PHP print variable from input - php

I have following code and it doesn't works for me.
I need print variable from input.
<html>
<head></head>
<body>
<form action='#' method="post">
<input type="text" name="name">
<input type="submit" name="submit">
</form>
<?php
$name = name
echo $name;
?>
</form>
</body>
</html>
Thanks.

you have to use POST to get text from input.
This is right code:
<html>
<head></head>
<body>
<form action='#' method="post">
<input type="text" name="name">
<input type="submit" name="submit">
</form>
<?php
if (isset($_POST['submit']))
{
$name = $_POST['name'];
echo $name;
}
?>
</form>
</body>
</html>

Related

Input from form to url?

So the goal is to take the input from a form append it to the end of the URL and then return the HTML from that page.
I am not entirely sure how to take the forms value (in this case $2) and attach it to the URL.
<html>
<head>
<title>Metar Test</title>
</head>
<body>
<form action="" method="POST">
<p>IACO Code: <input type="text" name="$2" value=""></p>
<input type="submit" name="submit" value="Submit">
</form>
<?php $html=file_get_contents("http://www.metar.mysite.net/metar?id=$2")?>
<?php echo $_POST ["$html"];?>
</body>
</html>
On submit any input from a form with method="POST" will be stored in the $_POST global. You can access it from there and append it to your URL string.
<html>
<head>
<title>Metar Test</title>
</head>
<body>
<form action="" method="POST">
<p>IACO Code: <input type="text" name="iacoCode" value=""></p>
<input type="submit" name="submit" value="Submit">
</form>
<?php
if (isset($_POST["iacoCode"])) {
$html = file_get_contents("http://www.metar.mysite.net/metar?id=" . $_POST["iacoCode"]);
echo $html;
}
?>
</body>
</html>
Using the IF statement to check if it is set will prevent it from loading the URL with no variable.
The way we extract data via php from a form is like below
<html>
<head>
<title>Metar Test</title>
</head>
<body>
<form action="" method="POST">
<p>IACO Code: <input type="text" name="$2" value=""></p>
<input type="submit" name="submit" value="Submit">
</form>
<?php $html=file_get_contents("http://www.metar.mysite.net/metar?id=$_POST['$2']")?>
<?php echo $_POST ["$html"];?>

How do you get user inpup and display it in php

I am trying to get user input from a text box and then echo it using php. Here is my code, and it is not seeming to work.
<html>
<body>
<?php
echo $_POST['value'];
?>
<form method="post" action="">
<input type="text" name="value">
<input type="submit">
</form>
</body>
</html>
<html>
<body>
<?php
if(!empty($_POST['value']))
{
echo filter_var($_POST['value'], FILTER_SANITIZE_STRING);}
?>
<form method="post" action="">
<input type="text" name="value">
<input type="submit">
</form>
</body>
</html>
First check if form posted.

PHP - form action calling itself, how to display everything in 1 page?

I have two .php files as such:
test1.php
<html>
<head>
<title>Sample</title>
</head>
<body>
<form action="test2.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
</body>
</html>
test2.php
<html>
<head>
<title>Sample Display Page</title>
</head>
<body>
<?php
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
?>
</body>
</html>
I want to know how can I get it all to display on a single page, i.e. on the test1.php, without having to have two files.
Check if $_POST["userNumber"] isset, and echo the form if it's not.
<html>
<head>
<title>Sample</title>
</head>
<body>
<?php
if(isset($_POST["userNumber"]){
echo "You have chosen ".$_POST["userNumber"];
}else{
echo '<form action="test1.php" method="post">';
echo 'Please enter a number <input type="number" name="userNumber"><br>';
echo '<input type="submit">';
echo '</form>';
}
?>
</body>
</html>
Your test1.php will need to look like this
<html>
<head>
<title>Sample</title>
</head>
<body>
<form action="test1.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
<?php
if(isset($_POST["userNumber"])) {
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
}
?>
</body>
</html>
You simply need to remove the action="test2.php" section and combine both pages into one just like the following:
<html>
<head>
<title>Sample</title>
</head>
<body>
<form method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
<?php
if(isset($_POST['userNumber'])){
$user_number = $_POST['userNumber'];
echo "You have chosen $user_number";
}
?>
</body>
</html>
Pretty simple actually:
<html>
<head>
<title>Sample</title>
</head>
<body>
<?php
if(isset($_POST) && isset($_POST['userNumber'])) {
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
}
?>
<form method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
</body>
</html>
<html>
<head>
<title>Sample</title>
</head>
<body>
<?php
if (isset($_POST['submitted'])) {
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
}
else {
?>
<form method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit" name="submitted">
</form>
<?php
}
?>
</body>
</html>
Just use isset() for submit button, note that i added name="submit" to your button since it was missing
<html>
<head>
<title>Sample</title>
</head>
<body>
<?php
if(isset($_POST['submit'])
{
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
} else {
?>
<form action="test2.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit" name="submit">
</form>
<?php } ?>
</body>
</html>
This one is fully tested and allows you to revise the value after submission. You still have to worry about cross-site scripting attacks, etc... but that is outside of the scope of the question.
<html>
<head>
<title>Sample</title>
</head>
<body>
<?php
$user_number = '';
if (isset($_POST['userNumber'])) {
$user_number=$_POST['userNumber'];
?>
<p>You have chosen <?php echo $user_number ?></p>
<?php
}
?>
<form method="post">
Please enter a number <input type="number" name="userNumber" value="<?php echo $user_number ?>"><br>
<input type="submit">
</form>
</body>
</html>
set the variables up before the form as placeholders and check for isset
<?php
$user_number = '';
if(isset($_POST["userNumber"])) {
$user_number = $_POST["userNumber"];
}
?>
<form action="test2.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit" name="submit">
</form>
<p>You have chosen: <?php echo $user_number ?></p>
Sample
<form action="<?php $_PHP_SELF ?>" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
</body> </html>

Can't access post in php

Here's my code in a short way. This code can't access post and I don't know why.
if ($_POST) {
echo 'post a girdim';
}
?>
<html>
<head>
</head>
<body>
<form action="" method="post">
<input type="submit">
</form>
</body>
You don't have any elements in your form with a name. Therefore no values are submitted to the server, the $_POST array is empty and if ($_POST) is false.
If you don't want any named elements, you can check whether the request method in $_SERVER['REQUEST_METHOD'] is "POST".
set input name like:
<input name="submit" type="submit" />
Give your input tags names.
if (isset($_POST["submit"])) {
echo 'post a girdim';
}
?>
<html>
<head>
</head>
<body>
<form action="" method="post">
<input type="submit" name="submit">
</form>
</body>
you should follow this
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
and further
you can give name for input
for example
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="submit" name="submit">
</form>
and then you must check this
<?php
extract($_POST);
if (isset($_POST['submit'])) {
echo $_POST['submit'];
}
?>

Printing text box value on page in PHP

I am new to PHP (in-fact I am learning it). I am trying to get the value of textbox defined in HTML and print it on HTML page through PHP code. I am able to send the value of textbox to another page using form post and getting it using $_POST['name'] on other page.
I am not able to print the value of textbox on same page.
Here is my code:
<html>
<head>
<?php
function myfunction()
{
printf($_POST['fname']);
}
?>
</head>
<body>
Name: <input type="text" name="fname" />
<input type="button" onClick="myfunction()" value="Click" />
</body>
</html>
First of all it's not required to write php function within <head> tags. Second,
you can't POST data without <form> tags. try this
<?php
if(isset($_POST['submit']) && $_POST['submit']=='Submit'){
$name=$_POST['fname'];
echo $name;
}
else {
?>
<html>
<head>
</head>
<body>
<form method="POST" action="<?=$_SERVER["PHP_SELF"]?>">
Name: <input type="text" name="fname" />
<input type="submit" name="submit" value="Submit"/>
</form>
</body>
</html>
<?php } ?>
you can not call PHP function by button's click event...
if you want to print on same page then you can do by below code..
<html>
<?php
if(count($_POST)>0){
echo $_POST['fname'];
}
?>
<body>
<form method='post'>
Name: <input type="text" name="fname" />
<input type="submit" value="Click" />
</form>
</body>
</html>
hope you will get what you want by my answer..

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