通常在 Gulp 中的任务是这样的:
Normally in Gulp tasks look like this:
gulp.task('my-task', function() {
return gulp.src(options.SCSS_SOURCE)
.pipe(sass({style:'nested'}))
.pipe(autoprefixer('last 10 version'))
.pipe(concat('style.css'))
.pipe(gulp.dest(options.SCSS_DEST));
});
是否可以将命令行标志传递给 gulp(这不是任务)并让它有条件地运行任务?比如
Is it possible to pass a command line flag to gulp (that's not a task) and have it run tasks conditionally based on that? For instance
$ gulp my-task -a 1
然后在我的 gulpfile.js 中:
And then in my gulpfile.js:
gulp.task('my-task', function() {
if (a == 1) {
var source = options.SCSS_SOURCE;
} else {
var source = options.OTHER_SOURCE;
}
return gulp.src(source)
.pipe(sass({style:'nested'}))
.pipe(autoprefixer('last 10 version'))
.pipe(concat('style.css'))
.pipe(gulp.dest(options.SCSS_DEST));
});
Gulp 没有为此提供任何类型的实用程序,但您可以使用许多命令 args 解析器之一.我喜欢 yargs.应该是:
Gulp doesn't offer any kind of util for that, but you can use one of the many command args parsers. I like yargs. Should be:
var argv = require('yargs').argv;
gulp.task('my-task', function() {
return gulp.src(argv.a == 1 ? options.SCSS_SOURCE : options.OTHER_SOURCE)
.pipe(sass({style:'nested'}))
.pipe(autoprefixer('last 10 version'))
.pipe(concat('style.css'))
.pipe(gulp.dest(options.SCSS_DEST));
});
您也可以将它与 gulp-if 结合起来,以有条件地管道流,对于开发与产品构建非常有用:
You can also combine it with gulp-if to conditionally pipe the stream, very useful for dev vs. prod building:
var argv = require('yargs').argv,
gulpif = require('gulp-if'),
rename = require('gulp-rename'),
uglify = require('gulp-uglify');
gulp.task('my-js-task', function() {
gulp.src('src/**/*.js')
.pipe(concat('out.js'))
.pipe(gulpif(argv.production, uglify()))
.pipe(gulpif(argv.production, rename({suffix: '.min'})))
.pipe(gulp.dest('dist/'));
});
并使用 gulp my-js-task 或 gulp my-js-task --production 调用.
And call with gulp my-js-task or gulp my-js-task --production.
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