what's wrong with this PHP function

Question

I am trying to extract a value from an array if it exists in the array but i always get null even when I know the value does exist

this works

$data['key']

this doesn't work. getVal1 returns null

$sql2="INSERT INTO test (text, date) VALUES ('".getVal1('key',$data)."', NOW())";
function getVal1($name,$d){
 if ($d !=null && in_array($name, $d) ) {
 return $d[$name];
 }
 return null;
 }

Is there something wrong in my getVal1() function?

• machineaddict: First $d !=null should be !is_null($d). And in_array($name, $d) should be isset($d[$name]).
• Smuuf: It should be $d !== null, if anything. One does not need an unnecessary overhead.
• Jonathon: In fact you could just do isset since it returns false on null as well.
• Mark Baker: Are you confusing keys with values? in_array() is searching for a value; but you're return assumes that it's a key
• deceze: Please read what in_array actually does.

Answer 1

Your problem is in_array searches for array values, yet you are passing it an array key.

You code can be simplified with isset and ternary if:

function getVal1($name,$d){
 return isset($d[$name])?$d[$name]:null;
}

Answer 2

You want:

if (isset($d[$name]))
 return $d[$name];
else
 return null;