Gulp task for multiple files

Question

I'm trying to create a gulp task that creates (in this case minifies) a css file for every project css there is. So for Project A it should combine and minify the following files:

global.projecta.css tool1.projecta.css tool2.projecta.css

It needs a kind of loop or something...? If there is a new Project D, it should build it too. If i need to add something like "projectd" to an array, that would be ok too.

This is the directory structure for the css

./global.projecta.css ./global.projectb.css ./global.projectc.css ./addons/tool1.projecta.css ./addons/tool2.projecta.css ./addons/tool1.projectb.css ./addons/tool2.projectb.css ./addons/tool1.projectc.css ./addons/tool2.projectc.css

This is the not working task

// not working.... gulp.task('css', function() { return gulp.src(['global.*.css', 'addons/tool*.css', '!*.min.css']) .pipe(concat('build.css')) .pipe(rename({suffix: '.min'})) .pipe(minifycss()) .pipe(gulp.dest('build/css')); });

Brian Glaz · Accepted Answer · 2015-06-05 15:02:28Z

You could try something like this, although it requires manually entering an array of your projects.

gulp.task('css', function() { var projects = ['a','b','c']; projects.forEach( function(proj) { gulp.src(['global.' + proj + '.css', 'addons/tool*.' + proj + '.css', '!*.min.css']) ... }); });

had to write forEach but thanks for the simple and working tip

mraiur · Accepted Answer · 2015-06-05 13:35:40Z

This outputs different files for each global project. For each global.project*.css will output global.project*.min.css:

var gulp = require('gulp'); var css = require('gulp-minify-css'); var concat = require('gulp-concat'); var rename = require('gulp-rename'); var glob = require('glob'); gulp.task('css', function() { glob('global.\*.css', {}, function (er, files) { for(var i = 0, cnt = files.length; i < cnt; i++){ var fileName = files[i].replace('.css', ''); gulp.src(['global.\*.css', 'addons/tool\*.css', '!*.min.css']) .pipe(concat('build.css')) .pipe(rename({basename: fileName, suffix: '.min'})) .pipe(css()) .pipe(gulp.dest('build/css')); } }); }); gulp.task('default', ['css']);

but that will build only one file build.css containing all css. my question was about one build file per project.

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