需要帮助完成加入

Question

当前的SELECT查询

SELECT a.*, ROUND(((a.avg-b.avg)/b.avg)*100,2) pct_change
FROM
  ( 
     SELECT x.*, COUNT(*) rank
     FROM blavg x 
         JOIN blavg y 
             ON y.bline_id = x.bline_id 
            AND y.id >= x.id 
     GROUP BY x.id
) a
JOIN
( 
     SELECT x.*, COUNT(*) rank
     FROM blavg x 
         JOIN blavg y 
             ON y.bline_id = x.bline_id 
            AND y.id >= x.id 
     GROUP BY x.id
 ) b
     ON b.bline_id = a.bline_id
     AND b.rank = a.rank+1
WHERE a.rank = 1;

我也想

SELECT bl_id, bline FROM bl
bl.bl_id = blavg.bline_id

我试过了

SELECT bl_id, bline 
FROM bl bl 
    LEFT JOIN ON bl.bl_id = blavg.bline_id

但不确定我是否使用了正确的语法，也不确定在何处添加ON以及parens（）在何处。

任何帮助将不胜感激。

Answer 1

因此，您正在执行某种查询以两次两次检索行并进行一些数学运算。
幸运的是，您的查询已经返回了需要链接到bl表的bline_id 。

因此，您只需要在结果上bl.bline上一个即可检索所需的bl.bline字段：

SELECT bl.bline, a.*, ROUND(((a.avg-b.avg)/b.avg)*100,2) pct_change
FROM
  ( 
     SELECT x.*, COUNT(*) rank
     FROM blavg x 
         JOIN blavg y 
             ON y.bline_id = x.bline_id 
            AND y.id >= x.id 
     GROUP BY x.id
) a
JOIN
( 
     SELECT x.*, COUNT(*) rank
     FROM blavg x 
         JOIN blavg y 
             ON y.bline_id = x.bline_id 
            AND y.id >= x.id 
     GROUP BY x.id
 ) b
     ON b.bline_id = a.bline_id
     AND b.rank = a.rank+1
     LEFT OUTER JOIN bl
         ON bl.bl_id = a.bline_id
WHERE a.rank = 1;