Question

我的问题是正确的联接不起作用，我猜是由于使用了别名。

我已使用-注释了这些行，以指示最有可能导致此问题的行。如果我单独运行代码的交叉连接部分，则可以正常运行。

如果我添加如下所示的交叉联接，然后使用final_leg表进行右联接a我收到一条错误消息，说“关系“ k”不存在“。

k是我在最后一个select语句中用来调用表的CROSS连接部分的别名。

amqmdnetstd.dll

Answer 1

该问题很可能是因为您没有为list_legs_ds子查询中的列加上别名，这意味着您至少有两列具有相同的名称。

这是一个很好的例子，说明了为什么通常不应该使用select *...以及为什么要在要选择的列中明确显示。请尝试以下操作：

WITH final_leg AS
 (SELECT y.*
  FROM   (SELECT y.shipment_id,
                 y.route_id,
                 MAX(leg_sequence_id) max_leg_sequence_id
          FROM   posimorders.sc_execution_eu.o_detailed_routes_v2 y
          GROUP  BY 1,
                    2) AS x
  INNER  JOIN posimorders.sc_execution_eu.o_detailed_routes_v2 y
  ON     x.route_id = y.route_id
  AND    x.shipment_id = y.shipment_id
  AND    y.leg_sequence_id = x.max_leg_sequence_id),
dest_leg AS
 (SELECT y.*
  FROM   (SELECT y.shipment_id,
                 y.route_id,
                 MIN(leg_sequence_id) max_leg_sequence_id
          FROM   posimorders.sc_execution_eu.o_detailed_routes_v2 y
          LEFT   JOIN warehouse_attributes w -- Joining to add origin country of origin FC
          ON     w.warehouse_id = y.leg_warehouse_id

          GROUP  BY 1,
                    2) x
  INNER  JOIN posimorders.sc_execution_eu.o_detailed_routes_v2 y
  ON     x.route_id = y.route_id
  AND    x.shipment_id = y.shipment_id
  AND    y.leg_sequence_id = x.max_leg_sequence_id),
list_legs_ds AS
 (SELECT t1.leg_warehouse_id         AS leg_ware,
         t1.total_packages,
         t2.leg_warehouse_id         AS last_ds,
         t2.destination_country_code
  FROM   (SELECT leg_warehouse_id AS leg_ware,
                 SUM(pck_count) AS total_packages
          FROM   posimorders.sc_execution_eu.o_detailed_routes_v2
          GROUP  BY 1
          HAVING SUM(pck_count) > 50000) t1
  CROSS  JOIN (SELECT DISTINCT leg_warehouse_id AS last_ds,
                               destination_country_code
               FROM   posimorders.sc_execution_eu.o_detailed_routes_v2) t2)
SELECT a.route_warehouse_id,
       k.leg_ware,
       k.last_ds lm_ds
       -- should there be something to aggregate here?
FROM   final_leg a
INNER  JOIN dest_leg b
ON     a.shipment_id = b.shipment_id
AND    a.route_id = b.route_id
RIGHT  JOIN list_legs_ds k
ON     a.leg_warehouse_id = k.leg_ware -- and a.leg_ship_method = k.lm_ds
GROUP  BY 1,
          2,
          3;

如何为多个选择语句分配别名

1 个答案: