Change values in the innermost part of PHP associative array

Question

I would like to append a string to the values of a nested associative array in PHP.

The following is my array

$styles = array(
    'Screen' => array(
        "master.css",
        "jquery-jvectormap-1.0.css"),
    'handheld' => array("mobile.css") 
     );

Looping over it to change it in the following way fails.

foreach($media as $medium => $filename)
foreach($filenames as &$filename)
    $filename = "/styles/".$filename;

Prefixing $medium with & just causes a syntax error.

This also fails.

function prepend($prefix,$string)
{
return $prefix.$string;
}

foreach($media as &$medium)
  $medium = array_map(prepend("/styles"),$medium);

What is the simplest way to prefix "/styles/" to those css filenames given this data structure?

Ravi · Accepted Answer · 2012-12-11 14:56:15Z

2

Try this:

$styles = array(
    'Screen' => array(
        "master.css",
        "jquery-jvectormap-1.0.css"),
    'handheld' => array("mobile.css") 
     );

foreach($styles as $k=>$v){
    foreach($v as $a=>$b){
        $styles[$k][$a] = '/styles/'.$b;
    }
}

print_r($styles);

answered Dec 11, 2012 at 14:56

Ravi

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Comments

Peter Körner · Accepted Answer · 2012-12-11 14:58:12Z

2

Try this:

foreach($styles as $medium => &$filenames) {
  foreach($filenames as &$filename) {
    $filename = "/styles/".$filename;
  }
}

Some tips:

Use braces to structure your nested code. This will give you (and php) a better understanding of what is you meant
Use Indenting. This will give you a better overview.
Name the Iterating-Variable using the singular of the Array-Variable (ie. foreach($apples as $apple)). This will give you an extra hint with which variable you're dealing with.

Good Luck!

answered Dec 11, 2012 at 14:58

Peter Körner

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1 Comment

mac389 Over a year ago

Ah, my error with & was in prepending it to the key (here, medium) and not the value (here, filenames). Thanks for pointing that out.

MAXIM · Accepted Answer · 2012-12-11 14:58:13Z

0

foreach($media as $medium => $filenames) {
    foreach($filenames as $key => $filename) {
        $media[$medium][$key] = "/styles/" . $filename;
    }
}

answered Dec 11, 2012 at 14:58

MAXIM

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1 Comment

mac389 Over a year ago

Won't this modify a copy of $filename?

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