我已经有了一个方法,可以用gulp-minify-css将文件的内容压缩到一个临时文件夹中,如果可能的话,我希望以后用gulp删除它,但我不知道如何首先替换这些文件,我有:
var revReplace = require('gulp-rev-replace'); // -> this folder is not recognized
gulp.task('revrep', () => {
return gulp.src('app/Styles/CssArqMincss')
.pipe(gulp.dest('app/Content/CSSArq/'));
})之后如何删除文件夹/Styles/CssArqMincss?谢谢
发布于 2021-06-29 08:38:34
要从文件夹位置删除文件,可以使用此任务并对其进行调整。
const { src, task } = require( 'gulp' );
const clean = require("gulp-clean");
const logger = require('node-color-log');
function cleanFolderTask() {
const yourPathRoot = "pathToYourFolder"; //I would recommend using config file to store paths, such as : config.path.artifact.root;
logger.color('yellow').log(`Cleaning ${yourPathRoot}`);
return src(yourPathRoot ,{allowEmpty: true},{read: false}).pipe(clean({force:true}));
};
const cleanFolderTask= task('clean:folder', cleanFolderTask);
exports.cleanFolderTask= cleanFolderTask;要从一个文件夹复制到另一个文件夹,我还将使用前面的任务首先清理目标,然后:
const { src, dest, task } = require( 'gulp');
const logger = require('node-color-log');
function copyFilesTask(callback) {
let filesLoc = "path/to/your/files";
let filesDest = "path/to/destination"; // I would recommend using config file to store paths, such as : config.path.artifact.files and config.path.solution.files;
logger.color('yellow').log(`Copy ${filesDest }`);
src(`${filesLoc}/**/*`)
.pipe(dest(filesDest),
callback());
};
const copyFiles = task('copy:files', copyFilesTask);
exports.copyFiles = copyFiles;https://stackoverflow.com/questions/68005408
复制相似问题