C# 受信時刻と今日の時差を計算する方法の例を共有します
/// <summary> /// 计算传入的时间距离今天的时间差 /// </summary> /// <param name="dt"></param> /// <param name="yy"></param> /// <param name="mm"></param> /// <param name="dd"></param> public void GetCriminalYX(DateTime dt, out int yy, out int mm, out int dd) { DateTime now = DateTime.Now; yy = mm = dd = 0; if (dt.Year > 9000 || dt.Year == 1900) { return; } if (dt <= now) { return; } StringBuilder str = new StringBuilder(); int dt_Y = dt.Year; int dt_M = dt.Month; int dt_D = dt.Day; int now_Y = DateTime.Now.Year; int now_M = DateTime.Now.Month; int now_D = DateTime.Now.Day; yy = dt_Y - now_Y; mm = dt_M - now_M; dd = 0; int dt_M_SY = 0; if (dt_D < now_D) { mm -= 1; dt_M_SY = dt_M - 1; if (dt_M_SY == 0) { dt_M_SY = 12; } if (dt_M_SY == 2) { dt_M_SY = dt_Y % 4 == 0 ? 29 : 28; } else { dt_M_SY = dt_M_SY == 2 || dt_M_SY == 4 || dt_M_SY == 6 || dt_M_SY == 9 || dt_M_SY == 11 ? 30 : 31; } dt_D += dt_M_SY; } dd = dt_D - now_D; if (mm < 0) { yy -= 1; mm += 12; } }
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