没有时间部分的DateTime比较？

Question

我写了这段代码。但是我想忽略时间，我只想比较一天。

from s in sayac_okumalari
where s.okuma_tarihi == startDate && s.sayac_id == sayac_id
group s by new { date = new DateTime(((DateTime)s.okuma_tarihi).Year, ((DateTime)s.okuma_tarihi).Month, ((DateTime)s.okuma_tarihi).Day, ((DateTime)s.okuma_tarihi).Hour, 1, 1) } into g
select new
{
     okuma_tarihi = g.Key.date,
     T1 = g.Sum(x => x.toplam_kullanim_T1),
     T2 = g.Sum(x => x.toplam_kullanim_T2),
     T3 = g.Sum(x => x.toplam_kullanim_T3)
};

例如：

25.02.1987 == 25.02.1987

Answer 1

使用s.okuma_tarihi.Value.Date == startDate.Date。这应该只允许您比较Date组件。

注释中的讨论中的更新看起来像是用户正在使用NullableType。因此更新了NullableType的解决方案。

Answer 2

使用DateTime的Date属性。对于ex来说，

var date= DateTime.Now.Date;

Answer 3

因为您可以将s.okuma_tarihi转换为DateTime，所以我认为您可以这样做：

var sonuc = from s in sayac_okumalari
        where (DateTime)s.okuma_tarihi.Date == startDate.Date && s.sayac_id == sayac_id
        group s by new { date = new DateTime(((DateTime)s.okuma_tarihi).Year, ((DateTime)s.okuma_tarihi).Month, ((DateTime)s.okuma_tarihi).Day, ((DateTime)s.okuma_tarihi).Hour, 1, 1) } into g
        select new
        {
             okuma_tarihi = g.Key.date,
             T1 = g.Sum(x => x.toplam_kullanim_T1),
             T2 = g.Sum(x => x.toplam_kullanim_T2),
             T3 = g.Sum(x => x.toplam_kullanim_T3)
        };

希望能有所帮助。