CODEWITHSUNDEEP
phpless
qaharmdz
qaharmdz

Reputation: 232

Overwriting LessPHP variable with PHP

I use LessPHP (v0.3.4-2), read the documentation and search around this -basic- question but still not found the correct and straight answer.

Here is my .less file:

@bgcolor :#ccc; 
.color(@color:#222) { color:@color; }

body {
  margin:10px auto;
 .color(@color);
 background:@bgcolor;
}

My php code:

<?php
include 'lessc.inc.php';

$style_setting = array (
  'color'       => '#f00',
  'bgcolor'    => '#fee'
);

$cssFile    = "stylesheet.css";
$lessFile    = new lessc("less/stylesheet.less");
file_put_contents($cssFile, $lessFile->parse(null, $style_setting));
?>

Stylesheet result:

body {
  margin:10px auto;
  color:#ff0000;
  background:#cccccc;
}

As you see on example above, the LesPHP arguments (.color(@color:#222)) is overwrited but the @bgcolor variable is not.

Just wondering is it correctly how LessPHP work or do I miss something here? Why does the LessPHP variable is not overwrited?

Any answer or hint is appreciated.

Upvotes: 1

Views: 648

Answers (1)

cchana
cchana

Reputation: 5000

Why not have the same implementation of @bgcolor as .color?

If the following works for @color:

.color(@color:#222) { color:@color; }

then the following should work for @bgcolor:

.bgcolor(@bgcolor:#ccc) { background:@bgcolor; }

Update, an explanation of what is happening:

Let's assume this was all written in PHP, this is what a .color function would look like:

function color($color = null) {
    if(!$color) {
        return 'color: #222;';
    }
    return 'color: $color;';
}
echo color(); //outputs "color: #222222;"
echo color('#f00'); //outputs "color: #ff0000;"
echo color(); //outputs "color: #222222;"

It takes the colour value passed OR outputs a default value. Now, this is what is happening with the @bgcolor variable:

$bgcolor = '#fee';
$bgcolor = '#ccc';
echo 'color: '.$bgcolor.';'; //outputs "background: #cccccc;"

They key difference is that .color is a function that can take values and @bgcolor is a variable.

You've effectively defined what you want, instantly overwritten it and then called the variable. The only way to make it dynamic is to turn it into a 'function' like you have with color.

Upvotes: 2

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