C# 具有聚合和分组依据的LINQ查询

我有以下SQL查询

    select  seaMake AS Make,
        seaModel AS Model,
        COUNT(*) AS [Count],
        MIN(seaPrice) AS [From],
        MIN(seaCapId) AS [CapId]
from tblSearch 
where seaPrice >= 2000
and seaPrice <= 7000
group by seaMake, seaModel
order by seaMake, seaModel

选择seaMake作为Make，
seaModel作为模型，
计数（*）为[计数]，
最低（海运价格）为[自]，
最小值（seacpid）为[CapId]
来自tblSearch
其中海运价格>=2000
而seaPrice=2000
&&s.seaPrice x.seaMake），
PriceFrom=g.Min（s.seaPrice）
};

---

我哪里出错了？

这应该是SQL的简单翻译：

from s in db.tblSearches
where
    s.seaPrice >= 2000 &&
    s.seaPrice <= 7000
group s by new {s.seaMake, s.seaModel} into g
orderby g.Key
select new
{
    Make =  g.Key.seaMake,
    Model = g.Key.seaModel,
    Count = g.Count(),
    From =  g.Min(x => x.seaPrice),
    CapId = g.Min(x => x.seaCapId)
}

来自数据库中的s.tblsearch
哪里
s、 海运价格>=2000&&
s、 seaPrice x.seaPrice），
CapId=g.Min（x=>x.seacpid）
}

当分组到g时，您将该集合转换为IEnumerable>，因此当前范围内的集合为

g

。因此，以下内容是有效的

from s in db.tblSearches
where s.seaPrice >= 2000
   && s.seaPrice <= 7000
orderby s.seaMake
group s by s.seaMake into g // the collection is now IEnumerable<IGrouping<TypeOfSeaMake, TypeofS>>
select new {
    make = g.Key, // this was populated by s.seaMake
    model = g.First().seaModel, // get the first item in the collection
    count = g.Max(x => x.seaMake), // get the max value from the collection
    PriceFrom = g.Min(x => x.seaPrice), // get the min price from the collection
};

来自数据库中的s.tblsearch
其中s.seaPrice>=2000
&&s.seaPrice x.seaMake），//从集合中获取最大值
PriceFrom=g.Min（x=>x.seaPrice），//从集合中获取最小价格
};

---

现在，每个分组将返回一个项目

如果您取出orderby和GroupBy子句，您可以选择make/model吗？是的，因为分组将使用var g，而var g与var不同，因此我猜您可以使用

g.Key

访问

seaMake

。试试看，让我知道它是否适合你。

from s in db.tblSearches
where s.seaPrice >= 2000
   && s.seaPrice <= 7000
orderby s.seaMake
group s by s.seaMake into g // the collection is now IEnumerable<IGrouping<TypeOfSeaMake, TypeofS>>
select new {
    make = g.Key, // this was populated by s.seaMake
    model = g.First().seaModel, // get the first item in the collection
    count = g.Max(x => x.seaMake), // get the max value from the collection
    PriceFrom = g.Min(x => x.seaPrice), // get the min price from the collection
};