mysql多表查询实训题(MYSQL多表联合查询问题)

本文目录
MYSQL多表联合查询问题
这种结构要一次查出来没有什么高效的方法,只能按一楼的方法去做,如果要提高效率,应该要分两步,第一步把结果集得到,然后遍历得到各节车厢的货物名称。
假设你使用adodb:
$rows=$db-》getAll("select * from A ");
for($i=0;$i《count($rows);$i++)
{
for($j=1;$j《=30;$j++)
{
$index1="NU".$j;
$index2="item".$j."_name";
if($rows》-1)
{
$rows);
}
}
}
结果集得到后,可以根据要求显示的格式做具体处理。
mysql多表查询问题
用左外连接查询:
SELECT xm ,je1, je2 ,je3 ,je4 ,je5,je12
FROM 表 12
lEFT JOIN 表1 ON (表12.xm=表1.xm)
LEFT JOIN 表3 ON (表3.xm=表2.xm)
……
第四题怎么做mysql 怎么用多表查询where来做
你的图片看不怎么清楚,我给你写个栗子
select b.classname,count(a.sno) as 总人数 from student a,classes b where a.sno=b.sno group by b.classname
mysql 多表查询问题
SELECT *
FROM ( SELECT *
FROM
table1 AS a ,
table2 AS b
WHERE a.title LIKE CONCAT(’%’,’火龙果’,’%’) AND b.status=2 )d ,
table3 AS c where c.cid=d.content_id GROUP BY title
ORDER BY release_date DESC
LIMIT 0,5
MySQL 多表查询
我不是很了解你想要的是怎么样。以下是我的思路:select A.survey_id, A.topic, A.qid, B.options, B.description from survey_single_choice as A left join survey_single_choice_option as B on A.id = B.single_choice_id 查出单选
select A1.survey_id, A1.topic, A1.qid, B1.options, B1.description from survey_multiple_choices as A1 left join survey_multiple_choices_option as B1 on A1.id = B1.single_choice_id 查出多选
select S.name, S.description, S.status, C1.topic, C1.qid, C1.options, C1.description from survey as S left join (select A.survey_id, A.topic, A.qid, B.options, B.description from survey_single_choice as A left join survey_single_choice_option as B on A.id = B.single_choice_id) as C1 on S.id = C1.survey_id 单选合并到问卷
select S2.name, S2.description, S2.status, C2.topic, C2.qid, C2.options, C2.description from survey as S2 left join (select A1.survey_id, A1.topic, A1.qid, B1.options, B1.description from survey_multiple_choices as A1 left join survey_multiple_choices_option as B1 on A1.id = B1.single_choice_id) as C2 on S2.id = C2.survey_id 多选合并到问卷
select S3.name, S3.description, S3.status, B2.topic, B2.qid from survey as S3 left join survey_short_answer as B2 on S3.id = B2.survey_id 简答表合并到问卷
如果你是要多行列出 问卷名 题号 题目select S.name, S.description, S.status, C1.topic, C1.qid from survey as S left join (select A.survey_id, A.topic, A.qid, B.options, B.description from survey_single_choice as A left join survey_single_choice_option as B on A.id = B.single_choice_id) as C1 on S.id = C1.survey_idUNION ALLselect S2.name, S2.description, S2.status, C2.topic, C2.qid from survey as S2 left join (select A1.survey_id, A1.topic, A1.qid, B1.options, B1.description from survey_multiple_choices as A1 left join survey_multiple_choices_option as B1 on A1.id = B1.multiple_choices) as C2 on S2.id = C2.survey_idUNION ALLselect S3.name, S3.description, S3.status, B2.topic, B2.qid from survey as S3 left join survey_short_answer as B2 on S3.id = B2.survey_idORDER BY name, qid ASC
如果是一条列出select * from (select * from (select S.name, S.description, S.status, C1.topic, C1.qid, C1.options, C1.description from survey as S left join (select A.survey_id, A.topic, A.qid, B.options, B.description from survey_single_choice as A left join survey_single_choice_option as B on A.id = B.single_choice_id) as C1 on S.id = C1.survey_id) as D left join (select S2.name, S2.description, S2.status, C2.topic, C2.qid, C2.options, C2.description from survey as S2 left join (select A1.survey_id, A1.topic, A1.qid, B1.options, B1.description from survey_multiple_choices as A1 left join survey_multiple_choices_option as B1 on A1.id = B1.single_choice_id) as C2 on S2.id = C2.survey_id) as D1 on D.name = D1.name) as E left join (select S3.name, S3.description, S3.status, B2.topic, B2.qid from survey as S3 left join survey_short_answer as B2 on S3.id = B2.survey_id) E1 on E.name = E1.name
(注意:要修改*号列出你想列的列名,并改一下选项里面的列名)

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