Question

我在匹配表中有以下数据：

5;{"Id":1,"Teams":[{"Name":"TeamA","Players":[{"Name":"AAA"},{"Name":"BBB"}]},{"Name":"TeamB","Players":[{"Name":"CCC"},{"Name":"DDD"}]}],"TeamRank":[1,2]}
6;{"Id":2,"Teams":[{"Name":"TeamA","Players":[{"Name":"CCC"},{"Name":"BBB"}]},{"Name":"TeamB","Players":[{"Name":"AAA"},{"Name":"DDD"}]}],"TeamRank":[1,2]}

我想按名称选择表格中每个最后一个不同的团队。即我想要一个将返回的查询：

6;{"Name":"TeamA","Players":[{"Name":"CCC"},{"Name":"BBB"}
6;{"Name":"TeamB","Players":[{"Name":"AAA"},{"Name":"DDD"}

所以每个团队从上次那个团队出现在表中我一直在使用以下（来自here）：

WITH t AS (SELECT id, json_array_elements(match->'Teams') AS team FROM matches)
SELECT MAX(id) AS max_id, team FROM t GROUP BY team->'Name';

但是这会回来：

ERROR: could not identify an equality operator for type json
SQL state: 42883
Character: 1680

我理解Postgres doesn't have equality for JSON。我只需要团队名称（一个字符串）的平等，该团队中的球员不需要进行比较。

有人可以建议另一种方法吗？
供参考：

SELECT id, json_array_elements(match->'Teams') AS team FROM matches

返回：

5;"{"Name":"TeamA","Players":[{"Name":"AAA"},{"Name":"BBB"}]}"
5;"{"Name":"TeamB","Players":[{"Name":"CCC"},{"Name":"DDD"}]}"
6;"{"Name":"TeamA","Players":[{"Name":"CCC"},{"Name":"BBB"}]}"
6;"{"Name":"TeamB","Players":[{"Name":"AAA"},{"Name":"DDD"}]}"

编辑：我转向text并跟随this question，我使用DISTINCT ON代替GROUP BY。这是我的全部问题：

WITH t AS (SELECT id, json_array_elements(match->'Teams') AS team
           FROM matches ORDER BY id DESC)
SELECT DISTINCT ON (team->>'Name') id, team FROM t;

返回上面我想要的内容。有没有人有更好的解决方案？

Answer 1

LATERAL加入更短，更快，更优雅：

SELECT DISTINCT ON (t.team->>'Name') t.team
FROM   matches m, json_array_elements(m.match->'Teams') t(team);
ORDER  BY t.team->>'Name', m.id DESC;  -- to get the "last"

如果你只想要不同的团队，那么ORDER BY就可以了。相关：

JSON和相等

Postgres中的json数据类型没有相等运算符，但jsonb有一个（Postgres 9.4 +）：

How to query a json column for empty objects?

Postgres中的GROUP BY - JSON数据类型不相等？

1 个答案:

JSON和相等