CODEWITHSUNDEEP
javascriptgulp
Max Grishaev
Max Grishaev

Reputation: 25

Gulp exits too early

There's simple gulpfile but it doesn't executes as expected:

var gulp = require('gulp');
var rename = require('gulp-rename');
var mapStream = require('map-stream');
var fs = require('fs');

gulp.task('default', function () {
    gulp.src('123.js')
        .pipe(log('before'))
        .pipe(rename({prefix: '_'}))
        .pipe(gulp.dest('.'))
        .pipe(log('after'))
    ;
});

function log (txt) {
    return mapStream(function (file, cb) {
        console.log(txt, fs.statSync(file.path).size, !!file.contents.toString());
        cb();
    });
}

logs only "before", but no "after". why?

Upvotes: 0

Views: 998

Answers (2)

Shuhei Kagawa
Shuhei Kagawa

Reputation: 4777

In addition to returning the stream as Sindre suggested, you need to pass 'file' to the callback function of 'map-stream' in order to pass the data through the stream.

function log (txt) {
    return mapStream(function (file, cb) {
        console.log(txt, fs.statSync(file.path).size, !!file.contents.toString());
        cb(null, file);
    });
}

Upvotes: 1

Sindre Sorhus
Sindre Sorhus

Reputation: 63487

You need to return the stream:

gulp.task('default', function () {
    return gulp.src('123.js')
        .pipe(log('before'))
        .pipe(rename({prefix: '_'}))
        .pipe(gulp.dest('.'))
        .pipe(log('after'));
});

Upvotes: 1

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