Gulp 吞咽部分删除目录结构
我的目录结构与此类似:Gulp 吞咽部分删除目录结构,gulp,gulp-rename,Gulp,Gulp Rename,我的目录结构与此类似: app/ directory1/ assets/ images/ js/ directory2/ assets/ images/ dist/ assets/ images/ js/ 我尝试使用Gulp实现的是从目录1、2、…“收集”资产。。。并将它们放入dist/assets/中,因此我写了以下内容: gulp.task('gather-assets', function() { gul
app/
directory1/
assets/
images/
js/
directory2/
assets/
images/
dist/
assets/
images/
js/
我尝试使用Gulp实现的是从目录1、2、…“收集”资产。。。并将它们放入dist/assets/中,因此我写了以下内容:
gulp.task('gather-assets', function() {
gulp.src('app/*/assets/**').pipe(gulp.dest('dist/assets/'));
});
问题是,运行此函数后,它将创建如下路径:
dist/assets/directory1/assets/images
gulp.task('gather-assets', function() {
gulp.src('app/*/assets/**').pipe(rename({dirname: ''})).pipe(gulp.dest('dist/assets/'));
});
按照来自的建议,我尝试使用gulp rename,但我的情况不同,如果我像这样使用gulp rename:
dist/assets/directory1/assets/images
gulp.task('gather-assets', function() {
gulp.src('app/*/assets/**').pipe(rename({dirname: ''})).pipe(gulp.dest('dist/assets/'));
});
它肯定会删除*星号处不必要的路径,但也会删除**路径。因此,images/和js/中的文件都将复制到assets/中,而不包含子目录。对于这种情况,我有什么解决方案?适合您
var flatten = require('gulp-flatten');
gulp.task('gather-assets', function() {
return gulp.src('app/*/assets/**')
// .pipe(rename({ dirname: '' }))
// -2 will keep the last two parents : assets/images or assets/js
.pipe(flatten({ includeParents: -2 }))
.pipe(gulp.dest('dist'));
});
如果您想使用gulp rename:[gulp flatte和gulp rename都只是对每个文件的目录结构进行字符串操作]
// .pipe(flatten({ includeParents: -2 }))
.pipe(rename(function (file) {
let tempArray = file.dirname.split(path.sep);
// remove the first array item : directory1 or directory2
// rejoin the remaining array items into a directory string
let temp = tempArray.slice(1).join(path.sep);
file.dirname = temp;
}))