我使用一些在线转换器将MySQL中的视图转换为oracle,结果是这样的:
CREATE VIEW actor_info
AS
SELECT
a.actor_id,
a.first_name,
a.last_name,
GROUP_CONCAT(DISTINCT ||(c.name, ': ',
(SELECT GROUP_CONCAT(f.title FROM dual ORDER BY f.title SEPARATOR FROM dual ', ')
FROM film f
INNER JOIN film_category fc
ON f.film_id = fc.film_id
INNER JOIN film_actor fa
ON f.film_id = fa.film_id
WHERE fc.category_id = c.category_id
AND fa.actor_id = a.actor_id
)
)
ORDER BY c.name SEPARATOR '; ')
AS film_info
FROM actor a
LEFT JOIN film_actor fa
ON a.actor_id = fa.actor_id
LEFT JOIN film_category fc
ON fa.film_id = fc.film_id
LEFT JOIN category c
ON fc.category_id = c.category_id
GROUP BY a.actor_id, a.first_name, a.last_name;
转换器转换得不太好,所以我不得不修改查询并最终得到如下结果:
CREATE VIEW actor_info
AS
SELECT
a.actor_id,
a.first_name,
LISTAGG(c.name || ': ',
(SELECT LISTAGG(f.title, ', ') within group (order by f.title)
FROM film f
INNER JOIN film_category fc
ON f.film_id = fc.film_id
INNER JOIN film_actor fa
ON f.film_id = fa.film_id
WHERE fc.category_id = c.category_id
AND fa.actor_id = a.actor_id
)
) WITHIN group (order by c.name)
AS film_info
FROM actor a
LEFT JOIN film_actor fa
ON a.actor_id = fa.actor_id
LEFT JOIN film_category fc
ON fa.film_id = fc.film_id
LEFT JOIN category c
ON fc.category_id = c.category_id
GROUP BY a.actor_id, a.first_name, a.last_name, c.name;
但是我仍然收到多个错误,例如我无法在第一个
distinct
或那个 listagg
中使用 argument should be a constant or a function of expression in GROUP BY
。我没有办法尝试解决这个问题,对于错误是什么或执行此视图的其他方法有什么建议吗?
Oracle的
listagg()
内部不支持distinct,这有点蛋疼。您遇到的另一个错误 ORA-30497 是因为您不小心让第二个 listagg()
调用了第一个的分隔符。
如果没有样本数据和预期结果,有点难以判断,但看起来您正在寻找类似的东西:
SELECT a.actor_id, a.first_name, a.last_name,
LISTAGG(c.name || ': ' || (
SELECT LISTAGG(f.title, ', ') WITHIN GROUP (ORDER BY f.title)
FROM film f
INNER JOIN film_actor fa
ON fa.film_id = f.film_id
INNER JOIN film_category fc
ON f.film_id = fc.film_id
WHERE fc.category_id = c.category_id -- from main query
AND fa.actor_id = a.actor_id -- from main query
), '; ') WITHIN GROUP (ORDER BY c.name)
AS film_info
FROM actor a
LEFT JOIN (
SELECT DISTINCT fa.actor_id, c.category_id, c.name
FROM film_actor fa
LEFT JOIN film_category fc
ON fc.film_id = fa.film_id
LEFT JOIN category c
ON c.category_id = fc.category_id
) c
ON c.actor_id = a.actor_id
GROUP BY a.actor_id, a.first_name, a.last_name;
“独特”部分是通过内联视图实现的,而内部
listagg
只需要找到该类别/演员的所有电影。