CODEWITHSUNDEEP
variablesgulpglob
Sekhemty
Sekhemty

Reputation: 1532

How to correctly get gulp.src globs from a variable?

Given the following Gulp setup

const { src, dest } = require('gulp');
var zip = require('gulp-zip');

var pkgDist = 'packages/';

function pkg(done) {
    src(['./**', '!node_modules/**', '!vendor/**', '!.gitignore', '!*.json', '!*.lock'], {base: '..'})
        .pipe(zip('archive.zip'))
        .pipe(dest(pkgDist))
    done();
};

exports.pkg = pkg;

how can I modify it in order to get src globs from a variable, i.e. pkgSrc, something like this:

[...]

var pkgSrc = <what to put here?>;

[...]

    src(pkgSrc)

[...]

I've tried to use this var pkgSrc = " ['./**', '!node_modules/**', '!vendor/**', '!.gitignore', '!*.json', '!*.lock'], {base: '..'} "; but it doesn't work.

If it's easier, I'm also open to solutions that result into this src([pkgSrc], {base: '..'})

Upvotes: 0

Views: 173

Answers (1)

Mark
Mark

Reputation: 180885

You can go with just:

var pkgSrc = ['./**', '!node_modules/**', '!vendor/**', '!.gitignore', '!*.json', '!*.lock']

gulp.src first argument can be a string or an array, so now it is an array above.

The second argument is an object of options. Include the options sepearately: {base: '..'} so

src(pkgSrc, {base: '..'})

Upvotes: 2

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