Cannot echo the values from a simple php form

Question

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?

Have it echo something inside the if(){} just to make sure we're making it in to there. Either way, check for typos in names. — below9k
– below9k, Commented Sep 16, 2014 at 16:38
did you enter value in text box :-) this code works for me . check just echo or phpinfo() — zod
– zod, Commented Sep 16, 2014 at 16:43

Niet the Dark Absol · Accepted Answer · 2014-09-16 16:40:50Z

1

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)

answered Sep 16, 2014 at 16:40

Niet the Dark Absol

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1 Comment

Panagiotis Iliakopoulos Over a year ago

It is uploaded on a server and i tried right clicking and i only see my html

putvande · Accepted Answer · 2014-09-16 16:48:16Z

0

Use !empty($_POST['newitem'] instead:

if(!empty($_POSΤ["newitem"])){
    echo $itemnew = $_POSΤ["newitem"];
}

empty()

answered Sep 16, 2014 at 16:48

putvande

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Comments

SuperDJ · Accepted Answer · 2014-09-16 16:48:31Z

0

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.

answered Sep 16, 2014 at 16:48

SuperDJ

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3 Comments

SuperDJ Over a year ago

@AdamSinclair yes, but why would you use 2 functions if 1 is enough. What I mean is I see a lot of if(isset() && !empty()). While !empty() is actually enough to do both

Panagiotis Iliakopoulos Over a year ago

tried that as well and the same output.I think it has to do with my server

SuperDJ Over a year ago

@PanagiotisIliakopoulos I would guess so. Since this code should work

user4038382 · Accepted Answer · 2014-09-16 17:02:46Z

0

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.

answered Sep 16, 2014 at 17:02

user4038382

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