输出 css 文件只有在它首先为空时才会编译。必须删除 css 内容，保存 css，然后在 scss 上触发监视

Question

我有以下 package.json 但是在我的环境中 (windows 10/vs code/vagrant homestead php) compile:development 只有当我清空我的首先输出 build.css，然后点击保存并触发手表。否则 css 文件永远不会改变。这显然适用于其他机器（可能是转移注意力；不相信任何人）。

"scripts": {
    "autoprefixer": "postcss -u autoprefixer --no-map -r assets/css/*.css",
    "compile:production": "node-sass --omitSourceMapUrl true --output-style compressed -o assets/css assets/scss",
    "compile:development": "node-sass --omitSourceMapUrl true --output-style expanded -o assets/css assets/scss",
    "watch:css": "onchange \"assets/scss\" -- run-s development:css",
    "watch:js": "run-s development:js",
    "development:css": "run-p compile:development autoprefixer",
    "development:js": "rollup -c -w",
    "production:css": "run-p compile:production autoprefixer",
    "production:js": "rollup -c --compact",
    "build": "run-p production:*",
    "serve": "run-p watch:*"
},
"devDependencies": {
    "@babel/cli": "^7.12.8",
    "@babel/core": "^7.12.9",
    "@babel/preset-env": "^7.12.7",
    "@rollup/plugin-babel": "^5.2.2",
    "@rollup/plugin-commonjs": "^17.0.0",
    "@rollup/plugin-node-resolve": "^11.0.0",
    "autoprefixer": "^6.3.6",
    "core-js": "^3.8.0",
    "fs": "0.0.1-security",
    "node-sass": "^4.11.0",
    "npm-run-all": "^2.1.1",
    "onchange": "^5.2.0",
    "postcss-cli": "^6.1.3",
    "regenerator-runtime": "^0.13.7",
    "rollup": "^2.34.1",
    "rollup-plugin-gzip": "^2.5.0",
    "rollup-plugin-multi-input": "^1.1.1",
    "rollup-plugin-node-polyfills": "^0.2.1",
    "rollup-plugin-uglify": "^6.0.4",
    "uikit": "^3.6.16"
},

Answer 1

重写编译脚本如下，让它工作

    "compile:production": "node-sass --omitSourceMapUrl true --output-style compressed assets/scss -o assets/css",
    "compile:development": "node-sass -w assets/scss -o assets/css",

输出 css 文件只有在它首先为空时才会编译。必须删除 css 内容，保存 css，然后在 scss 上触发监视

Output css file only compiles if it's empty first. Have to delete css contents, save css, then trigger watch on scss

compilation

sass

npm