How do I prepend a stream to another?

Hi!

I’m still getting used to gulp’s processing model and it’s not entirely clear to me how to achieve something like this:

src1 | filter concat src2 | filter > output

The recipe here results in an unordered stream, but I’d like concat to be really concat and not just merge chucks arbitrarily.

Thanks!

[darsain]

Shit. I forgot an option flag 😃 You need to specify { objectMode: true }:

var gulp = require('gulp');
var concat = require('gulp-concat');
var streamqueue = require('streamqueue');

gulp.task('default', function () {
    return streamqueue({ objectMode: true },
        gulp.src('foo/*'),
        gulp.src('bar/*')
    )
        .pipe(concat('result.txt'))
        .pipe(gulp.dest('build'));
});

// ... or ...

gulp.task('default', function () {
    var stream = streamqueue({ objectMode: true });
    stream.queue(gulp.src('foo/*'));
    stream.queue(gulp.src('bar/*'));
    return stream.done()
        .pipe(concat('result.txt'))
        .pipe(gulp.dest('build'));
});

[chrismarx]

+1 for objectMode, that’s critical!