Gulp缩小并且不复制原始文件

Question

I want to minify some JS files with Gulp, but can't seem to get control over the process. I want only the minified version in the destination, and am currently getting copies of the originals as well.

I'm thinking I may need the rename package, but am not sure how to use it for this task - I would presumably need some kind of variable to hold the current file name for each script.

Any help much appreciated. The code is below:

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

//script paths
var jsFiles = 'js/**/*.js',
    jsDest = 'js/dist/';

gulp.task('scripts', function() {
    return gulp.src(jsFiles)
        .pipe(minify())
        .pipe(gulp.dest(jsDest));
});

Answer 1

像下面这样设置noSource配置

  .pipe(minify({noSource: true})