Question

我有2张桌子。一个看起来像这样（它包含更多字段，但这不重要）：

TABLE 'containers';
INT id;
VARCHAR(20) name;
VARCHAR(20) uid;

TABLE 'sync';
INT id;
VARCHAR(20) containerUid;
VARCHAR(20) deviceUid;

现在我需要获取所有容器，在一个特定的deviceUid的'sync'表中没有条目。我试过这种方式，但结果似乎不正确。

SELECT * FROM (SELECT a.*, (SELECT COUNT(*) FROM sync AS b WHERE a.uid = b.containerUid AND b.deviceUid = 'DEVICEUID') AS anzahlSync FROM containers AS a) AS A WHERE anzahlSync = 0

我想在mysql中这样做，因为containers-table包含很多条目。所以我尽量不让它们回来，但是之前过滤它们。

Answer 1

使用not exists：

select *
from containers c
where not exists (select 1 from sync s where s.containerUid = c.id and s.deviceUid = 'DEVICEUID')

Answer 2

当您想要覆盖其中一个表中的NULL条目时，您可以使用此查询：

SELECT *
FROM
    (
        SELECT *
        FROM containers
        WHERE name LIKE 'DEVICEUID'
    UNION ALL
        SELECT id, containerUid as name, deviceUid as uid
        FROM sync
        WHERE name LIKE 'DEVICEUID'
    ) AS diff
GROUP BY name, uid
HAVING COUNT(*) = 1

从一个表中获取所有不在另一个特定表中的mysql条目

2 个答案: