当前位置:  开发笔记 > 数据库 > 正文

在LINQ to Entities中优化Group By

如何解决《在LINQtoEntities中优化GroupBy》经验,为你挑选了1个好方法。

我在LINQ to Entities中有这个查询.

        var query = (from s in db.ForumStatsSet
                     where s.LogDate >= date1 && s.LogDate <= date2
                     group s by new { s.Topic.topicID, s.Topic.subject, s.Topic.Forum.forumName, s.Topic.datum, s.Topic.Forum.ForumGroup.name, s.Topic.Forum.forumID } into g
                     orderby g.Count() descending
                     select new TopicStatsData
                     {
                         TopicId = g.Key.topicID,
                         Count = g.Count(),
                         Subject = g.Key.subject,
                         ForumGroupName = g.Key.name,
                         ForumName = g.Key.forumName,
                         ForumId = g.Key.forumID
                     });

我知道这是一种"邪恶"查询,但它仅用于管理界面.但它生成的SQL绝对令人恐怖.看看这个宝贝.

exec sp_executesql N'SELECT TOP (50) 
[Project6].[C1] AS [C1], 
[Project6].[TopicId] AS [TopicId], 
[Project6].[C4] AS [C2], 
[Project6].[subject] AS [subject], 
[Project6].[name] AS [name], 
[Project6].[forumName] AS [forumName], 
[Project6].[C2] AS [C3]
FROM ( SELECT 
    [Project5].[TopicId] AS [TopicId], 
    [Project5].[subject] AS [subject], 
    [Project5].[forumName] AS [forumName], 
    [Project5].[name] AS [name], 
    1 AS [C1], 
     CAST( [Project5].[forumID] AS int) AS [C2], 
    [Project5].[C1] AS [C3], 
    [Project5].[C2] AS [C4]
    FROM ( SELECT 
        [Project4].[TopicId] AS [TopicId], 
        [Project4].[forumID] AS [forumID], 
        [Project4].[subject] AS [subject], 
        [Project4].[forumName] AS [forumName], 
        [Project4].[name] AS [name], 
        [Project4].[C1] AS [C1], 
        (SELECT 
            COUNT(cast(1 as bit)) AS [A1]
            FROM        [dbo].[tForumStats] AS [Extent14]
            LEFT OUTER JOIN [dbo].[tTopic] AS [Extent15] ON [Extent14].[TopicId] = [Extent15].[topicID]
            LEFT OUTER JOIN [dbo].[tForum] AS [Extent16] ON [Extent15].[forumID] = [Extent16].[forumID]
            LEFT OUTER JOIN [dbo].[tForum] AS [Extent17] ON [Extent15].[forumID] = [Extent17].[forumID]
            LEFT OUTER JOIN [dbo].[tForum] AS [Extent18] ON [Extent15].[forumID] = [Extent18].[forumID]
            LEFT OUTER JOIN [dbo].[tForumGroup] AS [Extent19] ON [Extent18].[forumGroupID] = [Extent19].[forumGroupID]
            LEFT OUTER JOIN [dbo].[tForum] AS [Extent20] ON [Extent15].[forumID] = [Extent20].[forumID]
            LEFT OUTER JOIN [dbo].[tForumGroup] AS [Extent21] ON [Extent20].[forumGroupID] = [Extent21].[forumGroupID]
            WHERE ([Extent14].[LogDate] >= @p__linq__25) AND ([Extent14].[LogDate] = @p__linq__25) AND ([Extent6].[LogDate] = @p__linq__25) AND ([Extent1].[LogDate] 

我不像任何人那样解释该查询,但是如果能够优化查询以便它只是进行简单的常规连接,那将会很棒.如果我自己编写SQL,这样的东西就可以了.

SELECT COUNT(*) AS NumberOfViews, s.topicid AS topicId, t.subject AS TopicSubject, g.[name] AS ForumGroupName, f.forumName AS ForumName 
FROM tForumStats s
join tTopic t on s.topicid = t.topicid
join tForum f on f.forumid = t.forumid
JOIN tForumGroup g ON f.forumGroupID = g.forumGroupID
WHERE s.[LogDate] between @date1 AND @date2
group by s.topicid,  t.subject, f.Forumname, t.Datum, g.[name]
order by count(*) desc

顺便说一句,我喜欢这个网站.惊人的设计和可用性!希望能得到一些帮助:)



1> Ali Ersöz..:

您可以自己加入给定的表,而不是加入组中的所有表.你能尝试一下吗?

from s in db.ForumStatsSet
join t in db.Topics on t.TopicId == s.TopicId
join f in db.Forums on f.ForumId == t.ForumId
join fg in db.ForumGroups on fg.ForumGroupId == f.ForumGroupId
where s.LogDate >= date1 && s.LogDate <= 
group s by new { t.TopicId, t.subject, f.forumName, t.datum, fg.name, f.forumID } into g
orderby g.Count() descending
select new TopicStatsData
{
 TopicId = g.Key.topicID,
 Count = g.Count(),
 Subject = g.Key.subject,
 ForumGroupName = g.Key.name,
 ForumName = g.Key.forumName,
 ForumId = g.Key.forumID
 });

ps:可能有一些错误,但逻辑上它应该是正确的!

推荐阅读
mobiledu2402852357
这个屌丝很懒,什么也没留下!
DevBox开发工具箱 | 专业的在线开发工具网站    京公网安备 11010802040832号  |  京ICP备19059560号-6
Copyright © 1998 - 2020 DevBox.CN. All Rights Reserved devBox.cn 开发工具箱 版权所有