November 21, 2016 —John Koster
elixir($file, $buildDirectory = 'build')
The elixir
helper function can be used to get the path to a versioned Elixir file. The function accepts two arguments, the name of the $file
to find the path for and the $buildDirectory
to look in. By default, the $buildDirectory
is set to build
. The build directory is relative to the applications public directory (app/public
).
Assuming the following simple gulpfile:
1var elixir = require('laravel-elixir'); 2 3elixir(function(mix) { 4 mix.sass('app.scss'); 5 6 // Version using the default configured build path. 7 mix.version('css/app.css'); 8 9 // Version using a custom build path.10 mix.version('css/app.css', 'public/build2');11});
The path to the file versioned using the default configured build directory can be determined like this:
1<?php2 3elixir('css/app.css');
and would produce output similar to the following:
1/build/css/app-aa3c9d985b.css
Likewise, the path to the versioned file with the custom build directory can be determined by supplying an argument for $buildDirectory
:
1<?php2 3elixir('css/app.css', 'build2');
which would produce output similar to the following:
1/build2/css/app-64f4edce94.css
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