Can I use a Gulp task with multiple sources and multiple destinations?
Solution 1:
I guess that the running tasks per folder recipe may help.
Update
Following the ideas in the recipe, and oversimplifying your sample just to give the idea, this can be a solution:
var gulp = require('gulp'),
path = require('path'),
merge = require('merge-stream');
var folders = ['httpdocs-site1', 'httpdocs-site2', 'httpdocs-site3'];
gulp.task('default', function(){
var tasks = folders.map(function(element){
return gulp.src(element + '/media/sass/**/*.scss', {base: element + '/media/sass'})
// ... other steps ...
.pipe(gulp.dest(element + '/media/css'));
});
return merge(tasks);
});
Solution 2:
you are going to want to use merge streams if you would like to use multiple srcs but you can have multiple destinations inside of the same one. Here is an example.
var merge = require('merge-stream');
gulp.task('sass', function() {
var firstPath = gulp.src(sass_paths[0])
.pipe(sass({errLogToConsole: true}))
.pipe(autoprefixer('last 4 version'))
.pipe(minifyCSS({keepBreaks:true}))
.pipe(rename({ suffix: '.min'}))
.pipe(gulp.dest('./httpdocs-site1/media/css'))
.pipe(gulp.dest('./httpdocs-site2/media/css'));
var secondPath = gulp.src(sass_paths[1])
.pipe(sass({errLogToConsole: true}))
.pipe(autoprefixer('last 4 version'))
.pipe(minifyCSS({keepBreaks:true}))
.pipe(rename({ suffix: '.min'}))
.pipe(gulp.dest('./httpdocs-site1/media/css'))
.pipe(gulp.dest('./httpdocs-site2/media/css'));
return merge(firstPath, secondPath);
});
I assumed you wanted different paths piped here so there is site1 and site2, but you can do this to as many places as needed. Also you can specify a dest prior to any of the steps if, for example, you wanted to have one dest that had the .min file and one that didn't.
Solution 3:
You can use gulp-rename to modify where files will be written.
gulp.task('sass', function() {
return gulp.src(sass_paths, { base: '.' })
.pipe(sass({errLogToConsole: true}))
.pipe(autoprefixer('last 4 version'))
.pipe(minifyCSS({keepBreaks:true}))
.pipe(rename(function(path) {
path.dirname = path.dirname.replace('/sass', '/css');
path.extname = '.min.css';
}))
.pipe(gulp.dest('.'));
});
Important bit: use base
option in gulp.src
.
Solution 4:
For the ones that ask themselves how can they deal with common/specifics css files (works the same for scripts), here is a possible output to tackle this problem :
var gulp = require('gulp');
var concat = require('gulp-concat');
var css = require('gulp-clean-css');
var sheets = [
{ src : 'public/css/home/*', name : 'home.min', dest : 'public/css/compressed' },
{ src : 'public/css/about/*', name : 'about.min', dest : 'public/css/compressed' }
];
var common = {
'materialize' : 'public/assets/materialize/css/materialize.css'
};
gulp.task('css', function() {
sheets.map(function(file) {
return gulp.src([
common.materialize,
file.src + '.css',
file.src + '.scss',
file.src + '.less'
])
.pipe( concat(file.name + '.css') )
.pipe( css() )
.pipe( gulp.dest(file.dest) )
});
});
All you have to do now is to add your sheets as the object notation is constructed.
If you have additionnal commons scripts, you can map them by name on the object common
, then add them after materialize for this example, but before the file.src + '.css'
as you may want to override the common files with your customs files.
Note that in the src
attribute you can also put path like this :
'public/css/**/*.css'
to scope an entire descendence.