我使用的是其他人制作的现有Gulp脚本,有趣的是,类似的JS脚本可以很好地工作,但是这种样式却不能。 编码:
'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
css-compressor.js的摘录:
makeCache(sources, dest = './', additionalDirsToWatch=[]) {
Object.entries(sources).forEach(([cacheFileName, files]) => {
gulp.task('compress ' + cacheFileName, () => {
this.compress(files, cacheFileName, dest);
});
});
你能帮我吗? 谢谢!
最后,您需要声明一个默认任务。
使用Gulp 3.x:
使用Gulp 4.x:
用您的任务替换样式,脚本和图像。