甲骨文。 如果第一条语句存在记录，则忽略第二条

Question

我有一个问题，如何解决我的问题。

我有程序在里面我有这个SELECT里面LEFT OUTER JOIN女巫正在返回一些值：

SELECT * FROM database1 data1
                        JOIN database2 data2 ON data2.id = data1.attr_id
                        JOIN database3 data3 ON data3.attr_id = data2.id
                        JOIN database4 data4 ON data4.objt_attr_id = data3.id
                        JOIN database5 data5 ON data5.stya_id = data4.id
                            AND data5.value = 1
        JOIN database6 data6 ON data6.id = data5.sero_id
        JOIN database7 data7 ON srv.id = data6.srv_id
        JOIN database8 data8 ON data8.code IN ('CALC1','CALC2')
        WHERE data1.ordet_id IN (data8.id)

正如你所看到的，他正在寻找CALC1和CALC2然后在data1.ordet_id通过他们的ID's进行搜索。 自然，他还我2个条目女巫是不正确的。

如何进行检查。 当脚本将使用CALC1找到记录时，他会跳过对CALC2的检查，因此脚本仅通过CALC1返回 1 条记录（两者都不像现在）。 反之亦然，如果没有找到CALC1记录，他检查CALC2

Answer 1

我认为您可以利用分析函数row_number如下：

select * from
(SELECT data1.*, data2.*, .. data8.*, 
        row_number() over (partition by data1.ordet_id order by data8.code ) as rn 
        FROM database1 data1
        JOIN database2 data2 ON data2.id = data1.attr_id
        JOIN database3 data3 ON data3.attr_id = data2.id
        JOIN database4 data4 ON data4.objt_attr_id = data3.id
        JOIN database5 data5 ON data5.stya_id = data4.id
                            AND data5.value = 1
        JOIN database6 data6 ON data6.id = data5.sero_id
        JOIN database7 data7 ON srv.id = data6.srv_id
        JOIN database8 data8 ON data8.code IN ('CALC1','CALC2')
        WHERE data1.ordet_id IN (data8.id))
where rn = 1

- 更新

根据您的以下评论，您还可以使用NOT EXISTS和CASE..WHEN组合，如下所示：

SELECT  *
FROM
    DATABASE1 DATA1
    JOIN DATABASE2 DATA2 ON DATA2.ID = DATA1.ATTR_ID
    JOIN DATABASE3 DATA3 ON DATA3.ATTR_ID = DATA2.ID
    JOIN DATABASE4 DATA4 ON DATA4.OBJT_ATTR_ID = DATA3.ID
    JOIN DATABASE5 DATA5 ON DATA5.STYA_ID = DATA4.ID
                            AND DATA5.VALUE = 1
    JOIN DATABASE6 DATA6 ON DATA6.ID = DATA5.SERO_ID
    JOIN DATABASE7 DATA7 ON SRV.ID = DATA6.SRV_ID
    JOIN DATABASE8 DATA8 ON DATA8.CODE IN (
        'CALC1',
        'CALC2'
    )
WHERE
    CASE
        WHEN DATA8.CODE = 'CALC1' THEN DATA1.ORDET_ID
        WHEN DATA8.CODE = 'CALC2' THEN CASE
            WHEN NOT EXISTS (
                SELECT 1 FROM
                    DATABASE8 D8
                WHERE DATA1.ORDET_ID = D8.ID
                  AND D8.CODE = 'CALC1'
            ) THEN DATA1.ORDET_ID
        END
    END = DATA8.ID

干杯！！

Answer 2

使用OUTER JOIN加入两次（即，如果匹配的代码不存在，则不会丢失任何记录，但不会复制结果），并从第一次或第二次加入中使用COALESCE获取优先代码。

SELECT 
   data1.ordet_id ,
   COALESCE(data81.code, data82.code) code
FROM data1.ordet_id 
LEFT OUTER JOIN data81 ON data1.ordet_id = data81.id and data81.code = 'CALC1'
LEFT OUTER JOIN data82 ON data1.ordet_id = data82.id and data82.code = 'CALC2'