我有这个 node.js 代码,它试图将多个 js 文件缩小并合并到一个 js 文件中.
I have this node.js code that tries to minify and combine multiple js files to a single js file.
var concat = require('gulp-concat');
var gulp = require('gulp');
gulp.task('scripts', function() {
//gulp.src(['./lib/file3.js', './lib/file1.js', './lib/file2.js'])
gulp.src(['./js/*.js'])
.pipe(concat('all.js'))
.pipe(uglify())
.pipe(gulp.dest('./dist/'))
});
我所有的 js 文件都位于 js 文件夹中.我的 node.js 文件位于 js 文件夹上方.我期望单个缩小文件出现在 dist 文件夹中.运行代码时,我什么也看不到,也没有收到任何错误消息.可能出了什么问题?
All my js files are located in js folder. My node.js file is above the js folder. I am expecting the single minified file to appear in dist folder. I see nothing and get no error message when I run the code. What could have gone wrong?
"use strict";
var concat = require('gulp-concat');
var gulp = require('gulp');
var uglify = require('gulp-uglify'); // Add gulp-uglify module to your script
gulp.task('scripts', function() {
gulp.src('./js/*.js')
.pipe(concat('all.js'))
.pipe(uglify())
.pipe(gulp.dest('./dist/'));
});
运行 npm install
以验证是否正确加载了所有依赖项.我认为这是您的问题:
Run npm install
to verify that all dependencies correctly loaded. I think this was your issue:
{
"dependencies": {
"gulp-concat": "2.x",
"gulp": "3.x",
"gulp-uglify": "1.x"
}
}
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