我想从我的angularjs项目中获得一些特定的js。 我不想要uglify al js path。
我试过这个
src : 'www/js/app.js', 'www/js/controllers.js', 'www/js/directives.js', 'www/js/factories.js', 'www/js/filters.js', 'www/js/services.js',
和这个
src : 'www/js/app.js, www/js/controllers.js, www/js/directives.js, www/js/factories.js, www/js/filters.js, www/js/services.js',
而不是运气。
module.exports = function(grunt){
grunt.loadNpmTasks('grunt-contrib-clean');
grunt.loadNpmTasks("grunt-contrib-uglify");
grunt.initConfig({
clean : [ 'www/dist/*' ],
uglify : {
options : {
report : 'min',
mangle : true
},
my_target : {
files : [ {
src : 'www/js/app.js, www/js/controllers.js, www/js/directives.js, www/js/factories.js, www/js/filters.js, www/js/services.js',
dest : 'www/dist/app.realease.min.js'
} ]
}
}
})
grunt.registerTask('default', ['clean', 'uglify']);
}
如果有人可以帮助我,我会过分帮助你。
感谢。
答案 0 :(得分:1)
您的问题是您提供了一个简单的字符串列表文件,您应该为每个要定位的文件提供一个字符串数组。
src : 'www/js/app.js', 'www/js/controllers.js', 'www/js/directives.js', 'www/js/factories.js', 'www/js/filters.js', 'www/js/services.js'
应该成为
src : ['www/js/app.js', 'www/js/controllers.js', 'www/js/directives.js', 'www/js/factories.js', 'www/js/filters.js', 'www/js/services.js']
但对我来说,你应该将你的文件汇总到一个,然后像这样对它进行uglify:
module.exports = function(grunt) {
grunt.initConfig({
clean : [ 'www/dist/*' ],
concat: {
options: {
separator: ';'
},
dist: {
src: ['www/js/app.js', 'www/js/controllers.js', 'www/js/directives.js', 'www/js/factories.js', 'www/js/filters.js', 'www/js/services.js'],
dest: 'www/dist/app.release.js'
}
},
uglify: {
options: {
mangle: false
},
dist: {
files: {
'dist/app.release.min.js': ['<%= concat.dist.dest %>']
}
}
}
});
grunt.loadNpmTasks('grunt-contrib-clean');
grunt.loadNpmTasks('grunt-contrib-uglify');
grunt.loadNpmTasks('grunt-contrib-concat');
grunt.registerTask('default', ['clean', 'concat', 'uglify']);
};