How do I turn a string reference into a variable

Question

I want $masking_x to act as $masking_y (i.e. dynamically insert $radiovalue but still act as a varibale).

My code is

while($field_radio = mysql_fetch_assoc($result_radio)) {
    $radiovalue = $field_radio[radiovalue];
    echo "Radio value: ".$radiovalue."</br>";   
    $masking = "field_masking[checkbox_1001_".$radiovalue."]";
    $masking_x = '$'.$masking;
    $masking_y = "$field_masking[checkbox_1001_2]";
    echo "Masking: ".$masking."</br>";
    echo "Masking_x: ".$masking_x."</br>";
    echo "Masking_y: ".$masking_y."</br>";
    die;

I'm getting the following output:

Radio value: 2
Masking: field_masking[checkbox_1001_2]
Masking_x: $field_masking[checkbox_1001_2]
Masking_y: 2

Any guidance would be greatly appreciated. Thanks in advance.

Does field_masking have to be dynamic as well? What's wrong with echo $field_masking["checkbox_1001_".$radiovalue];? — Naltharial
– Naltharial, Commented Jul 18, 2012 at 13:09

Community · Accepted Answer · 2020-06-20 09:12:55Z

2

You are using variable variables slightly wrong. Try this instead:

$radiovalue = $field_radio['radiovalue'];
echo "Radio value: ".$radiovalue."</br>";   
$masking_x = "field_masking['checkbox_1001_".$radiovalue."']";
$masking_y = "field_masking['checkbox_1001_2']";
echo "Masking_x: ".$$masking_x."</br>";
echo "Masking_y: ".$$masking_y."</br>";

The double dollar-sign must be in the code, not in the string.

It's also possible to do this:

$radiovalue = $field_radio['radiovalue'];
echo "Radio value: ".$radiovalue."</br>";   
echo "Masking_x: ".${"field_masking['checkbox_1001_".$radiovalue."']"}."</br>";
echo "Masking_y: ".${"field_masking['checkbox_1001_2']"}."</br>";

Note also that I added single quotes into your string to properly quote the strings used for the array keys, associative array keys these are regular strings and should be quoted as such.

However it is important to note that variable variables are rarely the right solution to a given problem. It's usually possible to come up with something better using arrays or references instead.

EDIT

After playing around with it, it seems you cannot use a variable variable to reference an array key when the key definition is in the string. It's such a horrible thing to do, I'm not surprised I've never come across this limitation.

The long of the short of it is that you cannot do exactly what you are trying to do, and the best solution for you is probably to use references instead:

$radiovalue = $field_radio['radiovalue'];
echo "Radio value: ".$radiovalue."</br>";   
$masking = &$field_masking['checkbox_1001_'.$radiovalue];
echo "Masking: ".$masking."</br>";
$masking = "A different value";
echo "Masking: ".$masking."</br>";
// Original value has been updated as well
echo "\$field_masking['checkbox_1001_'.\$radiovalue]: ".$field_masking['checkbox_1001_'.$radiovalue]."</br>";

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jul 18, 2012 at 13:09

DaveRandom

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3 Comments

Naltharial Over a year ago

I get undefined errors on all of those. What PHP version supports that?

DaveRandom Over a year ago

@Naltharial After playing around with it, it seems you cannot use a variable variable to reference an array key when the key definition is in the string. It's such a horrible thing to do, I'm not surprised I've never come across this limitation. I will edit my answer to reflect this.

Rob Williams Over a year ago

Thank you; The field_masking did not have to be dynamic so Naltharial's $field_masking["checkbox_1001_".$radiovalue]; solution solved my problem. I also learned quite a bit from the other comments and I'm gratefully for everyone's help. Thanks again.

Naltharial · Accepted Answer · 2012-07-18 13:27:05Z

1

If you want to dynamically access an array variable, split it in two parts, single variable won't work.

$name = 'field_masking';
$key = 'checkbox_1001_'.$radiovalue;

echo $$name[$key];

answered Jul 18, 2012 at 13:27

Naltharial

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1 Comment

Rob Williams Over a year ago

Thank you; The field_masking did not have to be dynamic so Naltharial's $field_masking["checkbox_1001_".$radiovalue]; solution solved my problem. I also learned quite a bit from the other comments and I'm gratefully for everyone's help. Thanks again.

mogelbrod · Accepted Answer · 2012-07-18 13:11:50Z

0

Take a look at variable variables. Do note that this is a potential security risk if you're working with user input.

In your case: $masking_x = $$masking;

answered Jul 18, 2012 at 13:11

mogelbrod

2,33419 silver badges20 bronze badges

1 Comment

Rob Williams Over a year ago

Thank you; The field_masking did not have to be dynamic so Naltharial's $field_masking["checkbox_1001_".$radiovalue]; solution solved my problem. I also learned quite a bit from the other comments and I'm gratefully for everyone's help. Thanks again.

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