Question

我试图从表1（报告）中获取所有结果，并将其他两个表连接到它（用户）和（工作单位）

报告包含与用户和工作人员相关的密钥，但它们存储在csv值中。我试图执行类似于这个伪代码的东西

`SELECT * 
FROM reports 
LEFT JOIN users ON reports = (WHERE users.userID IN (reports.users))
LEFT JOIN workorders ON reports = (WHERE workorder.status IN (reports.filters) 
AND reports.reportid = 10
`

reports.users和reports.filters看起来像“1,2,3,4,5,6”

Answer 1

如果我理解正确（意味着reports.users和reports.filters是逗号分隔值的字符串），则需要FIND_IN_SET函数：

SELECT * from reports 
LEFT JOIN users ON FIND_IN_SET(users.userID, reports.users)
LEFT JOIN workorders ON FIND_IN_SET(workorder.status, reports.filters) 
  AND reports.reportid = 10

Mysql使用一对多关系加入非匹配表

1 个答案: