RE:

From: Christine Turner <christine.turner_at_ips-sendero.com>
Date: Tue, 19 Sep 2000 08:44:18 -0000
Message-Id: <10624.117356@fatcity.com>

THANK YOU, THANK YOU, THANK YOU. That definately worked! Sincerely appreciate the help!

Thanks Again,
Christine :o)

-----Original Message-----

From:	Rachel Carmichael [SMTP:carmichr_at_hotmail.com]
Sent:	Monday, September 18, 2000 6:30 PM
To:	Multiple recipients of list ORACLE-L
Subject:	Re: 

Christine,

I'm hoping you mean find all rows, not all columns. If so, it's fairly easy.

There is a 'replace' command in SQL, you can use that.

update <tablename> set <column>=

         replace(<column>,'E:\Jun30backup\','\\ServerName\backup\')\
      where <column> like '%E:\Jun30backup\%';

should do the trick.

Rachel

>From: Christine Turner <christine.turner_at_ips-sendero.com>
>Reply-To: ORACLE-L_at_fatcity.com
>To: Multiple recipients of list ORACLE-L <ORACLE-L_at_fatcity.com>
>Date: Mon, 18 Sep 2000 13:46:51 -0800
>
>Hello All.........
>
>I have an interesting request from a co-worker of mine. Listed below is a
>sub-set of data from a table in one of our databases. The request was a
>basic search and replace. Find all columns with the string "E:\Jun30
>backup\" in them and replace it with "\\ServerName\backup\".
>
>PATH_NAME
>------------------------------------------------------------------------
>-----------
>E:\April30 backup\daily_gl_baleom.prn
>E:\May30 backup\dailyglbal.prn
>E:\Jun30 backup\BORROWALS30.LST
>E:\Jun30 backup\dda30.LST
>E:\Jun30 backup\loans30.LST
>
>Anyone have ANY ideas???? I'm a little stumped so any help is appreciated.
>
>Thanks!!!
>Christine Turner
>DBA
>IPS Sendero
>Scottsdale, Arizona
>
>--
>Author: Christine Turner
> INET: christine.turner_at_ips-sendero.com
>
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-- 
Author: Rachel Carmichael
  INET: carmichr_at_hotmail.com

Fat City Network Services    -- (858) 538-5051  FAX: (858) 538-5051
San Diego, California        -- Public Internet access / Mailing Lists
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Received on Tue Sep 19 2000 - 03:44:18 CDT