Question

我无法让PostgreSQL识别匹配表是空的，因此应该为下面视图中的胜利和匹配列输出0。相反，我一直在看空视图。当我应该获得“玩家ID”，“玩家ID”，每行0,0。可能是什么问题？

CREATE VIEW player_standings AS
SELECT 
    players.id,
    players.name,
    CASE WHEN EXISTS (SELECT * FROM matches) THEN COUNT(matches.winner) ELSE 0 END AS wins,
    CASE WHEN EXISTS (SELECT * FROM matches) THEN COUNT(matches.winner) + COUNT(matches.loser) ELSE 0 END AS matches
FROM players
INNER JOIN matches
    ON players.id = matches.winner
GROUP BY players.id
ORDER BY
    wins DESC;

Answer 1

您不需要case，因为count(null)给出了0。
如果某位玩家没有匹配，您应left join匹配。
您应该join两次匹配才能获得胜负次数。

CREATE OR REPLACE VIEW player_standings AS
SELECT 
    p.id,
    p.name,
    COUNT(m1.winner) AS wins,
    COUNT(m1.winner) + COUNT(m2.loser) AS matches
FROM players p
LEFT JOIN matches m1
    ON p.id = m1.winner
LEFT JOIN matches m2
    ON p.id = m2.loser
GROUP BY p.id, p.name
ORDER BY
    wins DESC;

PostgreSQL：如果表存在于视图中的其他0，则返回函数

1 个答案: