What is the process for generating a blank stream in Gulp?

Question

What is the process for generating a blank stream in Gulp?

Snippet:

gulp.task('foo', [], function() {
    var barFiles = getBarFiles();
    var bazFiles = getBazFiles();

    var barStream, bazStream;
    if (barFiles && barFiles.length > 0) {
        barStream = gulp.src(barFiles);
    }
    if (bazStream && bazStream.length > 0) {
        bazStream = gulp.src(bazStream);
    }
    return eventStream
        .merge(barStream, bazStream)
        .pipe(g.concat('bar-and-baz'))
        .pipe(gulp.dest('./foo-output/'));
});

getBarFiles() or getBazFiles() may potentially return an empty array, which won't work with gulp.src(): resulting in Error: Invalid glob argument, hence the need to create a stream only when files are present.

The main query is, how can an empty stream be constructed, to be combined with another empty or non-empty stream?

javascript stream gulp

Answer 1

Answer №1

Have you considered using this syntax?

return gulp.src('.');

If you are using a newer version of Gulp, you may need to include the allowEmpty: true option like this:

return gulp.src('.', {allowEmpty: true});

I believe this code simply passes the current directory into the stream without performing any actions on it. It has worked well for me.

Answer 2

Have you considered using this syntax?

return gulp.src('.');

If you are using a newer version of Gulp, you may need to include the allowEmpty: true option like this:

return gulp.src('.', {allowEmpty: true});

I believe this code simply passes the current directory into the stream without performing any actions on it. It has worked well for me.

Answer 3

Answer №2

One way to create an empty stream for gulp is by using a function found in the source code of vinyl-fs. Here is the function:

var through2 = require('through2');

function createEmptyStream() {
    var pass = through2.obj();
    process.nextTick(pass.end.bind(pass));
    return pass;
}

If you assign barFiles to createEmptyStream() in your code, it will have the same effect.

Testing done with the latest versions of gulp (3.9.1) and vinyl-fs (0.3.0) on 2016-03-05 shows that an empty src array is allowed. A similar method to the one mentioned above is used to create an empty stream. Your code should work with the current versions:

var eventStream = require('event-stream');
var gulp = require('gulp');
var g = require('gulp-load-plugins')();

function getBarFiles() {
    return [
    ];
}

function getBazFiles() {
    return [
        'baz.txt'
    ];
}

gulp.task('foo', [], function() {
    var barFiles = getBarFiles();
    var bazFiles = getBazFiles();
    var barStream = gulp.src(barFiles);
    var bazStream = gulp.src(bazFiles);

    return eventStream
        .merge(barStream, bazStream)
        .pipe(g.concat('bar-and-baz.txt'))
        .pipe(gulp.dest('./'));
});

gulp.task('default', [ 'foo' ]);

The resulting file, bar-and-baz.txt, will contain the concatenated contents of all files returned by the functions. An empty list of files is also accepted.

Answer 4

One way to create an empty stream for gulp is by using a function found in the source code of vinyl-fs. Here is the function:

var through2 = require('through2');

function createEmptyStream() {
    var pass = through2.obj();
    process.nextTick(pass.end.bind(pass));
    return pass;
}

If you assign barFiles to createEmptyStream() in your code, it will have the same effect.

Testing done with the latest versions of gulp (3.9.1) and vinyl-fs (0.3.0) on 2016-03-05 shows that an empty src array is allowed. A similar method to the one mentioned above is used to create an empty stream. Your code should work with the current versions:

var eventStream = require('event-stream');
var gulp = require('gulp');
var g = require('gulp-load-plugins')();

function getBarFiles() {
    return [
    ];
}

function getBazFiles() {
    return [
        'baz.txt'
    ];
}

gulp.task('foo', [], function() {
    var barFiles = getBarFiles();
    var bazFiles = getBazFiles();
    var barStream = gulp.src(barFiles);
    var bazStream = gulp.src(bazFiles);

    return eventStream
        .merge(barStream, bazStream)
        .pipe(g.concat('bar-and-baz.txt'))
        .pipe(gulp.dest('./'));
});

gulp.task('default', [ 'foo' ]);

The resulting file, bar-and-baz.txt, will contain the concatenated contents of all files returned by the functions. An empty list of files is also accepted.

Answer 5

Answer №3

Upon encountering the same query as the original poster, I found that a simple solution was available without the need for any extra modules:

if (platform != 'android') {
    // No action required for Android
    return gulp.src([]); // empty stream
}

To clarify, gulp.src() did not yield results, but gulp.src([]) proved effective. Am I overlooking something?

UPDATE 18 Mar 2020: It appears that Gulp 4.0 now mandates the use of the allowEmpty parameter for this functionality:

if (platform != 'android') {
    // No action required for Android
    return gulp.src([], {allowEmpty: true}); // empty stream
}

Refer to Artur Stępień's response to this inquiry, or check out this related question on SO.

Answer 6

Upon encountering the same query as the original poster, I found that a simple solution was available without the need for any extra modules:

if (platform != 'android') {
    // No action required for Android
    return gulp.src([]); // empty stream
}

To clarify, gulp.src() did not yield results, but gulp.src([]) proved effective. Am I overlooking something?

UPDATE 18 Mar 2020: It appears that Gulp 4.0 now mandates the use of the allowEmpty parameter for this functionality:

if (platform != 'android') {
    // No action required for Android
    return gulp.src([], {allowEmpty: true}); // empty stream
}

Refer to Artur Stępień's response to this inquiry, or check out this related question on SO.

Answer 7

Answer №4

One effective solution I found to meet my requirements was using gulp-tap:

var config = require('./config');
var gulp = require('gulp');
var tap = require('gulp-tap');
var concat = require('gulp-concat');
var uglify = require('gulp-uglify');

var noop = function() {};

gulp.task('mytask', function() {
  return gulp.src('*.js')
    .pipe(config.production ? uglify() : tap(noop))
    .pipe(concat('code.min.js'))
    .pipe(gulp.dest('dist'));
});

With this method, you can easily create an empty stream to pass inside a .pipe() without the need for an external if statement.

Note: It's important to avoid storing the tap(noop) into a variable and using it across multiple tasks, as this can mix your streams and lead to unpredictable behavior.

Answer 8

One effective solution I found to meet my requirements was using gulp-tap:

var config = require('./config');
var gulp = require('gulp');
var tap = require('gulp-tap');
var concat = require('gulp-concat');
var uglify = require('gulp-uglify');

var noop = function() {};

gulp.task('mytask', function() {
  return gulp.src('*.js')
    .pipe(config.production ? uglify() : tap(noop))
    .pipe(concat('code.min.js'))
    .pipe(gulp.dest('dist'));
});

With this method, you can easily create an empty stream to pass inside a .pipe() without the need for an external if statement.

Note: It's important to avoid storing the tap(noop) into a variable and using it across multiple tasks, as this can mix your streams and lead to unpredictable behavior.

Answer 9

Answer №5

This is my preferred method:

https://github.com/mikeroberts3000/data-flow

df.combine([]);

It does the job effectively!

Answer 10

This is my preferred method:

https://github.com/mikeroberts3000/data-flow

df.combine([]);

It does the job effectively!

Answer 11

Answer №6

Another approach:

gulp.task('bar', [], function() {
    var fooFiles = getFooFiles();
    var bazFiles = getBazFiles();

    var fooStream, bazStream;
    if (fooFiles && fooFiles.length > 0) {
        fooStream = gulp.src(fooFiles);
    }
    if (bazStream && bazStream.length > 0) {
        bazStream = gulp.src(bazStream);
    }
    var mergedStream;
    if (fooStream && bazStream) {
        mergedStream = eventStream
            .merge(fooStream, bazStream);
    }
    else if (fooStream || bazStream) {
        mergedStream = fooStream || bazStream;
    }
    if (mergedStream) {
        return mergedStream
            .pipe(g.concat('foo-and-baz'))
            .pipe(gulp.dest('./bar-output/'));
    }
});

By checking for the presence of each stream before merging, we can avoid creating an empty stream unnecessarily.

Still, it would be beneficial to discover a method to generate an empty stream in gulp.

Answer 12

Another approach:

gulp.task('bar', [], function() {
    var fooFiles = getFooFiles();
    var bazFiles = getBazFiles();

    var fooStream, bazStream;
    if (fooFiles && fooFiles.length > 0) {
        fooStream = gulp.src(fooFiles);
    }
    if (bazStream && bazStream.length > 0) {
        bazStream = gulp.src(bazStream);
    }
    var mergedStream;
    if (fooStream && bazStream) {
        mergedStream = eventStream
            .merge(fooStream, bazStream);
    }
    else if (fooStream || bazStream) {
        mergedStream = fooStream || bazStream;
    }
    if (mergedStream) {
        return mergedStream
            .pipe(g.concat('foo-and-baz'))
            .pipe(gulp.dest('./bar-output/'));
    }
});

By checking for the presence of each stream before merging, we can avoid creating an empty stream unnecessarily.

Still, it would be beneficial to discover a method to generate an empty stream in gulp.

What is the process for generating a blank stream in Gulp?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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