本文介绍了如何使用 Gulp.js 将流保存到多个目的地?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!
问题描述
const gulp = require('gulp');
const $ = require('gulp-load-plugins')();
const source = require('vinyl-source-stream');
const browserify = require('browserify');
gulp.task('build', () =>
browserify('./src/app.js').bundle()
.pipe(source('app.js'))
.pipe(gulp.dest('./build')) // OK. app.js is saved.
.pipe($.rename('app.min.js'))
.pipe($.streamify($.uglify())
.pipe(gulp.dest('./build')) // Fail. app.min.js is not saved.
);
当前不支持当 file.contents 为流时通过管道传输到多个目标.有什么办法可以解决这个问题?
Piping to multiple destinations when file.contents is a stream is not currently supported. What is a workaround for this problem?
推荐答案
目前在使用 file.contents 作为流时,每个 dest 必须使用两个流.这可能会在未来得到解决.
Currently you have to use two streams for each dest when using file.contents as a stream. This will probably be fixed in the future.
var gulp = require('gulp');
var rename = require('gulp-rename');
var streamify = require('gulp-streamify');
var uglify = require('gulp-uglify');
var source = require('vinyl-source-stream');
var browserify = require('browserify');
var es = require('event-stream');
gulp.task('scripts', function () {
var normal = browserify('./src/index.js').bundle()
.pipe(source('bundle.js'))
.pipe(gulp.dest('./dist'));
var min = browserify('./src/index.js').bundle()
.pipe(rename('bundle.min.js'))
.pipe(streamify(uglify())
.pipe(gulp.dest('./dist'));
return es.concat(normal, min);
});
此错误现已在 gulp 中修复.您原始帖子中的代码应该可以正常工作.
This bug is now fixed in gulp. The code in your original post should work fine.
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