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我有以下查詢:Mysql的內部連接 - >如果返回null空,否則返回名字
SELECT
projects.id AS project_id,
projects.name AS project_name,
projects_tasks.id AS task_id,
projects_tasks.name AS task_name,
projects_tasks_timings.id AS timing_id,
projects_tasks_timings.registered_time AS timing_time,
projects_tasks_timings.edit_reason AS timing_description,
projects_tasks_timings.date AS timing_date,
projects_tasks_timings.created_at AS timing_date2,
projects_tasks_timings.updated_at AS timing_date3,
users.name AS timing_user_name,
users.surname AS timing_user_surname,
users2.name AS timing_user_name_update,
users2.surname AS timing_user_surname_update
FROM `projects_tasks_timings`
INNER JOIN `users`
ON users.id = projects_tasks_timings.user_id
INNER JOIN `users`
AS users2 ON users2.id = projects_tasks_timings.updated_by
INNER JOIN `projects_tasks`
ON projects_tasks.id = projects_tasks_timings.task_id
INNER JOIN `projects`
ON projects.id = projects_tasks.project_id
$task_clause
$user_clause
LIMIT $limit OFFSET $start
這將返回大量的數據,但它沒有。我認爲因爲users2
上的inner join
與任何列都不匹配,因爲它們都是null
值。
我想實現的是:
如果沒有匹配它應該返回null。如果有匹配,它應該返回用戶的名字。
我已經通過使用內部連接的where子句來查看一些問題,但迄今爲止沒有成功。
任何人都可以幫助我嗎?謝謝!
感謝,這對我的作品:) –