[英]Sql conditional query on table
我有2张桌子。
我正在尝试从articles
获取所有列-并将status = 0的状态数作为status0
并将status = 1的状态数作为status1
“从文章中获取所有内容,并为每个文章行获取status = 0作为status0的注释数,并获取status = 1作为status1的注释数”。 可能?
表:
articles
========
id name
---------
1 abc
2 def
3 ghi
comments
========
id article_id status
-------------------------
1 2 1
2 2 0
3 1 0
4 3 1
带有状态编号的文章预期结果:
id name status0 status1
------------------------------
1 abc 1 0
2 def 1 1
3 ghi 0 1
我正在使用Laravel的Eloquent,但足以看到原始sql语句。 我不知道如何查询和计算这些状态。
多亏了小提琴,我设法创建了这个查询,但是出现一个错误:请注意,( articles = db_surveys
和comments = db_answers
)
"SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.`status=1) status1` from `db_surveys` left join `db_answers` on `db_surveys`.`i' at line 1 (SQL: select `db_surveys`.`*, SUM(db_answers`.`status=0) status0, SUM(db_answers`.`status=1) status1` from `db_surveys` left join `db_answers` on `db_surveys`.`id` = `db_answers`.`surveyid` where `db_surveys`.`userid` = 6oGxr)"
完整查询:
"select `db_surveys`.`*, SUM(db_answers`.`status=0) status0, SUM(db_answers`.`status=1) status1` from `db_surveys` left join `db_answers` on `db_surveys`.`id` = `db_answers`.`surveyid` where `db_surveys`.`userid` = 123 group by `db_surveys`.`id`"
**
**
SELECT
`s`.*,
SUM(`a`.`status`='pending') `status0`,
SUM(`a`.`status`='confirmed') `status1`
FROM
`db_surveys` s
LEFT JOIN `db_answers` a
ON `s`.`id` = `a`.`surveyid`
WHERE `s`.`userid` = '6oGxr'
GROUP BY `s`.`id`
您可以将sum()与expression一起使用以根据您的条件获取计数,使用sum中的expression将得到布尔o或1
SELECT a.*
,SUM(`status` =0) status0
,SUM(`status` =1) status1
FROM articles a
LEFT JOIN comments c ON(a.id = c.article_id)
GROUP BY a.id
编辑在原始查询您不使用回蜱正确
SELECT
`s`.*,
SUM(`a`.`status`=0) `status0`,
SUM(`a`.`status`=1) `status1`
FROM
`db_surveys` s
LEFT JOIN `db_answers` a
ON `s`.`id` = `a`.`surveyid`
WHERE `s`.`userid` = 123
GROUP BY `s`.`id`
可能是这样的:
SELECT a.*, COUNT(c.*) AS status0, COUNT(c2.*) AS status1
FROM articles AS a
LEFT JOIN comments AS c ON c.article_id = a.id AND c.status = 0
LEFT JOIN comments AS c2 ON c.article_id = a.id AND c.status = 1
不是最漂亮的,但它可以工作:
SELECT articles.id, articles.name,
(SELECT COUNT(*) FROM comments WHERE article_id = articles.id AND status = 0),
(SELECT COUNT(*) FROM comments WHERE article_id = articles.id AND status = 1)
FROM articles;
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