笨拙地创建和运行较少的编译任务

Question

i am very new to grunt and i wanted to achieve following goal

• iterate every directory in pointed location
• compile less file to css

source:

• theme 1
  • file.less
• theme 2
  • file.less
• theme 3
  • file.less

after processing:

• theme 1
  • file.less
  • out.css
• theme 2
  • file.less
  • out.css
• theme 3
  • file.less
  • out.css

I read somewhere that grunt does that, this is my entire run.js file:

var grunt = require('grunt'),
    less = require('less');

  grunt.initConfig({
    less: {
      compileCore: {
        options: {
          strictMath: true
        },
        files: [{
          expand: true,
          src: ['themes/**/file.less'], 
          ext: '.css',
          extDot: 'first'
        }]
      }
  }
  });

  grunt.registerTask('default', ['less']);

  grunt.task.run('default');

it doesnt give any errors but it doesnt work either, how can i make this to work? (again, i have no exprience with grunt whatsoever, yet)

if its possible also to achieve this goal with just command line execution (without runing node.js file) that would be good too ;)

Answer 1

you can use grunt.file.expandMapping ( https://gruntjs.com/api/grunt.file#grunt.file.expandmapping ) to iterate through all the files in a directory and provide options to according to the requirements.

The below code will work according to your requirement.

 module.exports = function (grunt) { grunt.initConfig({ less: { development: { files: grunt.file.expandMapping(['themes/**/*.less'], '', { rename: function (destBase, destPath) { return destBase + destPath.replace('.less', '.css'); } }) } } }); grunt.loadNpmTasks('grunt-contrib-less'); grunt.registerTask('default', ['less']); }