将 gulp.start 函数迁移到 Gulp v4
我一直在将我所有的 gulp v3 代码库迁移到 v4.但是我被困在我有 gulp start 功能的地方，当我在 gulp v4 中运行 gulp start 时它会抛出一个错误.
I have been migrating all of my gulp v3 code bases into v4. However I'm stuck at a point where I have gulp start function and it throws me an error when I run gulp start in gulp v4.
这是我在版本 3 中的功能:
This is the function I had in version 3:
gulp.task('default', ['watch'], function () {
gulp.start('styles', 'scripts', 'images');
});当迁移到 gulp v4 时，我实现了这个功能:
When migrating to v4 of gulp I implemented this function:
gulp.task('default', gulp.parallel('watch', function(done) {
gulp.start('styles', 'scripts', 'images');
done();
}));如何使用新的 gulp 版本完成相同的过程?我需要在 gulp.series 中使用 gulp.parallel 吗?
How to accomplish the same process with the new gulp version? Do I need to use gulp.parallel inside gulp.series?
推荐答案
把你的任务改成函数后，简单使用；
After changing your tasks to functions, simply use;
gulp.task('default', gulp.series (watch, gulp.parallel(styles, scripts, images),
function (done) { done(); }
));
watch 函数将首先运行，然后是样式、脚本和图像函数并行运行.
The watch function will run first, then the styles, scripts and images functions in parallel.
来自 gulp.series 文档:
组合的嵌套深度没有强加限制使用 series() 和 parallel() 进行操作.
There are no imposed limits on the nesting depth of composed operations using series() and parallel().
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