php warning for while(list=each)

[Anon]

The following line in my code

while(list($key,$value) = each($ar))

sometimes generates a PHP warning:

Variable passed to each() is not an array or object

I'm guessing this is a result of $ar sometimes not being set (normal behavior as the variable is based on user input from an html form).

Do I need to enclose the while loop inside an if statement checking on whether $ar exists, or is there a more elegant solution?

[[deleted]]

Checking if $ar exists is not enough, you have to check if $ar exists as an array.

And FYI, that is a very elegant solution.

look at http://www.php.net/manual/ref.var.php

[babak]

try this
if(is_array($ar)){
while(list($k,$v)=each($ar)){

	}
}

//babak

[Anon]

Would the following also work, or is it bad form:

while(is_array($ar) && (list($k,$v)=each($ar))) { }

[[deleted]]

It is correct PHP, but you'll hate yourself in the morning :-)

Using this means that PHP will check if $ar is an arrayeach time it loop through the arrayitself. So if the array has 100 elements, the is_array() is done 100 times. That's not a big problem, but it mean 99 useless checks.
Plus, right now this loop may be the only thing that depends on $ar being an array, but in future that may well change. If you start with an IF statement now,
then later you will immediately see that there is something about $ar that needs checking.

As a wise man once said, don't sacrifice clarity for cleverness. :-)

[babak]

AMEN.
//babak

[Anon]

Great.
It works.