Gulp/Grunt Question

[Steven Morgan]

Pretty sure it’s the same in both packages.
How do I use variables for file names?

For instance, if I have:
file1.js
file2.js
file3.js

And I want to give that a destination of:
file1.min.js
file2.min.js
file3.min.js

return gulp.src('js/file1.js')
    .pipe(uglify())
    .pipe(concat('file1.min.js'))
    .pipe(gulp.dest('js/build'));

Without having to rewire this 3 different times with the 3 different file names.

[shaneisme]

Just use the * as a wild-card and you’re good.

return gulp.src('js/*.js)

[Steven Morgan]

I thought that would work, but it just takes all of the files and runs them into *.js

[Alen]

https://www.youtube.com/watch?v=6Jhgkp67GxI

[maxisix]

You can use an array like this

gulp.src([‘file1.js’, ‘file2.js’]).pipe(…)

If you want you can take a look on my gulpfile

https://github.com/maxisix/maxisix-gulp-template/blob/master/gulpfile.js

[garystorey]

If you haven’t already check out the gulp-rename plugin.

[Author]

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