无法创建SQL联接查询

表一

表2

ID  LinkId
1     0
2     1
3     2

表3

ID  IntersectionID  X   Y
1       5          100  200
2       6          300  400
3       7          800  500

我想要一个单一的查询给我以下信息

ID  IntersectionID  Sequence_number linkId  LinkDirection
1        5              0             0       Positive
1        5              1             1       negative
2        6              0             0       negative

对于表1中的每一行，从表3（使用链接字段）中获取其IntersectionID，（链接方向-正表示起点，负表示终点）。然后转到表2，填入x，y值

帮我实现它

已尝试此查询。无法得到它

ID  LinkId  X(start) Y(Start) X(End) Y(End)
1      0      100     200      300    400

---

通过这个查询，您应该能够完成它

如果是文本，请注意链接方向

select r1.id id, r1.linkid base_link_id, r2.X startX,r2.Y startY from  table1 r1, table2 r2 
  where  startX = (select X from table2 where id = r1.id and intersectionId =
  (select intersectionId from table3 where id = r1.id and linkId= r1.linkid  and LinkDirection = positive)) and startY
  = (select Y from table2 where id = r1.id and intersectionId  = 
  (select intersectionId from table3 where id = r1.id and linkId= r1.linkid  and LinkDirection = negative));

如果表2中没有行，您将在

t2\u start.X t2\u start.Y t2\u end.X t2\u end.Y中获得空字段。您可以使用将其更改为0或-1

如果t2_start.X不为空，则t2_start.X ELSE-1 END
如果需要澄清，希望下面的查询有用。请问

SELECT 
    t1.ID
    ,t1.LinkId
    ,t2_start.X as X_Start
    ,t2_start.Y as Y_Start
    ,t2_end.X as X_End
    ,t2_end.Y as Y_End
FROM    
    table1 t1
    LEFT JOIN table3 t3_start
        on t1.linkId = t3_start.linkId  
        and t3_start.LinkDirection = 'Positive'
    LEFT JOIN table2 t2_start
        on t2_start.IntersectionID = t3_start.IntersectionID 

    LEFT JOIN table3 t3_end
        on t1.linkId = t3_end.linkId  
        and t3_end.LinkDirection = 'negative'
    LEFT JOIN table2 t2_end
        on t2_end.IntersectionID = t3_end.IntersectionID 

这会回来的

SELECT ID,LinkID,MAX(StartX) AS StartX ,MAX(Starty) AS StartY,
MAX(EndX) AS EndX,MAX(EndY) AS EndY FROM
(
  SELECT a.ID,a.LinkID,X as StartX,Y as StartY,null as EndX,null as EndY  
   FROM 
          table1 a JOIN table3 b 
               ON a.Linkid=b.Linkid 
          JOIN table2 c 
               ON b.id=c.id WHERE LinkDirection='Positive'
  UNION ALL
  SELECT a.ID,a.LinkID,null as StartX,null as StartY,X as EndX,Y as EndYY 
   FROM 
          table1 a JOIN table3 b 
                ON a.Linkid=b.Linkid 
          JOIN table2 c 
                ON b.id=c.id WHERE LinkDirection='Negative'
) t
GROUP BY ID,LinkID

什么是shapepointtype，在表1中没有这样的列。@kaleel:忽略它。我已将其从查询中删除。道歉！你在使用哪种数据库管理系统？这就是为什么我喜欢ORM框架或工具。我一点也不喜欢sql命令。你应该看看我的答案：）@Ryx5我在发布我的答案之前做的。请更正以下更改，LinkDirection属于表3而不是表2，左侧联接返回空行，起始值和结束值将显示在单独的行中。@Ryx5它的工作方式。提示：若要删除带有null的记录，请使用Join而不是Left Join。我认为获取null值（并在需要时进行更改）很有趣。很高兴与您共享。@Ryx5:非常感谢。：）很好的一天！
ID  LinkID  StartX  StartY   EndX    EndY
1     0      100     200     300     400
2     1      NULL    NULL    100     200