摘要: function getResult(status) { if (status 0) { return 'offline' } else if (status 1) { return 'online'; } else if (status 2) { return 'deleted' } return 阅读全文
posted @ 2020-06-18 06:31 wzndkj 阅读(153) 评论(0) 推荐(0) 编辑