Why is this error happening?

[synergy]

The form works fine, but I get this error when the page displays:

Notice: Undefined index: submit in c:\apache\apache\htdocs\admin.php on line 28

Code:

if ($_POST['submit']) {
// process form
}

<input type="submit" name="submit" value="Submit">

Line 28 is the if statement above.. any ideas why I get the error? Thanks.

[sgoku01]

Yeah; It means the $_POST['x'] isn't set.

You have to check empty() if it's set, like this:

if (! empty($_POST['submit']) )
{
//process form
}

I think that has to work 🙂

[diego25]

If the variable is not set, then empty() will also complain that it isn't set. Use isset() instead:

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

Diego

[gtibok]

There is no need to worry ... it only says that it (the PHP parser) is expecting a index (number - numeric data) ... like

if($_POST['7']){
.....
}

...

[diego25]

Hm...I think "Undefined index: submit " means that the index 'submit' is undefined...🙂

Diego

[synergy]

Originally posted by diego25
Hm...I think "Undefined index: submit " means that the index 'submit' is undefined...🙂

Diego

Well no shit, but I didnt know how to fix it! I just started programming PHP about a month ago, give me a break here 😛

I'll try using the isset on Monday. Thanks! 😃

[diego25]

synergy, sorry, my reply was to gtibok, who said:

There is no need to worry ... it only says that it (the PHP parser) is expecting a index (number - numeric data)

Diego

[Weedpacket]

Beg to differ, diego25, but empty($unset_variable) won't trigger a notice about undefined variable use. The difference between isset() and empty() is that the latter also looks to see if the contents of the variable (if it is set) are equivalent to false; in other words !empty($var) is equivalent to (isset($var) && $var!=false) - something that it is often necessary to check.