多个gulp src

Question

如何简化此任务？

const merge = require('merge-stream')
gulp.task('templates', ()=>{
    return merge(
        gulp.src('./src/dashboard/views/**/*').pipe(gulp.dest('public/views/dashboard')),
        gulp.src('./src/core/views/**/*').pipe(gulp.dest('public/views/core')),
        gulp.src('./src/auth/views/**/*').pipe(gulp.dest('public/views/auth')),
        gulp.src('./src/templates/views/**/*').pipe(gulp.dest('public/views/templates'))
    )
})

Answer 1

您需要使用public/views/作为目标路径，并删除源路径中的/views部分：

• public / views / + dashboard / views = public / views / dashboard
• public / views / + core / views = public / views / core
• public / views / + auth / views = public / auth / auth
• public / views / + templates / views = public / auth / templates

您可以使用gulp-rename执行此操作：

var rename = require('gulp-rename');

gulp.task('templates', ()=>{
  return gulp.src('./src/*/views/**/*')
    .pipe(rename(function(file) {
      file.dirname = file.dirname.replace(/\/views/, '');
    }))
    .pipe(gulp.dest('public/views/'));
});

还有gulp-regex-rename更短（从未使用过那个，所以我不能担保）：

var rename = require('gulp-regex-rename');

gulp.task('default', ()=>{
  return gulp.src('./src/*/views/**/*')
    .pipe(rename(/\/views/, ''))
    .pipe(gulp.dest('public/views/'));
});