linq - C# Linq 在多列上连接 2 个表并使用 GROUP BY 进行计数

Question

我需要在五列上加入两个实体，我需要对 SELECT 中的每一列进行 GROUP BY，并在每个 GROUP BY 上获得一个 COUNT。虽然这对我来说是 SQL 的蛋糕，但我对 LINQ 所能找到的每一个例子都感到非常困惑。

我发现以下两个问题( C# Linq Group By on multiple columns [duplicate] 和 Group By Multiple Columns )与我的相似，但我仍然对如何做到这一点感到困惑。这是我的 SQL 语句:

SELECT  o.org_hq_name,
        o.org_command_name,
        o.org_region_name,
        o.org_installation_name,
        o.org_site_name,
        o.org_subsite_name,
        o.org_hq_id,
        o.org_command_id,
        o.org_region_id,
        o.org_installation_id,
        o.org_site_id,
        count(org_site_id) AS count

FROM    organization o, asset a

WHERE      o.org_hq_id = hq_org_id
AND        o.org_command_id = a.command_org_id
AND        o.org_region_id = a.region_org_id
AND        o.org_installation_id = a.installation_org_id
AND        o.org_site_id = a.site_org_id

GROUP BY o.org_hq_name,
        o.org_command_name,
        o.org_region_name,
        o.org_installation_name,
        o.org_site_name,
        o.org_subsite_name,
        o.org_hq_id,
        o.org_command_id,
        o.org_region_id,
        o.org_installation_id,
        o.org_site_id

我在下面加入:

var join1 =  from m in context.asset
             join o in context.organization
             on new {hqID = a.hq_org_id, commandID = a.command_org_id, regionID = a.region_org_id, installationID = a.installation_org_id, siteID = a.site_org_id}
             equals new {hqID = o.hq_id, commandID = o.command_id, regionID = o.region_id, installationID = o.installation_id, siteID = o.site_id}
             select new
             {
                 hqID = o.hq_id,
                 commandID = o.command_id,
                 regionID = o.region_id,
                 installationID = o.installation_id,
                 siteID = o.site_id
                 //hqId = o.count(org_site_id) AS count
             };

我在下面有一个 Group By:

var group1 = from a in context.asset
            group a by new
            {
                a.hq_org_id,
                a.command_org_id,
                a.region_org_id,
                a.installation_org_id,
                a.site_org_id

                // I am not sure how to get the count

            } into asset
            select new
            {
                // I cannot figure out how to join
            }

请帮助我在两个表上使用 JOIN GROUP BY 的语法，但在两个表的 MULTIPLE 列上并获取每个组的计数。

Answer 1

您的查询可以像这样在 Linq 中重写

var join1 =  from m in context.asset
         join o in context.organization
         on new {hqID = a.hq_org_id, commandID = a.command_org_id, regionID = a.region_org_id, installationID = a.installation_org_id, siteID = a.site_org_id}
         equals new {hqID = o.hq_id, commandID = o.command_id, regionID = o.region_id, installationID = o.installation_id, siteID = o.site_id}
         group new {m,o} by new {   o.org_hq_name,
                                    o.org_command_name,
                                    o.org_region_name,
                                    o.org_installation_name,
                                    o.org_site_name,
                                    o.org_subsite_name,
                                    o.org_hq_id,
                                    o.org_command_id,
                                    o.org_region_id,
                                    o.org_installation_id,
                                    o.org_site_id
                                } into gr
         select new
         {
            org_hq_name = gr.Key.org_hq_name,
            org_command_name = gr.Key.org_command_name,
            org_region_name = gr.Key.org_region_name,
            org_installation_name = gr.Key.org_installation_name,
            org_site_name = gr.Key.org_site_name,
            org_subsite_name = gr.Key.org_subsite_name,
            org_hq_id = gr.Key.org_hq_id,
            org_command_id = gr.Key.org_command_id,
            org_region_id = gr.Key.org_region_id,
            org_installation_id = gr.Key.org_installation_id,
            org_site_id = gr.Key.org_site_id,
            Count = gr.Count()
         };