Reputation: 21
I am using an existing Gulp script that someone else made and it's interesting that a similar one for JS works fine, but this one for styles, doesn't. The code:
'use strict';
const
path = require('path'),
pathToLibs = path.resolve('../../../../libs');
const
{CssCompressor} = require('css-compressor.js'),
compressor = new CssCompressor(false),
resolveLocalSrc = function (files) {
return files.map(item => path.resolve(__dirname, item));
},
resolveLibsSrc = function (files) {
return files.map(item => path.resolve(pathToLibs, item));
};
let sources = {
"cache_main.css": [].concat(
resolveLocalSrc([
"scss/design-core.scss"
]),
resolveLibsSrc([
"vendor/jquery/dropzone-4.3.0/dropzone.css"
]),
resolveLocalSrc([
"../scripts/jquery/whatever.css",
"scss/whatever.scss",
])
),
"cache_light.css": resolveLocalSrc(["scss/design-core.scss"])
};
compressor.makeCache(sources, './', ["scss/*.scss"]);
When running gulp
, it says:
Task never defined: default
Excerpt from css-compressor.js:
makeCache(sources, dest = './', additionalDirsToWatch=[]) {
Object.entries(sources).forEach(([cacheFileName, files]) => {
gulp.task('compress ' + cacheFileName, () => {
this.compress(files, cacheFileName, dest);
});
});
Can you help? Thank you!
Upvotes: 2
Views: 324
Reputation: 1128
You need to declare a default task in the end.
Using Gulp 3.x:
gulp.task('default', ['styles', 'scripts', 'images']);
Using Gulp 4.x:
gulp.task('default', gulp.parallel('styles', 'scripts', 'images'));
Replace styles, scripts and images by your tasks.
Upvotes: 1