Variable in Variable (?)- php loop

Question

I'm relatively new to PHP loops, been learning them and they sure do make my life easier. However, I came across a bit of a struggle when trying to create a new variable for the PHP loop.

Background:

I declared 21 variables such as:

$q1W = 5;
$q2W = 10;
$q3W = 2;

Then I grabbed the $_GET (q1,q2,q3) variables and put them into variables with their values:

foreach($_GET as $qinput => $value) {
    $$qinput  = $value ;
}

Now, essentially, I want to turn this code:

$q1final = $q1 * $q1W;
$q2final = $q2 * $q2W;
$q3final = $q3 * $q3W;

Into a loop so I don't need to type out that all the way to 21. This is what I have thus far:

<?php for ($i=1; $i<=21; $i++) { 
$q.$i.final = $q.$i * $q.$i.W
}

What am I missing?

You can't concatenate when making dynamic variables the same way as normal (periods). Also, generally if you have to use dynamic variables, you're doing something wrong. — John V.
– John V., Commented Jun 26, 2012 at 22:28
@AlexLunix You can concatenate strings to make dynamic variable names (see my comment on the answer below) but I agree that if "dynamic variable names" is the right answer, you're probably asking the wrong question. — DaveRandom
– DaveRandom, Commented Jun 26, 2012 at 22:32

Rusty Fausak · Accepted Answer · 2012-06-26 22:22:43Z

5

I would recommend using arrays instead of lots of variables. It makes relating your data more straightforward. For example:

$mults = array(
    'q1W' => 5, 
    'q2W' => 10,
    'q3W' => 2
);
$final = array();
foreach ($_GET as $qinput => $value) {
    $final[$qinput] = $mults[$qinput] * $value;
}
print_r($final);

answered Jun 26, 2012 at 22:22

Rusty Fausak

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1 Comment

DaveRandom Over a year ago

This is the correct answer (+1), but for reference to the OP/future visitors, the correct syntax for using a counter as above to define a variable variable name would be use strings in braces, e.g. ${'q'.$i.'final'} = ${'q'.$i} * ${'q'.$i.'W'};

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