oracle Where子句中的数字通配符
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Number Wildcard in Where Clause
提问by jmq
I'm trying to add a number wildcard to a query to look for a number in a specific position. The query looks something like this:
我正在尝试向查询添加数字通配符以在特定位置查找数字。查询如下所示：
SELECT SUBMITTER
FROM BASE_ELEMENT
WHERE SUBMITTER LIKE 'm_%';
The problem with this query is that it picks up everything that starts with "m" and has a character in the second position. I need something that work like a Unix wildcard:
此查询的问题在于它选取以“m”开头且在第二个位置有一个字符的所有内容。我需要一些像 Unix 通配符一样工作的东西：
'm[0-9]*'
I want it to include m0, m1, m2, etc, but exclude ma, mb, mc, etc.
我希望它包括 m0、m1、m2 等，但不包括 ma、mb、mc 等。
How would I accomplish this in Oracle 10g?
我将如何在 Oracle 10g 中完成此操作？
回答by Ben
In 10G you have the wonderof regular expressions. So, your query could be:
select submitter
from base_element
where regexp_like(submitter, '^m[[:digit:]]')
^anchors the expression to the start of the line and [[:digit:]]matches to any digit.
^将表达式锚定到行首并[[:digit:]]匹配任何数字。
回答by bruno.zambiazi
Try to use the REGEXP_LIKE function (http://docs.oracle.com/cd/B19306_01/appdev.102/b14251/adfns_regexp.htm).
尝试使用 REGEXP_LIKE 函数 (http://docs.oracle.com/cd/B19306_01/appdev.102/b14251/adfns_regexp.htm)。
Your case could be solved like this:
你的情况可以这样解决：
select submitter from base_element where regexp_like( submitter, '^m[0-9]' );
select submitter from base_element where regexp_like( submitter, '^m[0-9]' );
回答by Andomar
You could use regexp_like:
你可以使用regexp_like：
where regexp_like(submitter, '^m\d')
