Javascript 从单个流生成多个文件。吞咽

我对使用了

，但这是不对的

gulp.task('themes', function() {
    for (var color in config.themes) {
        for (var shine in config.shines) {
            gulp.src(['src/scss/_config.scss'])
                .pipe(rename('_' + config.themes[color] + '-' + config.shines[shine] + '.scss'))
                .pipe(gulp.dest('src/scss/themes'));
        }
    }
});

如何处理streams？
只需使用多个pipe gulp.dest（）函数：

gulp.task('themes', function() {
   for (var color in config.themes) {
        for (var shine in config.shines) {
            gulp.src(['src/scss/_config.scss'])
                .pipe(rename('_' + config.themes[color] + '-' + config.shines[shine] + '.scss'))
                .pipe(gulp.dest('src/scss/themes'));
                .pipe(gulp.dest('src/scss/themes2'));
        }
    }
});'src/scss/themes'));