SUBSTRING for a regular expression - Mailing list pgsql-general

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Subject SUBSTRING for a regular expression
Date
Msg-id 64617.206.53.65.243.1089074434.squirrel@$HOSTNAME
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Responses Re: SUBSTRING for a regular expression  (mike g <mike@thegodshalls.com>)
Re: SUBSTRING for a regular expression  ("Greg Sabino Mullane" <greg@turnstep.com>)
List pgsql-general
I can't seem to get right the regular expression for parsing data like
these four sample rows (names and addresses changed to ficticious values)
from a text-type column:

Yolanda Harris, 38, of 40 South Main St., Newtown City, was charged
Sunday with breach of peace and interfering with a police officer.

Allen K. George, 30, of 88 Beverly Court was charged Saturday with
possession of marijuana, third-degree criminal mischief, breach of peace,
evading responsibility, interfering with a police officer, driving with a
suspended license, driving under the influence of drugs or alcohol,
failure to drive right and failure to have proper insurance.

Brain T. Grafton, 18, of 97 Bristol Ave. was charged Sunday with
possession of marijuana, possession of alcohol by a minor and failure to
wear a seat belt.

Brian D. Sptizer Jr., 18, of 20 Walling Ave., Northford, was charged
Sunday with driving under the influence of drugs or alcohol, evading
responsibility and following too closely.

Into separate columns for: name, age, address, charge. For example the
first record would have

name='Yolanda Harris'
age=38
address='40 South Main St., Newtown City'
charge='was charged Sunday with breach of peace and interfering with a
police officer.'

To get the name, for instance, I tried

SELECT SUBSTRING(description FROM '^([:alnum:]*), \d{2}, .*$') FROM
police_log;

or the age value

SELECT SUBSTRING(description FROM '^[:alnum:]*, (\d{2}), .*$') FROM
police_log;

But return values are all NULL. Can anyone give me some RE help, please?

--Berend Tober




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