通常我们可以通过类似 gulp mytask 的方式从控制台运行 gulp 任务.无论如何,我可以将参数传递给 gulp 任务吗?如果可能,请举例说明如何做到这一点.
Normally we can run gulp task from console via something like gulp mytask. Is there anyway that I can pass in parameter to gulp task? If possible, please show example how it can be done.
推荐答案这是一个程序不能没有的功能.你可以试试yargs.
It's a feature programs cannot stay without. You can try yargs.
npm install --save-dev yargs你可以这样使用它:
gulp mytask --production --test 1234在代码中,例如:
var argv = require('yargs').argv; var isProduction = (argv.production === undefined) ? false : true;为了您的理解:
> 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希望你能从这里拿走它.
Hope you can take it from here.
你可以使用另一个插件,minimist.还有另一篇文章,其中有 yargs 和 minimist 的好例子:(是否可以将标志传递给 Gulp 让它以不同的方式运行任务?)
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?)
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将参数传递给 Gulp 任务
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