我有这个查询
SELECT
s.account_number,
a.id AS 'ASPIRION ID',
a.patient_first_name,
a.patient_last_name,
s.admission_date,
s.total_charge,
astat.name AS 'STATUS',
astat.definition,
latest_note.content AS 'LAST NOTE',
a.insurance_company
FROM
accounts a
INNER JOIN
services s ON a.id = s.account_id
INNER JOIN
facilities f ON f.id = a.facility_id
INNER JOIN
account_statuses astat ON astat.id = a.account_status_id
INNER JOIN
(SELECT
account_id, MAX(content) content, MAX(created)
FROM
notes
GROUP BY account_id) latest_note ON latest_note.account_id = a.id
WHERE
a.facility_id = 56
我的问题来自
(SELECT
account_id, MAX(content) content, MAX(created)
FROM
notes
GROUP BY account_id)
内容是一个varchar字段,我需要获取最新记录。我现在明白MAX不会像我想要的那样在varchar字段上工作。我不确定如何通过此加入中的 MAX ID 和组帐户ID 来获取相应的内容。
最好的方法是什么? 我的笔记表看起来像这样......
id account_id content created 1 1 This is a test 2011-03-16 02:06:40 2 1 More test 2012-03-16 02:06:40
答案 0 :(得分:0)
SELECT
s.account_number,
a.id AS 'ASPIRION ID',
a.patient_first_name,
a.patient_last_name,
s.admission_date,
s.total_charge,
astat.name AS 'STATUS',
astat.definition,
latest_note.content AS 'LAST NOTE',
a.insurance_company
FROM
accounts a
INNER JOIN services s ON a.id = s.account_id
INNER JOIN facilities f ON f.id = a.facility_id
INNER JOIN account_statuses astat ON astat.id = a.account_status_id
INNER JOIN
(SELECT account_id, MAX(created) mxcreated
FROM notes GROUP BY account_id) latest_note ON latest_note.account_id = a.id and
latest_note.mxcreated = --datetime column from any of the other tables being used
WHERE a.facility_id = 56
您必须join
max(created)
才会提供最新内容。
或者您可以将查询更改为
SELECT account_id, content, MAX(created) mxcreated
FROM notes GROUP BY account_id
mysql
允许您,即使您未在group by
子句中包含所有非聚合列。但是,除非您在最长日期join
,否则您将无法获得正确的结果。
答案 1 :(得分:0)
这是两个选择。如果您的content
不是很长并且没有时髦的字符,您可以使用substring_index()
/ group_concat()
技巧:
(SELECT account_id,
SUBSTRING_INDEX(GROUP_CONCAT(content ORDER BY created desc SEPARATOR '|'
), 1, '|') as content
FROM notes
GROUP BY account_id
) latest_note
ON latest_note.account_id = a.id
鉴于列和表的名称,这可能不起作用。然后,您需要from
子句中的其他连接或相关子查询。我认为在这种情况下这可能是最简单的:
select . . .,
(select n.content
from notes n
where n.account_id = a.id
order by created desc
limit 1
) as latest_note
from . . .
此方法的优点是它只获取所需行的注释。并且,您不需要left join
来保留所有行。为了提高性能,您需要notes(account_id, created)
上的索引。
答案 2 :(得分:0)
最后创建的记录是不存在更新的记录。因此:
SELECT
s.account_number,
a.id AS "ASPIRION ID",
a.patient_first_name,
a.patient_last_name,
s.admission_date,
s.total_charge,
astat.name AS "STATUS",
astat.definition,
latest_note.content AS "LAST NOTE",
a.insurance_company
FROM accounts a
INNER JOIN services s ON a.id = s.account_id
INNER JOIN facilities f ON f.id = a.facility_id
INNER JOIN account_statuses astat ON astat.id = a.account_status_id
INNER JOIN
(
SELECT account_id, content
FROM notes
WHERE NOT EXISTS
(
SELECT *
FROM notes newer
WHERE newer.account_id = notes.account_id
AND newer.created > notes.created
)
) latest_note ON latest_note.account_id = a.id
WHERE a.facility_id = 56;