Call an existing php variable by concatenating two strings

Question

I have these seven php variables:

$dataA = 1;
$targetA = 2;
$dataB = 3;
$targetB = 4;
$dataC = 5;
$targetC = 6;
results = array('A','B','C');

I would like to loop through the results array and pass the corresponding variables to a function.

foreach($results as $value) {
    $data = '$data'.$value;
    $target = '$target'.$value;
    buildOutput('$data'.$value,'$target'.$value); 
}

// example, first time thru, wish to pass $dataA variable and $targetA variable

function buildOutput($data,$target) {
    echo "data=$data,target=$target<br>"; // echo's strings "$dataA" and "$targetA"
}

I cannot figure out how to declare the variables. Or if this is even possible.

I have more than just $data and $target variables, but I simplified down to two for the question.

Please use arrays to store values: array('A' => ['data' => 1, 'target' => 2]); // etc — u_mulder
– u_mulder, Commented Nov 30, 2021 at 7:03
I didn't wish to make the example complicated, but the $data's are associative arrays. The $target's are not. But I will re-think this. — user2140857
– user2140857, Commented Nov 30, 2021 at 7:46

Schlotter · Accepted Answer · 2021-11-30 07:07:19Z

1

I would not recommend using it, but try this:

$dataA = 1;
$targetA = 2;
$dataB = 3;
$targetB = 4;
$dataC = 5;
$targetC = 6;

$results = array('A','B','C');

foreach($results as $value) {
  buildOutput(${'data'.$value}, ${'target'.$value}); 
}

function buildOutput($data, $target) {
  echo "data=$data,target=$target<br>";
}

answered Nov 30, 2021 at 7:07

Schlotter

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2 Comments

user2140857 Over a year ago

That worked. Thank you. But why "would not recommend"? I'm only using this on my localhost.

Schlotter Over a year ago

im not recommend it, because it could break your code with wrong data/values, or cause some unexpected behavior. I would suggest to use Arrays, if possible. If only using for yourself, its ok, i guess.

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