CODEWITHSUNDEEP
javascriptwindowsgulpwatch
Victor
Victor

Reputation: 73

gulp.watch() not running associated task

I have the following code in my gulpfile.js:

gulp.task('test-watch', function () {
    console.log('running test-watch...');

    return gulp.src('./views/layout.tmpl')
        .pipe(rename('layout.html'))
        .pipe(gulp.dest('./views/'));
});

gulp.task('watch', function () {
    var watcher = gulp.watch('./views/layout.tmpl', ['test-watch']);
    watcher.on('change', function (event) {
        console.log('File ' + event.path + ' was ' + event.type + ', running tasks...');
    });
});

It results in the following output (the last line occurs after I edit/save "layout.tmpl"):

[15:53:12] Using gulpfile C:\project1\gulpfile.js
[15:53:12] Starting 'watch'...
[15:53:12] Finished 'watch' after 6.72 ms
File C:\project1\views\layout.tmpl was changed, running tasks...

The file "layout.html" (which is supposed to be a copy of "layout.tmpl") does not get created. Note that in the above output while the "console.log" within the "watcher.on" invocation is shown in output, the "console.log" within the "test-watch" task is not shown. If I run the test-watch task directly, the file does get created.

Here is the output if I run the test-watch task directly:

[15:53:56] Using gulpfile C:\project1\gulpfile.js
[15:53:56] Starting 'test-watch'...
running test-watch...
[15:53:56] Finished 'test-watch' after 12 ms

Note that the "console.log" within the "test-watch" task is shown.

This is on a Windows 7 machine, running gulp tasks from cmd.exe.

Upvotes: 1

Views: 198

Answers (1)

Victor
Victor

Reputation: 73

Support for passing in watch tasks to the gulp.watch API as an array requires gulp version 3.5 at minimum. The gulp.watch API accepts an array of tasks only in gulp version 3.5 and up.

My local gulp was version 3.4.

I updated gulp to latest 3.8 version and my watch task worked correctly.

Upvotes: 1

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