将两个标签合并成一行

Question

我需要创建一个mysql查询，该查询应该从两个表中仅获取一行结果。

我有一个名为users的表和一个名为users_sports的表。

 let  context  = appdelegate.persistentContainer.viewContext
 let proj = Project() 
        let arrProj = dic.object(forKey: "Projects") as! NSArray
        for n in 0..<arrProj.count {
            let subDic = arrProj.object(at: n) as! NSDictionary
            let item = ProjectItem(dict: subDic)
            proj.arrProjs.append( item )


           let projects = NSEntityDescription.entity(forEntityName: "Projects", in: context)
            projects?.setValue(item.Project, forKey: "project1")
            projects?.setValue(item.Project2, forKey: "project2")
            projects?.setValue(item.ID, forKey: "projectid")
            projects?.setValue(item.radius, forKey: "radius")
            projects?.setValue(item.GeofenceType, forKey: "geo_Type")
            projects?.setValue(item.Geofence, forKey: "geofence")
            projects?.setValue(item.Coordinates, forKey: "coordinates")
        }

我只需要一行中包含表用户的所有用户值和特定用户的所有user_sport_sport值的查询结果。

例如：SELECT * FROM users WHERE id = 71 SELECT user_sport_sport FROM users.sports WHERE user_sport_user = 71

编辑tb_user_sports

表格用户

表tb_sports

Answer 1

您可以使用group_concat来获取该用户ID的所有体育项目

select u.name, GROUP_CONCAT(ts.sport_name) sports from users u
join tb_user_sports tus ON tus.user_sport_user = u.id
left join tb_sports ts ON ts.id_sport = tus.user_sport_sport
where u.id = 72
group by u.id

来源：https://dev.mysql.com/doc/refman/8.0/en/group-by-functions.html#function_group-concat

对于别名：link1和link2