两个DateTime类型的值相减之后取得时间差（以月份形式！）？

[guohua1205]

两个DateTime类型的值相减之后取得时间差之后，怎么得到其相隔的月份呢？

[Peter200694013]

DateTime dt1 = new DateTime(2010,4,1);
DateTime dt2 = new DateTime(2010, 8, 1);

TimeSpan diff = dt2 - dt1;
Console.WriteLine(diff.Days/30);

[dopsop110]

            DateTime dt = new DateTime();
            DateTime dt1 = new DateTime();
            dt.Month - dt1.Month;

[vrhero]

最简单的办法是TimeSpan.Days/30...

[lzsh0622]

DateTime dt1 = new DateTime(2009, 01, 01);
DateTime dt2 = new DateTime(2010, 04, 30);
int months = (dt2.Year * 12 + dt2.Month) - (dt1.Year * 12 + dt1.Month);

[baishi1987]

LZ正解，我比较推重这个写法

[zhanshenjsb]

sql sever数据库中有DateDiff函数求两个日期间的间隔数
DateDiff(mm,start_date,end_date)
mm表示两个日期间的间隔数是月
start_date开始时间，（时间类型）
end_date结束时间

[huming_h]

[Quote=引用 1 楼 peter200694013 的回复:]
DateTime dt1 = new DateTime(2010,4,1);
DateTime dt2 = new DateTime(2010, 8, 1);

TimeSpan diff = dt2 - dt1;
Console.WriteLine(diff.Days/30);
[/Quote]
up

[guyehanxinlei]

Select DateDiff(Month,GetDate(),DateAdd(Day,365,GetDate())) As 'DistantMonths'

[hai_yang_09]

lu guo~~~~~~~`

[singincloud]

[Quote=引用 4 楼 lzsh0622 的回复:]

DateTime dt1 = new DateTime(2009, 01, 01);
DateTime dt2 = new DateTime(2010, 04, 30);
int months = (dt2.Year * 12 + dt2.Month) - (dt1.Year * 12 + dt1.Month);
[/Quote]


顶！！

[aaseh]

楼上的全都在误导人啊。。除了8楼

[zhangdoublepan]

DateTime dt1 = new DateTime();
DateTime dt2 = new DateTime();
int i = dt2.Year - dt1.Year;
int j = 0;
if (dt2.Month > dt1.Month)
{
i = i * 12 + dt2.Month - dt1.Month;
}
else
{
j = i * 12 + dt1.Month - dt2.Month;
}

[wuyq11]

DateTime d1= DateTime.Parse("");
DateTime d2= DateTime.Parse("");
int month1 = d1.Year * 12 + d1.Month;
int month2 = d2.Year * 12 + d2.Month;
int month = month2 - month1;

[wengxiaoxi]

觉得1楼的好