c# - 如何在 C# 中将 Linq count 与 group by 一起使用

Question

我有一个看起来像这样的数据表

基本上我希望它按列 featurename(distinct) 进行分组，其中它应该在工作量和完整列中求和
指定所有 featureid,featurename 以逗号分隔
count of the featureid
分配给comman分隔
现在我希望数据表看起来像这样

不知道怎么用计数 代码

var result5= dtTaskandBugs.AsEnumerable().GroupBy(x => x["Functional Area"])
                               .Select(item => new
                               {
                                   FunctionalArea = item.Key,

                                   Completedsum = item.Sum(y => Convert.ToDecimal(y["Completed"])),
                                   effortsum = item.Sum(z => Convert.ToDecimal(z["effort"])),
                                   storyids = string.Join(",", item.Select(a => a["Storyid"]).Distinct()),
                                   storiesename= string.Join(",", item.Select(b => b["StoryName"]).Distinct()),
                                   Featureid = string.Join(",", item.Select(c => c["Featureid"]).Distinct()),
                                   Featurename= string.Join(",", item.Select(d => d["FeatureName"]).Distinct()),

                               });

Answer 1

使用 Count() 您将获得分组值的计数。 尝试像这样使用 Linq:

from s in dtTaskandBugs.AsEnumerable()
                        group s by s.Field<string>("Functional Area")
                            into grp
                            orderby grp.Key
                            select new { 
                                   FunctionalArea = grp.Key,
                                   Completedsum = grp.Sum(y => Convert.ToDecimal(y["Completed"])),
                                   effortsum = grp.Sum(z => Convert.ToDecimal(z["effort"])),
                                   storyids = string.Join(",", grp.Select(a => a["Storyid"]).Distinct()),
                                   storiesename= string.Join(",", grp.Select(b => b["StoryName"]).Distinct()),
                                   Featureid = string.Join(",", grp.Select(c => c["Featureid"]).Distinct()),
                                   Featurename= string.Join(",", grp.Select(d => d["FeatureName"]).Distinct()),
                                   Count = grp.Count()          // <----------------
 };