如何一次检索所有 grand parents

Question

我在 SQLITE3 中有两个简单的表（家庭关系）：

“人”（id、fname、lname）
“关系”(parent/child)

我想得到每个 grand children 以及他们所有的 grand parents（从 1 到 4 个取决于 grand child）这样 1 行结果=

Distinct Grand child, Grand parent 1, Grand parent 2, Grand parent 3, Grand parent 4

感谢 Caius Jard，我已经能够在另一个 star overflow 问题中得到每个 child 和它的大 parent。但是，到目前为止我有：
1 行 = Grand child、1 grand parent（因此最多需要 4 行才能获得 child 的所有 grand parent）。

SELECT c.fname, c.lname, gp.fname, gp.lname  
FROM relations p_c
INNER JOIN relationships gp_p ON gp_p.child = p_c.parent 
INNER JOIN persons gp ON gp.id = gp_p.parent 
INNER JOIN persons c ON c.id = p_c.child
ORDER BY c.id;

我该如何编辑它，以便在一行结果中得到每个大 child 以及所有大 parent？

如果可能，只使用 SELECT (+ count/distinct)、WHERE (+in/exists)、INNER/LEFT JOIN、GROUP BY (+having)、ORDER BY。

Answer 1

我们可以通过称为条件聚合的技术将行变成列：

假设你有一个结果集：

x, y
1, hello
2, goodbye

如果你这样写：

SELECT x, 
  CASE WHEN x = 1 THEN y END as x1, 
  CASE WHEN x = 2 THEN y END as x2
FROM ...

你会得到这个：

x, x1, x2
1, hello, NULL
2, NULL, goodbye

一列中只有一个值，其余为空。如果您然后使用例如MAX，丢弃空值：

SELECT
  MAX(CASE WHEN x = 1 THEN y END) as x1, 
  MAX(CASE WHEN x = 2 THEN y END) as x2
FROM ...

你会得到

x1, x2
hello, goodbye

我们不得不删除 x，因为我们不能将它放在那里，否则我们必须将它分组，这将防止行折叠

现在根据您的查询，我们可能必须添加其他信息，因为我们需要一些 CASE WHEN 的信息。您没有表示我们可以在其中任意输入行号。希望 persons 包含性别指示符，因为 4 祖父母是一男两女的结果（所以我们有 MM - 父亲的父亲，MF - 父亲的母亲，FM 和 FF 等）

SELECT ...  
FROM relations p_c
INNER JOIN relationships gp_p ON gp_p.child = p_c.parent 
INNER JOIN persons gp ON gp.id = gp_p.parent 
INNER JOIN persons p ON p.id = p_c.parent 
INNER JOIN persons c ON c.id = p_c.child

我们现在可以查询每一层家谱的性别：

SELECT c.name as child,
  CASE WHEN gp.gender = 'm' AND p.gender = 'm' THEN gp.name END AS fathersfather,
  CASE WHEN gp.gender = 'm' AND p.gender = 'f' THEN gp.name END AS fathersmother,
  CASE WHEN gp.gender = 'f' AND p.gender = 'm' THEN gp.name END AS mothersfather,
  CASE WHEN gp.gender = 'f' AND p.gender = 'f' THEN gp.name END AS mothersmother
FROM relations p_c
INNER JOIN relationships gp_p ON gp_p.child = p_c.parent 
INNER JOIN persons gp ON gp.id = gp_p.parent 
INNER JOIN persons p ON p.id = p_c.parent 
INNER JOIN persons c ON c.id = p_c.child

这将给出 child 然后是祖父母姓名的对角线数组。您可以按 child 将其压缩为一行：

SELECT c.name as child,
  MAX(CASE WHEN gp.gender = 'm' AND p.gender = 'm' THEN gp.name END) AS fathersfather,
  MAX(CASE WHEN gp.gender = 'm' AND p.gender = 'f' THEN gp.name END) AS fathersmother,
  MAX(CASE WHEN gp.gender = 'f' AND p.gender = 'm' THEN gp.name END) AS mothersfather,
  MAX(CASE WHEN gp.gender = 'f' AND p.gender = 'f' THEN gp.name END) AS mothersmother
FROM relations p_c
INNER JOIN relationships gp_p ON gp_p.child = p_c.parent 
INNER JOIN persons gp ON gp.id = gp_p.parent 
INNER JOIN persons p ON p.id = p_c.parent 
INNER JOIN persons c ON c.id = p_c.child
GROUP BY c.name

Answer 2

假设有四个爷爷奶奶，你可以枚举他们并合计：

SELECT fname, lname,
       MAX(CASE WHEN seqnum = 1 THEN grandparent_name END) as grandparent_1,
       MAX(CASE WHEN seqnum = 2 THEN grandparent_name END) as grandparent_2,
       MAX(CASE WHEN seqnum = 3 THEN grandparent_name END) as grandparent_3,
       MAX(CASE WHEN seqnum = 4 THEN grandparent_name END) as grandparent_4
FROM (SELECT c.id, c.fname, c.lname,
             gp.fname || ' ' || gp.lname as grandparent_name,
             ROW_NUMBER() OVER (PARTITION BY c.id ORDER BY gp.fname, gp.lanme) as seqnum 
      FROM relations p_c JOIN
           relationships gp_p
           ON gp_p.child = p_c.parent JOIN
           persons gp
           ON gp.id = gp_p.parent 
           persons c
           ON c.id = p_c.child
     ) cgp
GROUP BY fname, lname, id;

Answer 3

我建议使用 GROUP_CONCAT() 而不是 4 列的 grand parents 并得到 1 列所有的 grandparents:

SELECT p.fname, p.lname,
       GROUP_CONCAT(gp.fname || ' ' || gp.lname, '|')
FROM persons p
INNER JOIN relationships r1 ON r1.child = p.id
INNER JOIN relationships r2 ON r2.child = r1.parent
INNER JOIN persons gp ON r2.parent = gp.id
GROUP BY p.id

代码简单得多。
如果您希望返回所有人员，甚至是没有 grandparents 的人员，请使用 LEFT 联接而不是 INNER 联接。

如何一次检索所有 grand parents

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