
</form>

Executing PHP Function When a Form Submit is Clicked

I am having trouble understanding handling variables that are passed through pages when a form submit button is clicked. Basically i have a text area where a user writes an sql query. Then clicks submit. Then on the same page (x.php) , i have to display the results in a table. I figured, when the user clicks the button, i call a function that connects to the database, then runs the query, and outputs the result in a table. The code i have below is a mock, and isnt quite working.But above is essentially what i am trying to do.
In my code, I call the page, and check to see if the proper submit button has been clicked, is that how i am suppose to do it?
Also, I am trying to post the metadata in the code below, but how does the table replace what is already on the page?
<html>
<head>
<title>CNT 4714 - Project Five Database Client</title>
</head>
<body style="background-color:white">
<center>
<h1 style="color:red">CNT 4714 - Project Five Database Client</h1>
<hr>
<div style="float:left; text-align:left;padding-right:80px; padding-left:80px; ">
<font color="yellow">
<?php
?>
Welcome Back!<br>
<?php echo $_POST["user"];?>
</font>
</div>
<font color="yellow">
<div style="float:left; text-align:left">
<center>
<h2 style="color:green">Enter Query</h2><br><br>
Please enter a valid SQL Query or update statement. You may also just press<br>
"Submit Query" to run a defualt query against the database!
<form action="" id="sql" method="post">
<br>
<textarea rows="10" cols="50" name="query" form="sql">Enter text here...</textarea><br><br>
<input type="submit" name="submit" color="red">
<input type="submit" name="" color="red" value="Submit Update">
</form>
<?php
if(isset($_POST['submit'])){
echo "hello";
query(); //here goes the function call
}
function query()
{
echo "hello";
$conn = mysqli_connect("localhost:3306", "root", "*******", "project4");
$query = $_POST["query"];
$result = mysqli_query($conn, $query);
$metadata = mysqli_fetch_fields($result);
print("<tr>");
for($i=0; $i<count($metadata);$i++){
print("<tr");
printf("%s",$metadata[$i]->name);
print("</tr>");
}
}
?>
</center>
</div>
</font>
</center>
</body>
</html>
You are trying to get the values of the global variable $_POST while you are posing it to $_GET. The way to fix this is assigning the method into your form element.
Example:
<form id="sql" action="" method="POST">
There are many ways for checking or the form is submitted, one of this ways (the one I am always using) is checking or the $_SERVER['REQUEST_METHOD'] is equal to "POST". This way you can tell the different between a GET, POST, or PUT request.
Example:
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
if(isset($_POST['sql']))
{
....
}
}
If you're using $_POST, your form request method should be POST.
<form action="" id="sql" method="post">
Otherwise, it will submit it with a GET request by default.
In that case, you will have to access the variable using $_GET instead.

if isset submit failed

$jepse = $_POST['slj'];
if(isset($jepse)){
$sql43 = "UPDATE notifications SET seen='1' WHERE touser='$myid' ";
if(mysqli_query($con, $sql43)){
}}
?>
<center>
<form action="#" method="POST">
<input type="submit" name="slj" value="Seen" style="background: rgba(255,255,255, 0); ">
</form>
</center>
I have very stupid problem here... My submit button wont set ..... Don't know what is problem... I have many same things on different pages and with different names... But for this it wont work .... Anyone help?
:::::::::::::::::::UPDATE:::::::::::::::::::::::
I did it with moving update code to another file and at action set that file...
A submit button alone isn't enough to post a value. You should use a form field like a hidden input to post your data. I made this stupid mistake also.
<input type="hidden" name="slj" value="some-value">
You doesn't set any values to submit to the server.
You have to change the form like this:
<center>
<form action="" method="POST">
<input type="hidden" name="slj" value="1">
<input type="submit" value="Seen" style="background: rgba(255,255,255, 0); ">
</form>
</center>
A value in the submit button is just the text printed on the button. You have to set another hidden input value and that will be sent.
Also it could be heplful to change the action to "".
your codding is expected to work as is if there are no spelling mistakes somewhere: Although the
defining a POST element before checking if it is set, the program will see it as it is not defined ref: $jepse = $_POST['slj']; but that should not be the main problem why your database is not updating since once you clicked the button it was then defined.
Also in your form <form action="#"... remove the #
i.e. if you are on the same page <form action=""...
or
i.e. be direct <form action="the-php-page.php"...
<?php
if(isset($_POST['slj'])){
$sql43 = "UPDATE `notifications` SET `seen`='1' WHERE `touser`='$myid' ";
if(mysqli_query($con, $sql43)){
echo "updated";
}
else
{
echo "Error updating record: " . mysqli_error($con);
}
}
?>
<span>
<form action="" method="POST">
<input type="submit" name="slj" value="Seen" style="background: rgba(255,255,255, 0); ">
</form>
</span><br>
REGARDS
<center>
<form method="POST">
<input type="submit" name="slj" value="Seen" style="background: rgba(255,255,255, 0); ">
</form>
</center>
<?php
$jepse = $_POST['slj'];
if(isset($jepse)){
echo "Working fine";
}
?>
Let mw know whether this code return "Working fine" or not

HTML Button that saves and move to the next page

I have a small problem. I want to have a button in my html page that saves every data that is added in the textfields and also when I click it to move to the next page.
My code is the follow...
<input type=button onClick="location.href='education.php'" value='Next'>
but it only moves to next page it does not save the data in the database ...
Can you help me please?
Thanks.
Remove the JavaScript
Change the type to submit
Wrap it in a <form>
Set the action of the form to education.php
Set the method of the form to post
Then, in education.php, read the data from $_POST and use PDO (with bound variables) to insert it into the database.
Try this :
<?php
if(isset($_POST['submit']))
{
// Insert Query Put here
header('Location: education.php');
}
?>
<html>
<head>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="submit" value="Next" name="submit">
</form>
</body>
</html>
education.php :
<?php
echo "Successfully Updated.";
?>
You will have to set an action to your form like below because you are not submitting the form, but just redirection to another page without taking the form data.
<form action="education.php" method="post">
<!-- All your input fields here -->
<input type="submit" name="submit" value="Next">
</form>
and your education.php should be look like this:
<?php
//Get all parameters using $_POST
//Make A connection to database
//Choose a database in which you have to save the data
//Create a SQL query
// run query using mysql_query($query);
//Redirect to anywhere with header("Location:page.php");
?>

PHP form - on submit stay on same page

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)

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