一个表包含所有ID(项目ID)的行,另一表包含已查看的项目的“ project_ID”.因此,我需要获取该用户尚未查看但所有其他用户可能已查看过的所有ID.
SELECT p.id FROM projects p
LEFT JOIN projects_viewed pv ON p.id = pv.project_id
WHERE NOT pv.username = 'SomeOneElse';
它将起作用,但是我认为它可能会失败,因为p.username可以等于“ someOneElse”,也可以不等于“ SomeOneElse”,因为“ pv.project_id”不是前面提到的唯一.
假设初始值如下,这是一个示例:
INSERT INTO `projects` (`id`) VALUES
(1), (2), (3), (4), (5);
INSERT INTO `projects_viewed` (`project_id`, `username`,) VALUES
(1, 'Billy'),
(2, 'SomeOneElse'),
(2, 'Billy'),
(3, 'Billy'),
(4, 'SomeOneElse'),
(4, 'Billy'),
(5, 'Billy'),
如果用户名是“ SomeOneElse”,那么我想从项目表中返回所有ID,如果根本没有找到ID(来自项目表)和project_id(来自projects_viewed表)之间匹配,或者仅存在projects_viewed中的匹配项用户名不是我们要查询的用户名(本例中为“ SomeOneElse”).
因此,在此示例中,仅应返回id的1、3和5.
解决方法:
您正在否定外部联接.您可以将该条件移至联接,然后检查是否为空:
select p.id
from projects p
left join projects_viewed pv on
p.id = pv.project_id and
pv.username = 'SomeOneElse'
where pv.project_id is null
或者,您可以为此使用不存在:
select *
from projects p
where not exists (
select 1
from projects_viewed pv
where p.id = pv.project_id and
pv.username = 'SomeOneElse'
)