将gulp.start函数迁移到Gulp v4 [英] Migrate gulp.start function to Gulp v4

问题描述

我一直在将所有gulp v3代码库迁移到v4.但是我被卡在了gulp start函数的位置，在gulp v4中运行gulp start时会抛出错误.

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中拥有的功能:

This is the function I had in version 3:

gulp.task('default', ['watch'], function () {

    gulp.start('styles', 'scripts', 'images');

});

当迁移到gulp的v4时，我实现了以下功能:

When migrating to v4 of gulp I implemented this function:

gulp.task('default', gulp.parallel('watch', function(done) {

    gulp.start('styles', 'scripts', 'images');

    done();

}));

如何用新的gulp版本完成相同的过程?我需要在gulp.series内使用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

将任务更改为函数后，只需使用即可；

After changing your tasks to functions, simply use;

gulp.task('default', gulp.series (watch, gulp.parallel(styles, scripts, images),

    function (done) { done(); }    
));

首先运行watch函数，然后并行运行样式，脚本和图像函数.

The watch function will run first, then the styles, scripts and images functions in parallel.

从 gulp.series文档:

组合物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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