Use variable NAME as value in PHP

Question

Not sure how to ask this without rambling about preprocessors in C, but I will try:

Is there a way in PHP, to use the name of a variable as a string or value.

Such that instead of calling function like this,

$somevar = 'whatever';

fun('somevar', $somevar);

you can call it like this:

fun(<magic:$somevar>, $somevar);

or even, dare I say it, like this:

fun($somevar);

and the fun() would itself deduce both the value ('whatever'), and the name ('somevar') of the variable. (In C, it could be done with a preprocessor #define.)

_{"magic" above is of course a place holder for the magic syntax in PHP I am looking for which will help me do this. I guess it could be made somehow with eval()? But people keep telling me not to use eval.}

Someone hinted that introspection might somehow do the trick. Any leads there would be welcome.

I got some hints about ReflectionParameters by passers-by, but you deleted your comments. — Prof. Falken
– Prof. Falken, Commented Sep 6, 2012 at 7:24
@Amigable are you looking for something like returning the variable name as a value.? — ravi404
– ravi404, Commented Sep 6, 2012 at 7:29
@sberry, more succinct code, I have lots of variable names which are always the same as that string. Stupid to type the same thing twice. — Prof. Falken
– Prof. Falken, Commented Sep 6, 2012 at 7:29
@goldenparrot, from the description, technically yes, but actually not quite, what happens if I mistype, fun(ltrim('$somevarr', '$'), $somevar); is that the PHP gives no help at all. If the first argument was just a variable, the parser could warn that $somevarr was undefined. — Prof. Falken
– Prof. Falken, Commented Sep 6, 2012 at 7:58

ravi404 · Accepted Answer · 2012-09-06 07:45:18Z

2

Try this

function var_name(&$var, $scope=0)
{
    $old = $var;
    if (($key = array_search($var = 'unique'.rand().'value', !$scope ? $GLOBALS : $scope)) && $var = $old) return $key;  
}

edited Sep 6, 2012 at 7:45

answered Sep 6, 2012 at 7:22

ravi404

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13 Comments

Lucas Green Over a year ago

This isn't quite what he is looking for

ravi404 Over a year ago

@Mythril i think he wants the variable name to be as a string. This function returns that. or can u please explain where I'm wrong so that i correct it.

Lucas Green Over a year ago

Run this in the same scope as your function $x = 5; $y = 5; echo var_name($x), "\n"; echo var_name($y), "\n";

Lucas Green Over a year ago

also, this would utterly fail when trying to use it on a local inside of a function

Prof. Falken Over a year ago

If nothing better performance wise shows up, I will eventually accept this answer, because your function works. But I won't use it, it's too weird. :-)

|

RedHydra · Accepted Answer · 2012-09-06 07:51:35Z

2

http://php.net/manual/en/language.variables.variable.php

The probably explain it better than I can, I think this is what you're looking for?

Alright, before you call the function do the following:

$var= "variableName";
$$var = "variableValue";
fun($var);

Now within the function definition for fun:

function fun($parameter)
{
    global $$parameter;
    echo $parameter; //echos the variableName
    echo $$parameter; //echos the variableValue
}

edited Sep 6, 2012 at 7:51

answered Sep 6, 2012 at 7:24

RedHydra

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7 Comments

Prof. Falken Over a year ago

Either, no it's NOT what I am looking for, OR I don't understand what that page says.

maxdec Over a year ago

@RedHydra: I would have given the same answer

Prof. Falken Over a year ago

@Double so you would also have pointed to a page and not actually write an answer. ;-)

RedHydra Over a year ago

@AmigableClarkKant I've updated my answer, I'm pretty sure this will work for you, let me know if it doesn't.

Prof. Falken Over a year ago

How will that help me write: $clevername = 'something'; fun($clevername); ....... and the fun() knowing both 'clevername' and 'something'

|

troelskn · Accepted Answer · 2012-09-06 08:07:18Z

1

No. PHP's scoping rules prevent you from doing that. The only way would be through a pre processor (as you are aware) or by introspecting the code, which would be kind of the same thing. PHP has a built-in tokenizer, that you can use to parse a file with. You can use debug_backtrace to find out which files to parse through. It's going to be a really messy and inefficient thing to do though, so you would probably be better off rethinking what you're doing.

answered Sep 6, 2012 at 8:07

troelskn

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4 Comments

Prof. Falken Over a year ago

Elan - "I don't wanna sell you death sticks." troelskn - "You want to go home and rethink your life." Amigable - "I wanna go home and rethink my life."

troelskn Over a year ago

;) - Not knowing your use case, I may be off, but it sounds like you might want to use an array for this, somehow?

Prof. Falken Over a year ago

I think I will just use fun('x', $x); at least it's very readable and straightforward and easy to spot errors visually.

troelskn Over a year ago

If it works, then I concur. Verbose is usually better than clever.

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