Node.js/Gulp - 循环执行 Gulp 任务

2022-01-12 00:00:00 node.js stream build javascript gulp

我想遍历一个对象，并在每次迭代时将一组文件路径传递给 gulp.src，然后对这些文件进行一些处理.下面的代码仅用于说明目的，实际上并不能正常工作，因为 return 语句在第一遍就终止了循环.

I'd like to loop through an object and pass an array of file paths to gulp.src on each iteration and then do some processing on those files. The code below is for illustration purposes and won't actually work since the return statement kills the loop on the first pass.

gulp.task('js', function(){
    for (var key in buildConfig.bundle) {
        return gulp.src(bundleConfig.bundle[key].scripts)
            .pipe(concat(key + '.js'));
            // DO STUFF
    }
});

这是基本的想法.关于如何做到这一点的任何想法?

That's the basic idea. Any ideas on how to do this?

推荐答案

我能够使用合并流解决这个问题.如果有人有兴趣，这里是代码.这个想法是在循环中创建一个流数组，并在完成迭代时合并它们:

I was able to pull this off using merge-streams. If anyone's interested, here's the code. The idea is to create an array of streams inside your loop and merge them when finished iterating:

var merge = require('merge-stream');

gulp.task('js', function(){

    // Init vars
    var jsBundleStreams = [];
    var i = 0;

    // Create array of individual bundle streams
    for (var key in buildConfig.bundle) {
        jsBundleStreams[i] = gulp.src(bundleConfig.bundle[key].scripts)
            .pipe(concat(key + '.js'))
            .pipe(gulp.dest('./public/papasteftest/'));
        i++;
    }

    // Merge and return streams
    return merge.apply(this, jsBundleStreams);

});

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