Conditional return in a PHP function - php

Question

I have a return statement in my function:
function fun($result)
{
$counter = 0;
while($row=mysql_fetch_array($result))
{
/*if($counter==2)
{
return $name;
}*/
$user=$row['message'];
$sql2="select username from members where id='$user'";
$result2=mysql_query($sql2);
$row2=mysql_fetch_array($result2);
$name .= $row2['username']." and ";
$counter++;
if($counter==2)
{
return $name;
}
}
}
The return block (commented) doesn't work which immediately terminates the execution of the function irrespective of the condition. But the return block if placed in position as shown in the code just works fine. Why?
I want to return only 2 rows. It seems that it should work whether it is placed before or after,because it depends on the value of the counter. Check the condition before adding new row to $name variable.First row will be added to $name and $counter is incremented,then check the condition and so on.What is wrong?

Answer 1

Function doesn't return anything when there are less than 3 rows even if you want to return when you reach 2, you need 3 rows minimal.
The function checks $counter in the loop after you handled your 2nd row.
So there need to be at least 3 rows for the function to return the $ name. This is weird because you want it to return then there are 2 names found.
The function doesn't necessarily terminates, but simply doesn't return anything.

Answer 2

Your function may never reach the return statement. You have to check for both conditions to be false and then return $name if they are. A for loop should do what you need.
function fun($result)
{
for($cnt=0; ($row=mysql_fetch_array($result)) AND $cnt<2; $cnt++)
{
$name .= 'Your Data';
}
return $name;
}