Return value from PHP function

Question

what is wrong with the following code? Hope you understand what I'm trying to do. I'm not very familiar with functions.

function test ($variable) {
   $one = 3;
   if ($variable == 10) {
       $one = "2";
   }
   return $one;
}

foreach ($array as $arraypart) { 
   $part = explode(',',$arraypart);
   test($part[0]);
   echo $one;
}

PHP has documentation where you can learn more about functions: php.net/manual/en/language.functions.php — Felix Kling
– Felix Kling, Commented Jul 11, 2011 at 11:52
Learn about scope: php.net/manual/en/language.variables.scope.php — igorw
– igorw, Commented Jul 11, 2011 at 11:54

Felix Kling · Accepted Answer · 2011-07-11 11:52:08Z

5

You need to assign the result of function to a variable:

$one = test($part[0]);

edited Jul 11, 2011 at 11:52

Felix Kling

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answered Jul 11, 2011 at 11:50

Gedrox

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2 Comments

Felix Kling Over a year ago

You should also point out that this $one is different than the $one in the test function.

calumbrodie Over a year ago

Or just directly echo it... echo test($part[0]);

Rukmi Patel · Accepted Answer · 2011-07-11 11:56:41Z

0

here what happens, your function is being called but no variable is there to catch what test() is returning....

you need catch value which is returned by temp func like this

$val = temp($part[0]); or you can write direct echo temp($part[0]);

answered Jul 11, 2011 at 11:56

Rukmi Patel

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