PHP: Variable in array declaration

Question

I'm working in PHP and calling the following function from inside a foreach{} loop. This function needs to accept $subTheme as an option parameter because the purpose is to avoid unnecessary code repetition (DRY, right?).

So, first off - here's the function:

/*
 * CSS needed to apply the selected styles to text elements.
 */
function this_site_text_css( $subTheme = '' ) {

    $GLOBALS['this_theme_mod']; // this is in global scope already

    $themeVarSuffix = $subTheme['var_suffix'];
    $themeClassName = $subTheme['class_name'];

    $content_bg_color  = $this_theme_mod['content_bg_color' . $themeVarSuffix ];
    $page_bg_color     = $this_theme_mod['page_bg_color' . $themeVarSuffix ];
    $primary_color     = $this_theme_mod['primary_theme_color' . $themeVarSuffix ];

    $css = 'body' . $themeClassName . '{ /* special classes */ };'

    return $css
}

There's more happening but it's rather tedious and just concatenates CSS as a string to return.

It's being called like this

$data = '';
$data .= this_site_text_css();
$subThemeArray = array(
  'theme_a' => array( 
     'var_suffix' => '_theme_a',
     'class_name' => '.theme-a',
  ),
  'theme_b' => array( 
     'var_suffix' => '_theme_b',
     'class_name' => '.theme-b',
  ),
);
foreach( $subThemeArray as $key => $theme ) {
   $data .= this_site_text_css( $theme );
}

I'm getting a PHP warning Illegal string offset 'class_name' and I'm guessing this is because PHP doesn't want me to concatenate $themeVarSuffix inline when declaring it at $themeClassName. I'm pretty sure there's a way to do this, and I've searched all over, perhaps I haven't searched the right keywords, but any help would be appreciated.

hek2mgl · Accepted Answer · 2013-04-26 02:11:16Z

3

Illegal string offset 'class_name'

... means that $subTheme is actually a string and not an array. This happens because you have a default value for it in the function declaration $subTheme = '' and missed to pass a value once you call it, here:

$data .= this_site_text_css();

So $subTheme is an empty string which has of course no index 'class_name'.

answered Apr 26, 2013 at 2:11

hek2mgl

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1 Comment

Brian Over a year ago

Thanks! That was it... just needed to do function this_site_text_css( $subTheme = array() )

raidenace · Accepted Answer · 2013-04-26 02:12:50Z

1

In $subThemeArray array the index theme_class should be called class_name

answered Apr 26, 2013 at 2:12

raidenace

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3 Comments

hek2mgl Over a year ago

+1 Good catch! For others, note that the question was updated after this answer.

Brian Over a year ago

I appreciate the catch - but I was just writing some pseudo code so the incorrect code I posted initially is divergent from my working code and the working code was never wrong, only the pseudo code.

raidenace Over a year ago

np, thing is that when code is posted for debugging one can never be sure what is a typo and what is a real bug in code. Anyways glad @hek2mgl got it figured out for you

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