variable in a variable???

fifo

can this be done?

i have the following variables being pulled from a mysql database
as follows
$yyy_01_xxx
$yyy_02_xxx
$yyy_03_xxx
$yyy_04_xxx
$yyy_05_xxx
...
$yyy_10_xxx

if I echo $yyy_01_xxx it displays correctly what is in the db.

another variable now being updated by a for loop
$number is a number 01 - 10

now I want another variable

$variableXYZ = "$yyy".$number."xxx"; <- I THINK THIS IS MY PROBLEM IS

but when I echo $variableXYZ it displays the text ($number_01_xxx) instead of getting what is really in the variable $number_01_xxx

bpat1434

You want variable variables... try this:

<?php

function check($num)
{
    if($num<10 && strlen($num)<2)
    {
        $num = '0'.$num;
    }
    return $num;
}

for($i=1; $i<11; $i++)
{
    $var = 'yyy_'.check($i).'_xxx';
    echo ${$var}.'<br>';
}

?>

~Brett

fifo

It compiles and gives no errors but it prints nothing. Almost as if the variable is empty.

bpat1434

Sorry, the code should be:

?php

function check($num)
{
    if($num<10 && strlen($num)<2)
    {
        $num = '0'.$num;
    }
    return $num;
}

for($i=1; $i<11; $i++)
{
    $var = 'yyy_'.check($i).'_xxx';
    $$var = 'some value ['.check($i).']';
    echo ${$var}.'<br>';
}

?>

~Brett

fifo

Bret thanks again for helping me out. I tried that but Im not sure thats what I need. Or maybe I just dont quite understand it.

Assume I already pulled this variable $YYY_01_XXX from the db its value lets asume is 'ABCDEFG' so

$YYY_01_XXX = 'ABCDEFG';

basically I want to be able to end up with another variable that will hold the name of the variable pulled from the db. This variable will change in the for loop like you displayed.

$variable = "$YYY_01_XXX"
when in lopp will look like this

$variable = "$YYY_02_XXX"
$variable = "$YYY_03_XXX"
$variable = "$YYY_04_XXX"
...

on a side note i am trying this like so $variable = '$YYY'".$number."'XXX';

and

echoing $variable should give me the value that was pulled from the db.

Whats happening is $YYY_01_XXX is getting interpreted as a string instead of a php variable. and just displaying exactly that $YYY_01_XXX instead of ABCDEFG.

That last post might be correct I may have just misinterpreted it.

fifo

Hey Bret you are the man!!!

Thanks a million. I massaged it a bit more and finally understood what you were saying!. Disreguard my last message that worked!

all I needed was what you pointed out above

$variable = 'YYY'".number.'"XXX';

then call it like this ${$variable} it worked like a charm!