Question

我的要求是，我有两个表test_ref和test_data。

test_ref表记录：

Pattern
--------------
Delhi
Chennai
Bangalore
Calcutta
Hyderabad

test_data表recrods：

SlNo  descrpt
--------------------------------------------------

1    Very code and hot in Delhi, is the north part of india
2    Calcutta is famous for fish
3    This is first correct message to return
4    Very hot and nice in Chennai city
5    green and clean city is the Bangalore
6    Kochi is famous for travel
7    This is second correct message to return

现在我必须从test_ref表中搜索我的模式，其中descrpt列来自表格test_data，结果必须是

SlNo  descrpt
--------------------------------------------------

3    This is first correct message to return
6    Kochi is famous for travel
7    This is second correct message to return

这里

Delhi出现在test_data表的第一条记录中，因此应该消除。
Calcutta出现在test_data表的第三行，因此应该消除。
test_data表中找不到的模式应该返回。

请为此建议oracle查询。

Answer 1

您不需要使用正则表达式：

Oracle 11g设置：

CREATE TABLE test_ref ( pattern ) AS 
SELECT 'Delhi'     FROM DUAL UNION ALL
SELECT 'Chennai'   FROM DUAL UNION ALL
SELECT 'Bangalore' FROM DUAL UNION ALL
SELECT 'Calcutta'  FROM DUAL UNION ALL
SELECT 'Hyderabad' FROM DUAL;

CREATE TABLE test_data ( SlNo, descrpt ) AS
SELECT 1,    'Very code and hot in Delhi, is the north part of india' FROM DUAL UNION ALL
SELECT 2,    'Calcutta is famous for fish' FROM DUAL UNION ALL
SELECT 3,    'This is first correct message to return' FROM DUAL UNION ALL
SELECT 4,    'Very hot and nice in Chennai city' FROM DUAL UNION ALL
SELECT 5,    'green and clean city is the Bangalore' FROM DUAL UNION ALL
SELECT 6,    'Kochi is famous for travel' FROM DUAL UNION ALL
SELECT 7,    'This is second correct message to return' FROM DUAL;

<强>查询1 ：

SELECT *
FROM   test_data d
WHERE  NOT EXISTS ( SELECT 'X'
                    FROM    test_ref r
                    WHERE   INSTR( d.descrpt, r.Pattern  ) > 0 );

<强>结果：

      SLNO DESCRPT                                                
---------- ------------------------------------------------------
         3 This is first correct message to return                
         6 Kochi is famous for travel                             
         7 This is second correct message to return

你也可以使用：

WHERE d.descrpt LIKE '%'||r.Pattern||'%'或
WHERE REGEXP_LIKE( d.descrpt, r.pattern )（如果你真的想使用正则表达式）。

Answer 2

我不想用大数据集来做这件事，但以下工作：

with temp as
(
    select 'Delhi Chennai Bangalore Calcutta Hyderabad' Pattern  from dual
),
test_data as
(         select 1 sino,   'Very code and hot in Delhi, is the north part of india' descrpt from dual
union all select 2 sino,   'Calcutta is famous for fish' descrpt from dual
union all select 3 sino,   'This is first correct message to return' descrpt from dual
union all select 4 sino,   'Very hot and nice in Chennai city' descrpt from dual
union all select 5 sino,   'green and clean city is the Bangalore' descrpt from dual
union all select 6 sino,   'Kochi is famous for travel' descrpt from dual
union all select 7 sino,   'This is second correct message to return' descrpt from dual)
select sino, descrpt
from test_data 
where sino not in (select sino
                   from  test_data t2
                         -- the next query breaks the pattern into individual words
                        ,(select distinct
                          trim(regexp_substr(t.Pattern, '[^ ]+', 1, levels.column_value))  as word
                        from 
                          temp t,
                          table(cast(multiset(select level from dual connect by  level <= length (regexp_replace(t.pattern, '[^ ]+'))  + 1) as sys.OdciNumberList)) levels) wrds
                   where t2.descrpt like '%'||wrds.word||'%')

Oracle表中的模式匹配查询

2 个答案: