如何根据选择的列显示正确的记录>>

Question

LEFT OUTER JOIN (
   SELECT * FROM (
     SELECT soi.value, srv.osp_id, soi.stya_id, eoax.estpt_id, eoax.discount, int_sero.id AS sero_id, 2 as attr_number 
     FROM srv_obj_attr_intermediate soi
     JOIN estpt_objt_attr_xref eoax ON eoax.interest_rate = 1
     JOIN attribute_types attl ON attl.id = eoax.attr_id
     JOIN object_attribute_type_links oatl ON oatl.attr_id = attl.id
     JOIN service_type_attributes sta ON sta.objt_attr_id = oatl.id
     JOIN service_objects int_sero ON int_sero.id = soi.sero_id
     JOIN services srv ON srv.id = int_sero.srv_id
     JOIN order_event_types oet ON oet.code = 'CALC_INTERMEDIATE_ESTP'
     WHERE eoax.ordet_id = oet.id AND eoax.objt_attr_id = sta.objt_attr_id
        AND soi.stya_id = sta.id AND soi.value = 1

     UNION

    SELECT soa.value, srv.osp_id, soa.stya_id, eoax.estpt_id, eoax.discount, int_sero.id AS sero_id, 1 as attr_number
    FROM srv_obj_attributes soa
    JOIN estpt_objt_attr_xref eoax ON eoax.interest_rate = 1
    JOIN attribute_types attl ON attl.id = eoax.attr_id
    JOIN object_attribute_type_links oatl ON oatl.attr_id = attl.id
    JOIN service_type_attributes sta ON sta.objt_attr_id = oatl.id
    --LEFT JOIN srv_obj_attr_intermediate soi ON soi.stya_id = sta.id
          --AND soi.value = 1
    JOIN service_objects int_sero ON int_sero.id = soa.sero_id
    JOIN services srv ON srv.id = int_sero.srv_id
    JOIN order_event_types oet ON oet.code = 'CALC_INITIAL_ESTP'
    WHERE eoax.ordet_id = oet.id AND eoax.objt_attr_id = sta.objt_attr_id
      AND soa.stya_id = sta.id AND soa.value = 1 --AND soi.value IS NULL
  )
) intrate ON intrate.estpt_id = estpt.id 
 AND intrate.sero_id = esero.sero_id 
 AND intrate.osp_id = srv.osp_id --This parameters are served. For this example we replaced srv.osp_id to 619771
此选择向我返回此记录

|VALUE                                  |OSP_ID       |STYA_ID      |ESTPT_ID                               |DISCOUNT                               |SERO_ID      |ATTR_NUMBER                            |
|---------------------------------------|-------------|-------------|---------------------------------------|---------------------------------------|-------------|---------------------------------------|
|1                                      |619771       |34659        |1812                                   |30                                     |3933326      |1                                      |
|1                                      |619771       |34659        |1812                                   |30                                     |3933327      |1                                      |
|1                                      |619771       |34659        |1916                                   |30                                     |3933326      |1                                      |
|1                                      |619771       |34659        |1916                                   |30                                     |3933327      |1                                      |
|1                                      |619771       |34659        |1987                                   |30                                     |3933326      |1                                      |
|1                                      |619771       |34659        |1987                                   |30                                     |3933327      |1                                      |
|1                                      |619771       |34659        |2027                                   |30                                     |3933326      |1                                      |
|1                                      |619771       |34659        |2027                                   |30                                     |3933327      |1                                      |
|1                                      |619771       |34659        |2028                                   |30                                     |3933326      |1                                      |
|1                                      |619771       |34659        |2028                                   |30                                     |3933327      |1                                      |
|1                                      |619771       |34659        |2029                                   |30                                     |3933326      |1                                      |
|1                                      |619771       |34659        |2029                                   |30                                     |3933327      |1                                      |
|1                                      |619771       |34659        |1812                                   |40                                     |3933327      |2                                      |
|1                                      |619771       |34659        |1916                                   |40                                     |3933327      |2                                      |
|1                                      |619771       |34659        |1987                                   |40                                     |3933327      |2                                      |
|1                                      |619771       |34659        |2027                                   |40                                     |3933327      |2                                      |
|1                                      |619771       |34659        |2028                                   |40                                     |3933327      |2                                      |
|1                                      |619771       |34659        |2029                                   |40                                     |3933327      |2                                      |
我需要在每个选择上添加校验，并具有自己的attr_number。基于此列，我需要返回记录。

如果用exists记录attr_number = 2，那么我应该只获取那些记录。

如果用do not exists记录attr_number = 2，那么我应该仅在attr_number = 1处获得记录。

我正试着解决我的问题，您可以看到推荐的行（left join和soi.value IS NULL在哪里），但是它不能像我描述的那样工作。他不检查记录是否存在。

是，我们应该使用union删除重复的行。

此选择在left outer join中使用。

``

[LEFT OUTER JOIN（SELECT * FROM（选择soi.value，srv.osp_id，soi.stya_id，eoax.estpt_id，eoax.discount，int_sero.id AS sero_id，2 as attr_number from srv_obj_attr_intermediate soi ...

Answer 1

我知道这不能回答问题，但是您应该知道以下功能与您发布的代码相同：

Answer 2

如果attr_number = 2的记录存在，那么我应该只获取那些记录。如果attr_number = 2的记录不存在，那么我应该仅在attr_number = 1的位置获得记录。