基本的Gulp Uglify设置？

Question

I want to take all .js files in my project, and for each, save a minified version in the same directory.

That is, given this project directory structure:

project/
    gulpfile.js
    basic.js
    Project/
        Project.js
        Toolbelt.js
        Colors/
            RGBA.js
            HSLA.js

My gulpfile should create these minified files:

project/
    basic.js
    basic.min.js // ADDED BY GULP
    Project/
        Project.js
        Project.min.js // ADDED BY GULP
        Toolbelt.js
        Toolbelt.min.js // ADDED BY GULP
        Colors/
            RGBA.js
            RGBA.min.js // ADDED BY GULP
            HSLA.js
            HSLA.min.js // ADDED BY GULP

These seems like it should be relatively straightforward, but I seem to have missed it. I'm not sure what I'm doing wrong, but it only seems to work on basic.js ( basic.min.js is created successfully), but it's not working for any files within folders. I'm not sure if **.js isn't working as I'm expecting, or gulp.dest('') isn't pointing to the current file's directory.. I've done some experimenting off this, and I'm stumped.

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

gulp.task('js-minify', function(){
    return gulp.src(['**.js', '!gulpfile.js', '!**.min.js'])
        .pipe(uglify())
        .pipe(rename(function(path){
            path.extname = '.min.js';
         }))
        .pipe(gulp.dest(''));
});

Running gulp js-minify from the command line does not yield the results I'm looking for.

What am I doing wrong?

Answer 1

您的问题是您的gulp.src路径，您没有在目录中查找：

['**/*.js', '!gulpfile.js', '!**/*.min.js']