[RESOLVED] Help with creating function isset_var()!

[spiritssight]

Hello all,

I am trying to create a function that will take a var and see if it is defined if so then it displays that var value, if its undefined then it does not display any value.

This would allow me to have one form and have it be able to be used for new input and editing of old input, see when I get data from the db I convert it to the input names so they would be the same var?

this is what I got so far:

<?php
function isset_var($var)
	{
		if(isset($var))
		{
		return( $var );
		}
		else
		{
		$var = '';
		return($var);
		}
	}

// $test = 'test';

echo isset_var($test);
?>

if you notice I have a var called $test now when I don't have it blocked it works the way it should, but when I // the var $test then it gives me the following error:

Notice: Undefined variable: test in /home/dev/www/query/isset_var.php on line 18

So the question is how can I do this so that I don't have to have on every input field a if(isset($...)){echo $...;}

Thanks for any advice with this piece of code!

Sincerely,
christopher

[NogDog]

function isset_var($var)
{
   return (isset(${$var})) ? ${$var} : '';
}

// sample usage for variable $foobar:
echo "The value of \$foobar is: '" . isset_var('foobar') . "'.";

However, since (as shown above) you can do the check in a one-liner using the ternary operator, you could just forgo the function and do:

echo "The value of \$foobar is '" . (isset($foobar)) ? $foobar : '' . "'.";

[spiritssight]

Thanks very very much,
I could also do this:

'. (isset( $... )) .'

and not even need the rest of the line ? $... : '' as if there is no var set then I don't need to have any thing show so its either show the var or do nothing :-)

please correct me if I am wrong on that!

Sincerely,
Christopher

[NogDog]

[man]isset()[/man] returns a boolean true/false, not the value of the variable; so you still need to output the variable in some manner.

[spiritssight]

Thanks for the quick respone!

ok, I have changed my code to:

echo '
<!-- <label for="e_id">Entity Identification: </label> -->
<input type="hidden" value="'. (isset( $e_id )) ? $e_id : '' .'" id="e_id" name="e_id" />
';

and it gives the error of:

Notice: Undefined variable: e_id in /home/dev/www/query/general_form.php on line 8

so the question is what am I don't so wrong? 🙁

Thanks for your time and helping!

Sincerely,
Christopher

[NogDog]

Sorry, it's an operator precedence thing (the "." concatenation operator has higher precedence than the "?:" ternary operator), so you need to put parentheses around the entire ternary expression:

echo '
<!-- <label for="e_id">Entity Identification: </label> -->
<input type="hidden" value="'. ( (isset( $e_id )) ? $e_id : '' ) .'" id="e_id" name="e_id" />
'; 

[spiritssight]

Great that worked :-)

if I may asking another question, won't writting a function save on the type the if() over and over? the form has a number of pages and can have for 2 question on it to 6ish.

Thanks NogDog for your help and once again I have learnt some thing for how long it stays in the brain is another :-)

Sincerely,
Christopher