如何有效地将具有多个联接和计数的复杂 SQL 查询转换为 LINQ？

将包含多个连接、计数和左连接的SQL转换为LINQ

SQL 查询

考虑以下SQL查询：

<code class="language-sql">SELECT DISTINCT c.Id, 
       c.Title, 
       COUNT(v.Id) AS 'Nb_V2',
       COUNT(DISTINCT v.IdUser) AS 'Nb_V1',
       r.cnt AS 'Nb_R'
FROM TABLE_C c
JOIN TABLE_V v on c.Id = v.Id
LEFT JOIN ( 
    SELECT Id, COUNT(*)  AS cnt 
    FROM TABLE_R 
    GROUP BY Id
) r ON c.Id = r.Id
WHERE c.IdUser = '1234'
GROUP BY c.Id, c.Title, r.cnt</code>

LINQ 查询表达式

要将此SQL查询转换为LINQ查询表达式，请遵循以下准则：

1. 将子查询转换为单独声明的变量。
2. 按照LINQ子句顺序转换每个子句。
3. 将JOIN子句转换为导航属性或匿名对象。
4. 使用GroupJoin和DefaultIfEmpty()模拟LEFT JOIN。

应用这些规则会产生以下LINQ查询表达式：

<code class="language-csharp">var subrq = from r in Table_R
            group r by r.Id into rg
            select new { Id = rg.Key, cnt = rg.Count() };

var ansq = (from c in Table_C
            join v in Table_V on c.Id equals v.Id
            join r in subrq on c.Id equals r.Id into rj
            from r in rj.DefaultIfEmpty()
            where c.IdUser == "1234"
            group new { c, v, r } by new { c.Id, c.Title, r.cnt } into cvrg
            select new {
                cvrg.Key.Title,
                Nb_V2 = cvrg.Count(),
                Nb_V1 = cvrg.Select(cvr => cvr.v.IdUser).Distinct().Count(),
                Nb_R = (int?)cvrg.Key.cnt
            }).Distinct();</code>

Lambda表达式

对于lambda表达式转换，进一步将LEFT JOIN转换为GroupJoin...SelectMany：

<code class="language-csharp">var subr2 = Table_R.GroupBy(r => r.Id).Select(rg => new { Id = rg.Key, cnt = rg.Count() });
var ans2 = Table_C.Where(c => c.IdUser == "1234")
                  .Join(Table_V, c => c.Id, v => v.Id, (c, v) => new { c, v })
                  .GroupJoin(subr, cv => cv.c.Id, r => r.Id, (cv, rj) => new { cv.c, cv.v, rj })
                  .SelectMany(cvrj => cvrj.rj.DefaultIfEmpty(), (cvrj, r) => new { cvrj.c, cvrj.v, r })
                  .GroupBy(cvr => new { cvr.c.Id, cvr.c.Title, cvr.r.cnt })
                  .Select(cvrg => new { cvrg.Key.Title, Nb_V2 = cvrg.Count(), Nb_V1 = cvrg.Select(cvr => cvr.v.IdUser).Distinct().Count(), Nb_R = (int?)cvrg.Key.cnt });</code>

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