Checking if a user has submitted a form

[mark_aok]

Hi all,

 

I have a form that validates itself by putting "<?php $_SERVER['PHP_SELF'] ?>" as the form's action.

 

I have now written some simple validation, shown below.

$name = $_POST['name'];

if (!isset($name) || empty($name)) {

    echo "Error, please fill in the name textbox";

}

 

The problem is, this code shows up, even if the user has never submitted the form.

 

How do I only show the error AFTER the user has pressed my submit button??

 

Thanks,

Mark

[marcus]

Just do:

 

if($name){
//success
}else {
//failure
}

[BlueSkyIS]

"How do I only show the error AFTER the user has pressed my submit button??"

 

check to see if the form was POSTed. I use this:

 

if ($_SERVER['REQUEST_METHOD'] == "POST") {

  // The form was posted

 

}

 

[mark_aok]

@ mgallforever - The code you suggested never returns true.

 

@BlueSkyIS - $_SERVER['REQUEST_METHOD'] is always equal to post, because the form in the page is set to "post".  Even if the user never submits it.  But that code works for you?  How?  For me it still shows the error stuff...

 

Thanks for the suggestions so far though, I'm really not sure why I'm finding this so difficult to do.

[bache]

put an id on the submit button,

for example: <input type="submit" value="Click me" id="sbmtBtn">

 

and check:

 

if (isset($_POST['sbmtBtn']))

{

....

}

[chocopi]

You dont have to use isset and can just use

 

if($_POST)

[AndyB]

put an id on the submit button,

for example: <input type="submit" value="Click me" id="sbmtBtn">

 

and check:

 

if (isset($_POST['sbmtBtn']))

{

....

}

 

Not quite.  You need to NAME the input ... so:

 

<?php
if (isset($_POST['sbmtBtn'])) {
    $name = trim(strip_tags($_POST['name']));
    if (!isset($name) || empty($name)) {
        echo "Error, please fill in the name textbox";
    }
    // and more validation as you see fit for other inputs ...
}
?>
... your form here
<input type="submit" value="Click me" name="sbmtBtn"></form>