run-sequence doesn't run gulp tasks in order

你说的曾经没有我的故事 提交于 2019-12-05 14:44:10

The run-sequence documentation has the following to say about tasks with asynchronous operations:

make sure they either return a stream or promise, or handle the callback

Both your copy and replace tasks have more than one stream. You have to return all the streams, not just the last one. Gulp won't know anything about the other streams if you don't return them and therefore won't wait for them to finish.

Since you can only ever return a single stream you have to merge the streams [insert Ghostbusters reference here]. That will give you one merged stream that you can return from your task.

Here's how to do it using the merge-stream package :

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

gulp.task('copy', function() {
    var stream1 = gulp.src('node_modules/bootstrap/dist/**/*')
        .pipe(gulp.dest('dist/vendor'))
    //...
    var stream2 = gulp.src(['index.html', '404.html', '.htaccess'])
        .pipe(gulp.dest('dist/'));

    return merge(stream1, stream2);
});

gulp.task('replace', function(){
    var stream1 = gulp.src(['dist/index.php', 'dist/info.php'])
        .pipe(replace('fakedomain.com', 'realdomain.com'))
        .pipe(gulp.dest('dist'));

    var stream2 = gulp.src(['dist/config.php'])
        .pipe(replace('foo', 'bar'))
        .pipe(gulp.dest('dist'));

    return merge(stream1, stream2);
});
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