我知道我可以编写一个查询,该查询将返回给定列中包含任意数量值的所有行,如下所示:
I know I can write a query that will return all rows that contain any number of values in a given column, like so:
Select * from tbl where my_col in (val1, val2, val3,... valn)
但是,如果 val1
可以出现在 my_col
中的任何位置,它的数据类型为 varchar(300),我可能会改为:
but if val1
, for example, can appear anywhere in my_col
, which has datatype varchar(300), I might instead write:
select * from tbl where my_col LIKE '%val1%'
有没有办法将这两种技术结合起来.我需要搜索大约 30 个可能的值,这些值可能出现在列的自由格式文本中的任何位置.
Is there a way of combing these two techniques. I need to search for some 30 possible values that may appear anywhere in the free-form text of the column.
以下列方式组合这两个语句似乎行不通:
Combining these two statements in the following ways does not seem to work:
select * from tbl where my_col LIKE ('%val1%', '%val2%', 'val3%',....)
select * from tbl where my_col in ('%val1%', '%val2%', 'val3%',....)
这里有用的是 PostgreSQL 中可用的 LIKE ANY
谓词
What would be useful here would be a LIKE ANY
predicate as is available in PostgreSQL
SELECT *
FROM tbl
WHERE my_col LIKE ANY (ARRAY['%val1%', '%val2%', '%val3%', ...])
不幸的是,该语法在 Oracle 中不可用.您可以使用 OR
扩展量化比较谓词,但是:
Unfortunately, that syntax is not available in Oracle. You can expand the quantified comparison predicate using OR
, however:
SELECT *
FROM tbl
WHERE my_col LIKE '%val1%' OR my_col LIKE '%val2%' OR my_col LIKE '%val3%', ...
或者,使用 EXISTS
谓词和 辅助数组数据结构创建半连接(请参阅此详情的问题):
Or alternatively, create a semi join using an EXISTS
predicate and an auxiliary array data structure (see this question for details):
SELECT *
FROM tbl t
WHERE EXISTS (
SELECT 1
-- Alternatively, store those values in a temp table:
FROM TABLE (sys.ora_mining_varchar2_nt('%val1%', '%val2%', '%val3%'/*, ...*/))
WHERE t.my_col LIKE column_value
)
对于真正的全文搜索,您可能需要查看 Oracle Text:http://www.oracle.com/technetwork/database/enterprise-edition/index-098492.html
For true full-text search, you might want to look at Oracle Text: http://www.oracle.com/technetwork/database/enterprise-edition/index-098492.html
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