Can't get my array out of my function

[slancasterfreak]

I am new to functions, and have been doing ok until I tried to return an array. 

 

function WalkerArray () {
       require('connecttodatabasea.inc.php');
       // get Walkers names into an array
       $queryx=("SELECT * FROM walkers ORDER BY id ASC");
       $resultx=mysql_query($queryx)  or die(mysql_error());
       $numx=mysql_numrows($resultx);
       $k=0;
       while ($k<$numx) {
              $id=mysql_result($resultx,$k,"id");
              $firstname=mysql_result($resultx,$k,"firstname");
              $lastname=mysql_result($resultx,$k,"lastname");
              $walkerarray[$id]=$firstname.' '.$lastname;
              $k++;
              }
       print_r($walkerarray);
       return $walkerarray;
       }



// now try to use function and see if I have the array and can print it out

WalkerArray();
print_r($walkerarray);

$idnum=50;
echo $walkerarray[$idnum];

 

The print _ r command INSIDE the function outputs the complete array, so I know my Mysql etc. is good.

But the second print _ r OUTSIDE the function prints nothing

 

I've searched for answers online and cannot figure it out. Thanks for the help!!!!

[MadTechie]

your currently returning the array to nothing.. as has you want the returned results.. try this

$walkerarray = WalkerArray(); // <---
print_r($walkerarray);

 

thus $walkerarray is now the array

of course you could use any variable, ie

$MT = WalkerArray(); // <---
print_r($MT);

[recklessgeneral]

Hi,

 

The $walkerarray variable is local to the WalkerArray function, so it can only be used in the scope of that function. In order to use it (or more precisely, a copy of it) outside the function, you assign the result of the function to a new variable and that will then work:

 

$walkerarray = WalkerArray();
print_r($walkerarray);

 

This should now print the same results as inside the function. The point to bear in mind is that the $walkerarray in the code above is a different variable to the one declared in the function.

 

Cheers,

Darren.