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嗨，我正在使用 musicbrainz 数据库，我无法计算每个国家/地区的所有艺术家，但无论我尝试哪个国家/地区，我都会遇到相同的错误，即使我尝试使用 like。PLease 谁能告诉我我做错了什么？

错误：列“%u%”不存在第 7 行：WHERE AREA.NAME LIKE “%u%”

SELECT COUNT(artist.name)
FROM artist
JOIN area ON artist.area = area.id 
JOIN label ON area.id = label.area
JOIN country_area ON area.id = country_area.area
JOIN release_country ON country_area.area = release_country.country
WHERE AREA.NAME LIKE "%dom"
GROUP BY release_country.country
limit 5;

更新：

musicbrainz_db=> SELECT COUNT(artist.name)
musicbrainz_db-> FROM artist
musicbrainz_db-> JOIN area ON artist.area = area.id
musicbrainz_db-> JOIN label ON area.id = label.area
musicbrainz_db-> JOIN country_area ON area.id = country_area.area
musicbrainz_db-> JOIN release_country ON country_area.area = 
release_country.country
musicbrainz_db-> WHERE AREA.NAME LIKE '%dom'
musicbrainz_db-> GROUP BY release_country.country
musicbrainz_db-> limit 5;

错误：由于语句超时而取消语句

我的老师刚过来说没有子查询就不行？

select area.name, label_count
from area
where label_count in 
(
    select area.name, count(label.id) as "label_count"
    from area
    JOIN label on area.id = label.area
    group by area.name
);

子查询工作正常，但主查询失败？知道为什么。

4

2 回答 2

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SELECT COUNT(artist.name)
FROM artist
JOIN area ON artist.area = area.id 
JOIN label ON area.id = label.area
JOIN country_area ON area.id = country_area.area
JOIN release_country ON country_area.area = release_country.country
WHERE AREA.NAME LIKE '%u%'
GROUP BY release_country.country
limit 5;
于 2017-11-20T13:59:30.507 回答
1

见 MySQL 官方文档

https://dev.mysql.com/doc/refman/5.7/en/string-comparison-functions.html

并尝试使用单引号

SELECT COUNT(artist.name)
FROM artist
JOIN area ON artist.area = area.id 
JOIN label ON area.id = label.area
JOIN country_area ON area.id = country_area.area
JOIN release_country ON country_area.area = release_country.country
WHERE AREA.NAME LIKE '%dom'
GROUP BY release_country.country
limit 5;

双引号用于列名

于 2017-11-20T13:59:42.950 回答