Pass Parameter to Gulp Task
Solution 1:
It's a feature programs cannot stay without. You can try yargs.
npm install --save-dev yargs
You can use it like this:
gulp mytask --production --test 1234
In the code, for example:
var argv = require('yargs').argv;
var isProduction = (argv.production === undefined) ? false : true;
For your understanding:
> gulp watch
console.log(argv.production === undefined); <-- true
console.log(argv.test === undefined); <-- true
> gulp watch --production
console.log(argv.production === undefined); <-- false
console.log(argv.production); <-- true
console.log(argv.test === undefined); <-- true
console.log(argv.test); <-- undefined
> gulp watch --production --test 1234
console.log(argv.production === undefined); <-- false
console.log(argv.production); <-- true
console.log(argv.test === undefined); <-- false
console.log(argv.test); <-- 1234
Hope you can take it from here.
There's another plugin that you can use, minimist. There's another post where there's good examples for both yargs and minimist: (Is it possible to pass a flag to Gulp to have it run tasks in different ways?)
Solution 2:
If you want to avoid adding extra dependencies, I found node's process.argv
to be useful:
gulp.task('mytask', function() {
console.log(process.argv);
});
So the following:
gulp mytask --option 123
should display:
[ 'node', 'path/to/gulp.js', 'mytask', '--option', '123']
If you are sure that the desired parameter is in the right position, then the flags aren't needed.** Just use (in this case):
var option = process.argv[4]; //set to '123'
BUT: as the option may not be set, or may be in a different position, I feel that a better idea would be something like:
var option, i = process.argv.indexOf("--option");
if(i>-1) {
option = process.argv[i+1];
}
That way, you can handle variations in multiple options, like:
//task should still find 'option' variable in all cases
gulp mytask --newoption somestuff --option 123
gulp mytask --option 123 --newoption somestuff
gulp mytask --flag --option 123
** Edit: true for node scripts, but gulp interprets anything without a leading "--" as another task name. So using gulp mytask 123
will fail because gulp can't find a task called '123'.
Solution 3:
There's an official gulp recipe for this using minimist.
https://github.com/gulpjs/gulp/blob/master/docs/recipes/pass-arguments-from-cli.md
The basics are using minimist to separate the cli arguments and combine them with known options:
var options = minimist(process.argv.slice(2), knownOptions);
Which would parse something like
$ gulp scripts --env development
More complete info in the recipe.