在npm run-script中使用node-sass watch选项 [英] Using node-sass watch option with npm run-script

问题描述

所以我正在npm程序包脚本中运行任务，但是我想通过npm start中的watch选项.

So I'm running tasks in npm package scripts but I want to pass the watch option in npm start.

这有效:

"scripts": {
  "scss": "node-sass src/style.scss dist/style.css -w"
}

这不会编译，监视或引发任何错误:

This doesn't compile, watch, or throw any error:

"scripts": {
  "scss": "node-sass src/style.scss dist/style.css",
  "start": "parallelshell \"npm run scss -- -w\""
}

没有并行shell或没有简写都无法工作.

Doesn't work without parallelshell either or without shorthand.

我认为问题是运行脚本在引号中传递了额外的参数，因此命令如下所示:

I assume the problem is the run-script is passing the extra argument in quotes, so the command comes out like:

node-sass src/style.scss dist/style.css "-w"

我希望此方法无需添加任何依赖即可工作.我想念什么?

I'd like this to work without adding any dependencies. What am I missing?

顺便说一句，我在Windows 10中，使用命令提示符/git bash.

Btw, I'm in Windows 10 with command prompt/git bash.

推荐答案

这是我用于CSS构建的设置

This is my setup for css building

"scripts": {
  "css": "node-sass src/style.scss -o dist",
  "css:watch": "npm run css && node-sass src/style.scss -wo dist"
},
"devDependencies": {
  "node-sass": "^3.4.2"
}

-o标志设置目录以输出css. 我有一个非监视版本"css"，因为监视版本"css:watch"〜不会在运行后立即构建〜，它只能在更改时运行，所以我叫

The -o flag sets the directory to output the css. I have a non-watching version "css" because the watching version "css:watch" ~doesn't build as soon as it's run~, it only runs on change, so I call

npm run css 

致电

node-sass src/style.scss -wo dist

如果您只希望它在更改时运行，而不是在首次运行时运行，请使用

If you only want it to run on change, and not when first run, just use

"css:watch": "node-sass src/style.scss -wo dist"

这篇关于在npm run-script中使用node-sass watch选项的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！