我想根据传递给函数的int对查询进行不同的分组。目前,我有这个非常棘手的解决方案:
`.GroupBy(occurrence => new { date =
// Bucket by day.
timeBucket == 0 ? DbFunctions.TruncateTime(occurrence.occurrenceDate) :
// Bucket by week.
timeBucket == 1 ? DbFunctions.AddDays(DbFunctions.CreateDateTime(occurrence.occurrenceDate.Year, 1, 1, 0, 0, 0), 7*(occurrence.occurrenceDate.DayOfYear/7)) :
// Bucket by month.
timeBucket == 2 ? DbFunctions.TruncateTime(DbFunctions.CreateDateTime(occurrence.occurrenceDate.Year, occurrence.occurrenceDate.Month, 1, 1, 1, 1)) :
// Bucket by year.
DbFunctions.TruncateTime(DbFunctions.CreateDateTime(occurrence.occurrenceDate.Year, 1, 1, 1, 1, 1)),
type = occurrence.type })`
我如何计算日期的细节对我来说并不重要(但随时可以提供帮助)。我想避免不必对数据库中的每一行都执行此case语句。无论如何要避免这样做?我已经尝试了多种解决方案,例如使用表达式,但是我无法返回想要的对象,因为表达式树必须是无参数的...
如果有人有解决方案,将不胜感激。
谢谢!
定义一个石斑鱼类:
class TimeGrouper
{
public int Year { get; set; }
public int Month { get; set; }
public int Week { get; set; }
public int Day { get; set; }
}
还有一个返回表达式的函数:
using System.Data.Entity.SqlServer;
...
Expression<Func<Occurrence, TimeGrouper>> GetGrouper(int grouping)
{
switch (grouping)
{
case 1:
return o => new TimeGrouper
{
Year = o.occurrenceDate.Year
};
case 2:
return o => new TimeGrouper
{
Year = o.occurrenceDate.Year,
Month = o.occurrenceDate.Month
};
case 3:
return o => new TimeGrouper
{
Year = o.occurrenceDate.Year,
Week = SqlFunctions.DatePart("wk", o.StartDate).Value
};
default:
return o => new TimeGrouper
{
Year = o.occurrenceDate.Year,
Day = SqlFunctions.DatePart("dy", o.StartDate).Value
};
}
}
现在你可以打电话
db.Occurrences.GroupBy(GetGrouper(3));
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