Oracle。如果第一个语句存在记录，则忽略第二个]

Question

我在问如何解决我的问题。

[我有一个程序，其中我在SELECT女巫中有这个LEFT OUTER JOIN，正在向我返回一些值：

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)

您可以看到他正在寻找CALC1和CALC2，然后在ID's中按它们的data1.ordet_id进行搜索。自然，他还给我退回了2条女巫，这是不正确的。

如何进行支票。当脚本将找到带有CALC1的记录时，他将跳过对CALC2的检查，因此脚本在CALC1之前仅返回1条记录（这两个现在都不一样）。反之亦然，如果找不到CALC1记录，他会检查CALC2

Answer 1

我认为您可以利用分析函数row_number，如下所示：

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

干杯！

Answer 2

使用OUTER JOIN加入两次（即，如果不存在匹配的代码，则不丢失任何记录），并使用COALESCE获取优先级代码。

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'