Question

我想计算每个用户相册的照片数量，直到用户在一个相册中没有照片，然后（我认为）WHERE ap.del = 0该相册不会出现。请帮忙，DB是Postgresql！

 SELECT a.id_album, a.id_osoba, a.name, a.description, a.privacy, a.del, a.data_del, a.cover, a.date, count(ap.id_album_photo) AS ilosc
   FROM album a
   LEFT JOIN album_photo ap ON ap.id_album = a.id_album
  WHERE ap.del = 0
  GROUP BY a.id_album, a.id_osoba, a.name, a.description, a.privacy, a.del, a.data_del, a.cover, a.date
  ORDER BY a.name;

Answer 1

如果你想在没有album_photo时显示它，你可以这样做：

WHERE ap.del = 0 OR ap.id_album IS NULL

如果没有album_photo，则LEFT JOIN将返回所有album_photo值为NULL。检查NULL的连接键将显示没有加入任何行。

Answer 2

将WHERE ap.del = 0移至JOIN

LEFT JOIN album_photo ap ON ap.id_album = a.id_album AND ap.del = 0

当没有NULL时，

ap.del将为album_photo，这意味着WHERE子句将过滤掉ap.del为空的行。由于NULL = 0未知并返回false。

同样的事情适用于您想要在LEFT/RIGHT JOIN上过滤的任何时间。

Answer 3

怎么样：

SELECT
    a.id_album,
    a.id_osoba,
    a.name,
    a.description,
    a.privacy,
    a.del,
    a.data_del,
    a.cover,
    a.date,
    SUM(CASE
        WHEN ap.id_album_photo IS NOT NULL THEN 1
        ELSE 0
    END) AS liczba
FROM
    album AS a
LEFT JOIN
    album_photo AS ap
ON
    a.id_album = ap.id_album
AND
    ap.del = 0
GROUP BY
    a.id_album,
    a.id_osoba,
    a.name,
    a.description,
    a.privacy,
    a.del,
    a.data_del,
    a.cover,
    a.date
ORDER BY
    a.name;

或

SELECT
    a.id_album,
    a.id_osoba,
    a.name,
    a.description,
    a.privacy,
    a.del,
    a.data_del,
    a.cover,
    a.date,
    COALESCE(ap.liczba, 0) AS liczba
FROM
    album AS a
LEFT JOIN
    (
        SELECT
            ap.id_album,
            COUNT(1) AS liczba
        FROM
            album_photo AS ap
        WHERE
            ap.del = 0
        GROUP BY
            ap.id_album
    ) AS ap
ON
    ap.id_album = a.id_album
ORDER BY
    a.name;

SQL连接问题

3 个答案: