using regular expressions in Oracle

I needed to select rows of data from a column where some of the records begin with a date in the form:
YYYYMMDD

The other records in the column had random text strings. For example:
N MESSAGE
- -------
1 20080124: Hello. I have written to you today to compliment you
2 I have a problem with one of the products you sell.

I thought it would be perfect to use regular expressions to pluck out only the records I needed. Luckily, Oracle now supports for regular expressions as of version 10G:
Writing Better SQL Using Regular Expressions

I needed to first create the regular expression to match the date. In simple terms, I want to match the eight-digit string.

First, here is a regular expression that matches any digit, 0-9:
[0-9]

Secondly, we can expand this match to include the entire eight digit date:
[0-9]\{8\}

The escaped braces repeat the initial match for any number eight ("8") times. I will then put a carat (^) in front of the regular expression to match the beginning of the pattern buffer. The pattern buffer in this case is the MESSAGE column in our table:
^[0-9]\{8\}

OK! So we have our regular expression. Let's put it in a SELECT statement. We will use Oracle 10Gs new regexp_like condition to perform a regular expression match:
SELECT n,message
FROM message_table
WHERE regexp_like (message, '^[0-9]\{8\}');

N MESSAGE
- -------
1 20080124: Hello. I have written to you today to compliment you

Sweet! This works.

There is a slight problem. I did notice a substantial performance penalty for using regexp_like. For a very small table (<20,000 rows), it took about a minute to return the results of the search. Eeek. A normal WHERE clause came back in less than a second.

If I have any further information regarding performance tuning using regular expressions in Oracle, I will let you know.

Here are some more complex examples

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