Variables into array - $array[$var_name] = $var;

Question

I was wondering whether something like this is possible:

$var1 = 1; $var2 = 2;
function varsToArray($param1, $param2...);

and it returns array like this

array([var1] => 1, [var2] => 2).

Simply saying, I'd like arrays keys to be same as variable names. The problem is I don't know how to get variable name as string to put it as key(if possible of course...).

Well, since that's not valid PHP syntax it's obviously not possible to ascertain what you mean by it. Describe your goal. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Sep 19, 2011 at 15:29
possible duplicate of Create array names based on variable value ... please use the search before you ask a question. — Felix Kling
– Felix Kling, Commented Sep 19, 2011 at 15:30
Then you did not describe your problem properly... the accepted answer puts the parameter names in the array, not the variable names, and I wonder why one would use reflection in this case. — Felix Kling
– Felix Kling, Commented Sep 19, 2011 at 15:49
Nope, I meant varaibles. Noticed after a while that Yoshi's solutions puts parameters names as keys, not varaibles names. Sry, my fault. The example i gave is bad, I'll change it — f1ames
– f1ames, Commented Sep 19, 2011 at 16:08

zzzzBov · Accepted Answer · 2011-09-19 15:29:11Z

3

You want to use the compact function methinks.

answered Sep 19, 2011 at 15:29

zzzzBov

180k57 gold badges328 silver badges375 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

f1ames Over a year ago

compact works when you pass variables names as strings so compact("var1", "var2"); I wish to pass just as compact($var1, $var2). In this case @Yoshi's solution works fine:D

Felix Kling Over a year ago

@f1ames: It works because the function's parameters are called $var1 and $var2. If you call foo($bar, $baz), the output will still contain var1, var2. I think passing the variable names as string is the only useful solution.

Asaph · Accepted Answer · 2011-09-19 15:30:06Z

0

The built-in PHP function compact() does this.

BTW: I'm going to go ahead and assume that instead of:

function $varsToArray($var1, $var2...);

you actually meant:

function varsToArray($var1, $var2...);

Notice the dollar sign has been removed.

answered Sep 19, 2011 at 15:30

Asaph

163k25 gold badges204 silver badges204 bronze badges

Collectives™ on Stack Overflow

Variables into array - $array[$var_name] = $var;

2 Answers 2

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related