摘要: 1.开启进程 using System.Diagnostics; ProcessStartInfo process = new ProcessStartInfo(); process.FileName = "要开启的进程路径"; string arg1 = "进程参数1"; string arg2 阅读全文
posted @ 2019-06-21 11:24 Bridgebug 阅读(1626) 评论(0) 推荐(0)