Build a variable name in PHP out of two variables

Question

Depending on the current page, I would like to change the css of the chosen menu module and make others different, etc. All that while building dynamically the page before loading.

My problem is that I have a serie of variables going by this structure :

$css_(NAME_OF_MODULE)

To know what variable must be set , I have another variable I received in parameters of this functions, called

$chosen_menu

Say $chosen_Menu = "C1", I would like to add content to $css_C1. Thus, what I want is to create a variable name out of 2 variables, named $css_C1

I have tried :

${$css_.$chosen_menu} = "value";

But it doesnt seem to work. Any clue ?

Why not ? Is there any reason not to ? EDIT : There was an error in the question. — Alexandre Bolduc
– Alexandre Bolduc, Commented Nov 13, 2011 at 15:29

Aswin Kumar · Accepted Answer · 2014-03-14 08:50:03Z

5

That probably won't just work. PHP supports full indirection though, so something like this will work.

$varName = $css."_".$chosen_menu;
$$varName = "value";

If not, it will probably be attempting to interpret $css_ as a variable name in your second code sample, so just change that to $css."_".$chosen_menu.

edited Mar 14, 2014 at 8:50

Aswin Kumar

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answered Nov 13, 2011 at 15:29

slugonamission

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Dragos Custura · Accepted Answer · 2014-02-10 08:54:49Z

4

$nr = 1;        
       ${'name' . $nr} = 20 ;        
       var_dump($name1);

answered Feb 10, 2014 at 8:54

Dragos Custura

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Jim H. · Accepted Answer · 2011-11-13 15:32:53Z

2

Check out http://php.net/manual/en/language.variables.php

You should be able to use:

$menu = $css . '_' .$chosen_menu;
$$menu = 'some value';

or

${$menu} = 'some value';

answered Nov 13, 2011 at 15:32

Jim H.

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Comments

Niko · Accepted Answer · 2011-11-13 15:34:09Z

0

I'm not sure if I get you right, but how about this:

${$css."_".$chosen_menu} = "value";

edited Nov 13, 2011 at 15:34

answered Nov 13, 2011 at 15:28

Niko

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