我的grunt文件如下所示:
grunt.initConfig({
compress: {
foo: {
options: { archive: 'deploy/foo.zip', mode: 'zip' },
files: [{ expand: true, cwd: 'release/foo/', src: ['**'], dot: true }]
},
bar: {
options: { archive: 'deploy/bar.zip', mode: 'zip' },
files: [{ expand: true, cwd: 'release/bar/', src: ['**'], dot: true }]
},
baz: {
options: { archive: 'deploy/baz.zip', mode: 'zip' },
files: [{ expand: true, cwd: 'release/baz/', src: ['**'], dot: true }]
},...
我如何参数化,所以我不需要N个定义,我可以调用(伪代码)
> grunt compress "foo"
答案 0 :(得分:1)
如果我理解你的问题:
var initData = {
compress: { }
};
var compress = function(n){
initData.compress[n] = {
options: {archive: 'deploy/' + n + '.zip', mode: 'zip'},
files: [{expand: true, cwd: 'release/' + n + '/', src: ['**'], dot: true}]
};
}
compress('foo');
compress('bar');
compress('baz');
grunt.initConfig(initData);