我正在尝试设置style.scss文件以输出到style.css中。
我的package.json看起来像这样
{
"name": "soumghosh",
"version": "1.0.0",
"description": "",
"main": "index.js",
"dependencies": {
"grunt-sass": "^1.0.0"
},
"devDependencies": {},
"scripts": {
"test": "echo \"Error: no test specified\" && exit 1"
},
"author": "",
"license": "ISC"
}
我的gruntfile.js看起来像这样
module.exports = function(grunt) {
'use strict';
grunt.initConfig({
echo: {
one: { message: 'one has changed' },
two: { message: 'two has changed' },
wait: { message: 'I waited 2s', wait: 2000 },
interrupt: { message: 'I want to be interrupted', wait: 5000 },
fail: { fail: 1, message: 'This task should fail', wait: 1000 }
},
watch: {
one: {
files: ['lib/one.js', 'Gruntfile.js'],
tasks: 'echo:one',
},
two: {
files: ['lib/two.js'],
tasks: ['echo:two'],
},
wait: {
files: ['lib/wait.js'],
tasks: ['echo:wait'],
},
interrupt: {
files: ['lib/interrupt.js'],
tasks: ['echo:interrupt'],
options: { interrupt: true },
},
fail: {
files: ['lib/fail.js'],
tasks: ['echo:fail'],
},
},
});
// Load the echo task
grunt.loadTasks('../tasks');
// Load this watch task
grunt.loadTasks('../../../tasks');
grunt.registerTask('default', ['echo']);
};
我想我的困惑是如何启动进程以查看特定目录以编译scss文件。我应该在终端上做一些事情来启动脚本吗?我对节点很新,这就是为什么我对此有点困惑。