C# · 12月 23, 2021

c# – LINQ with subselect和groupby只能获取列表中每个项目的最新版本

对于LINQ来说,我是个新手

我有一个IEnumerable通用列表,其中包含不同版本的答案(每个都有一个FK的问题).从这个列表中我只需要获得最新版本的答案字典.

一个非常简化的类图:


-ID
-题
– …其他属性

回答
-ID
-版
-QuestionID
-值
– …其他属性

目前我已经得到以下几点:

IEnumerable<Answer> answers = GetAnswers();IDictionary<long,AnswerDTO> latestVersionAnswers = new Dictionary<long,AnswerDTO>();if (answers != null){latestVersionAnswers = answers .OrderBy(e => e.ID) .GroupBy(e => e.Question.ID) .Select(g => new AnswerDTO { Version = g.Last().Version,// g.Select(e => e.Version).Max(),QuestionID = g.Key,ID = g.Last().ID,Value = g.Last().Value }).ToDictionary(c => c.QuestionID); }

虽然这在很大程度上起作用,但您可以快速看到它需要一些严重的优化(并且有点脆弱,因为它取决于答案记录行顺序而不是“最大”逻辑).使用LINQ做什么最好的方式,或者最好是为每个循环做多个?

>如果我只需要版本(而不是ID,Value等),我不需要OrderBy,因为我可以去g.Select(e => e.Version).Max()(或我’现在已经看到了C# List<> GroupBy 2 Values的帖子,但是这又会返回key / s和一个属性:Version).
>最终,在这种特殊情况下,我更愿意只是“过滤”原始列表,并返回原始的答案而不是涉及到AnswerDTO.

任何指针或帮助将不胜感激!

解决方法 像…. private void button1_Click(object sender,EventArgs e) { List<Answer> list = GetAnswers(); var dict = (from a in list group a by a.QuestionID into grp from g in grp where g.Version == grp.Max(m => m.Version) select new { id = g.QuestionID,q = g }).ToDictionary( o => o.id,o => o.q); StringBuilder sb = new StringBuilder(); foreach (var elem in dict) { sb.AppendLine(elem.Key.ToString() + “-” + elem.Value.Version.ToString()); } Message@R_239_2419@.Show(sb.ToString()); } private List<Answer> GetAnswers() { List<Answer> result = new List<Answer>(); result.Add(new Answer() { ID = 1,QuestionID = 1,Version = 1 }); result.Add(new Answer() { ID = 2,Version = 2 }); result.Add(new Answer() { ID = 3,Version = 3 }); result.Add(new Answer() { ID = 4,QuestionID = 2,Version = 1 }); result.Add(new Answer() { ID = 5,Version = 2 }); result.Add(new Answer() { ID = 6,Version = 3 }); result.Add(new Answer() { ID = 7,QuestionID = 3,Version = 1 }); result.Add(new Answer() { ID = 8,Version = 2 }); result.Add(new Answer() { ID = 9,Version = 3 }); result.Add(new Answer() { ID = 10,Version = 4 }); return result; }