Question

我正在使用MEF并执行一项任务，该任务只有在返回多于1条记录时才需要与聚合函数分组。我需要将开始时间的最大值和结束时间的最小值分成单个记录，就像我的sql会导致重新生成的任务一样

var ohs = await Bl.UoW.Repositories.OperatingHours
    .FindInDataSourceAsync(oh => ((oh.ProductId == productTypeId 
        && oh.StateId == state) 
        || (oh.StateId == complianceHours.State)));

这是一个SQL，当我返回超过1条记录时，它基本符合我的要求

SELECT 
    StateId,
    MAX(ComplianceHourStart),
    MIN(ComplianceHourEnd)
FROM 
    OperatingHours 
GROUP BY
    StateId
HAVING 
    StateId = 'CA'

所以当超过1时我可以进一步过滤但不确定如何达到最大和最小？

if (ohs != null && ohs.Count() > 1)
{
    //
    ohs = ohs.GroupBy(x => x.State).Max(x => x.ComplianceHourStart?...

}

由于

Answer 1

这样的事情应该这样做：

ohs = ohs.GroupBy(x => x.State)
    .Select(g => new 
    { 
        //You need to make a choice on StateId, here... First one?
        StateId = g.First().StateId, 
        MaxComplianceHourStart = g.Max(o => o.ComplianceHourStart),
        MinComplianceHourEnd = g.Min(o => o.ComplianceHourEnd) 
    });

Answer 2

从SQL开始，这应该是关闭的：

var result = context.OperatingHours
                    .GroupBy(oh => oh.StateId)
                    .Select(oh => new {StateId = oh.Key, 
                                       MaxStart = oh.Max(x => x.ComplianceHourStart),
                                       MinEnd = oh.Min(x => x.ComplianceHourEnd)});

...虽然在限制状态ID列（组密钥）时我不确定为什么要分组。以下内容也应该足够：

var result = context.OperatingHours
                    .Where(oh => oh.StateId == 'CA')
                    .Select(oh => new {MaxStart = oh.Max(x => x.ComplianceHourStart),
                                       MinEnd = oh.Min(x => x.ComplianceHourEnd)});

将SQL语句转换为Linq / Lambda查询

2 个答案: