Variable variable, building an array variable

Question

I want to be able to pass a string ("['parent_array']['child_array']") to a function that then pulls this string and "my_array" to the front of it and then creates a variable variable.

Then inside the function i do print_r($$string) and nothing comes out. See the code below for a better explanation.

// DOES NOT WORK
$string1 = "my_array['parent_array']['child_array']";
print_r($$string1); //prints nothing.

// WORKS
$string2 = "test";
$test = "This will be printed!";
print_r($$string2); //prints "This will be printed!

// WORKS
print_r($my_array['parent_array']['child_array']);

May I ask why are you building the variables this way? I could bet a large sum of money for there being a more reasonable approach than (shudder) variable variables. — JJJ
– JJJ, Commented Mar 28, 2012 at 8:51
and / or eval, I bet one of the answers later would actually recommend to use eval — Andreas Wong
– Andreas Wong, Commented Mar 28, 2012 at 8:52
You seem to be asking for a bugfix of a patch from solution to a problem. That not the way to go. Instead you should describe the problem, show how you tried to sole it and ask for better solution or changes for the existing one. — tereško
– tereško, Commented Mar 28, 2012 at 9:54

David-SkyMesh · Accepted Answer · 2012-03-28 09:36:09Z

0

Variable variables are an ugly, ugly hack.

Use eval.

Example:

$expr = "\$my_array['parent_array']['child_array']";
$val = 'not assigned';
eval("\$val = $expr;");
print_r($val);

But please follow their warnings about the danger of evaling user-supplied code.

Updated: with example

edited Mar 28, 2012 at 9:36

answered Mar 28, 2012 at 8:51

David-SkyMesh

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8 Comments

halliewuud Over a year ago

Can I get a example how this would look like?

PeeHaa Over a year ago

Both are ugly hacks IMHO. Almost always there is a better solution. The real question to OP would be: why do you think you need it?

halliewuud Over a year ago

@RepWhoringPeeHaa look here: pastebin.com/bBS4iiMC I want to be able to pass a variable to the function that replaces "line29. $arrXml['Assignments']['Assignment']"

tereško Over a year ago

If eval() is the answer, you are asking the wrong question (or have an extreme lisp).

JJJ Over a year ago

@halliewuud And you can't do $arrXml[ $var1 ][ $var2 ] because...?

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