javascript - How to return all the streams at once from a gulp task with multiple source files? -

the gulp task wrote breaks runsequence in main gulp task. has run in between other tasks, this:

runsequence( ... ':build:prod:staticindex', ':build:prod:typedoctomd', // << task ':build:prod:staticdocumentation', ...etc 

and task after never gets executed. makes me think task doesn't return single stream or incorrectly.

here how task looks:

gulp.task(':build:prod:typedoctomd', (done: any) => { let tasknum: number = 0;   let streams = [];   let filespath = [     path.join(__config.tmpl_docs_path, '**/*.md')   ];   glob(filespath.tostring(), function (er, files) {     tasknum = files.length - 1;     files.foreach((filename) => {       let stream = gulp.src(filename)         .pipe(addtypedoc()) // << custom plugin wrote         .pipe(gulp.dest(function (file) {           return file.base;         }));       stream.on('end', () => {         streams.push(stream);         if (streams.length === tasknum) {           gutil.log('the end!');           return merge(streams);         }       });     });   }); }); 

so used glob recursively md files folder. , used merge-stream combine streams , return single one.

i 'the end!' message logged gulp never gets next task after that.

am missing here?

thanks in advance help!

ok, found solution this, , it's ridiculously simple: use task's callback argument this:

if (streams.length === tasknum) {   gutil.log('the end!');   done(); } 

now works! , didn't have use merge-stream after all.