如何使用 gulp 在一项任务中运行 uglify 然后剥离调试

Question

我想要的是缩小 index.html 中的所有 js，然后删除所有 console.logs

我尝试了两种选择：

我尝试了合并，但只执行了 uglify

// Command: gulp useref
gulp.task('useref', function(){
    var _uglify = gulp.src('app/index.html') // .src is the function that is very similar to locating or searching on that file or folder
    .pipe(useref())
    // Minifies only if it's a Javascript file
    .pipe(gulpIf('*.js', uglify()))
    // Minifies only if it's a CSS file
    .pipe(gulpIf('*.css', cssnano()))
    .pipe(gulp.dest('app/')) // .dest is the location where it will produce the output
    // set to app/, so it will automatically change the index and there's no need to move files 

    var _strip_debug = gulp.src('app/assets/js/scripts.js')
    .pipe(stripDebug())
    .pipe(gulp.dest('app/assets/js'));

    return merge(_uglify, _strip_debug);
});

我尝试返回两个，但只执行了 uglify：

    gulp.task('useref', function(){
        return gulp.src('app/index.html') // .src is the function that is very similar to locating or searching on that file or folder
        .pipe(useref())
        // Minifies only if it's a Javascript file
        .pipe(gulpIf('*.js', uglify()))
        // Minifies only if it's a CSS file
        .pipe(gulpIf('*.css', cssnano()))
        .pipe(gulp.dest('app/')) // .dest is the location where it will produce the output
        // set to app/, so it will automatically change the index and there's no need to move files 

        return gulp.src('app/assets/js/scripts.js')
        .pipe(stripDebug())
        .pipe(gulp.dest('app/assets/js'));
    });

Answer 1

我假设app/assets/js/scripts.js 是由gulp-useref 生成的串联JavaScript 文件。

在这种情况下，使用merge-stream 将不起作用，因为当您尝试gulp.src() 时，app/assets/js/scripts.js 文件可能还不存在。相反，只需在您的第一个流中添加另一个 gulpIf 阶段：

gulp.task('useref', function(){
   return gulp.src('app/index.html')
     .pipe(useref())
     .pipe(gulpIf('*.js', stripDebug()))
     .pipe(gulpIf('*.js', uglify()))
     .pipe(gulpIf('*.css', cssnano()))
     .pipe(gulp.dest('app/'))
});