基于正则表达式设置标志列
基于正则表达式设置标志列
提问于 2018-10-25 17:25:56
我开发了以下查询，但它不像预期的那样工作：
WITH TABLE1 AS
(
SELECT 613414473 as ID, 1706014200964 as P_NUM, 119539 as d_id, 'F20.0' AS CODE FROM DUAL UNION ALL
SELECT 613414473 as ID, 1706014200964 as P_NUM, 119539 as d_id, 'F22.0' AS CODE FROM DUAL UNION ALL
SELECT 613415801 as ID, 1707045167741 as P_NUM, 115182 as d_id, 'A94.0' AS CODE FROM DUAL UNION ALL
SELECT 613415801 as ID, 1707045167741 as P_NUM, 115182 as d_id, NULL AS CODE FROM DUAL UNION ALL
SELECT 613417084 as ID, 1702038456441 as P_NUM, 6541 as d_id, 'E79' AS CODE FROM DUAL UNION ALL
SELECT 613417084 as ID, 1702038456421 as P_NUM, 6541 as d_id, 'I10' AS CODE FROM DUAL UNION ALL
SELECT 613418372 as ID, 1706226211517 as P_NUM, 25727 as d_id, 'F32.9' AS CODE FROM DUAL )
SELECT T1.*
, CASE when regexp_like( CODE, 'C0[5-9]|' ||
'A0[0-9]|A1[0-9]|A2[0-9]|A3[0-9]|A4[0-9]|A5[0-9]|A6[0-9]|A7[0-9]|A8[0-9]|A9[0-7]|' )
THEN 1
ELSE 0 END AS FOUND_CODE
FROM TABLE1 T1;我希望使用值1标记类似于C0[5-9]%或A0[0-97]的代码，如果至少找到一个代码将该p_num的所有标志设置为1，则对相同的p_num进行标记。
上面的示例输出：
| 613414473|1706014200964|119539|F20.0|0|
| 613414473|1706014200964|119539|F22.0|0|
| 613415801|1707045167741|115182|A94.0|1|
| 613415801|1707045167741|115182|NULL |1|
| 613417084|1702038456441|6541 |E79 |0|
| 613417084|1702038456421|6541 |I10 |0|
| 613418372|1706226211517|25727 |F32.9|0|如何修改查询以获得输出？还有更好的正则表达式吗？
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-10-25 17:49:13
根据您的描述，正则表达式模式应该是
'^(C0[5-9]|A[0-8][0-9]|A9[0-7])'^锚定在值的开头，括号允许任何管道分隔的模式匹配；模式被简化，因为A00到A89可以一次处理。
标记与原始查询相同的一行。下一步是将其移到子查询中，然后使用由p_num分区的解析函数，您希望该函数是通用的：
max(found_code) over (partition by p_num)因此，这将成为(与另一条规则相匹配的附加行)：
with table1 (id, p_num, d_id, code) as
(
select 613414470, 1706014200960, 119530, 'D99' from dual union all
select 613414471, 1706014200960, 119531, 'C05' from dual union all
--
select 613414473, 1706014200964, 119539, 'F20.0' from dual union all
select 613414473, 1706014200964, 119539, 'F22.0' from dual union all
select 613415801, 1707045167741, 115182, 'A94.0' from dual union all
select 613415801, 1707045167741, 115182, null from dual union all
select 613417084, 1702038456441, 6541 , 'E79' from dual union all
select 613417084, 1702038456421, 6541 , 'I10' from dual union all
select 613418372, 1706226211517, 25727 , 'F32.9' from dual
)
select id, p_num, d_id, code, max(found_code) over (partition by p_num) as found_code
from (
select t1.*
, case when regexp_like( code, '^(C0[5-9]|A[0-8][0-9]|A9[0-7])' )
then 1
else 0
end as found_code
from table1 t1
);
ID P_NUM D_ID CODE FOUND_CODE
------------- ------------- ------------- ----- -------------
613414470 1706014200960 119530 D99 1
613414471 1706014200960 119531 C05 1
613414473 1706014200964 119539 F20.0 0
613414473 1706014200964 119539 F22.0 0
613415801 1707045167741 115182 A94.0 1
613415801 1707045167741 115182 1
613417084 1702038456441 6541 E79 0
613417084 1702038456421 6541 I10 0
613418372 1706226211517 25727 F32.9 0页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/52994930
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