Re: Use SQL to search for a substring anywhere in row, and then update that substring?

That is exactly what Elvis' example does, I think. The POSITION function 
returns the spot where the string is in the field - you don't need to know that 
position ahead of time.

-------------- Original message -------------- 
From: "Ron Adams" <rondadams@xxxxxxxxx> 

I'm curious, along these lines, can you replace a string of characters where 
you don't know what specific position they are in? 

I'm having an issue where the plus sign ("+") has been added to a field 
across numerous records. Not sure what's happenig, it's happened multiple 
times to different users. 



On 8/8/06, Elvis Budimlic wrote: 

REPLACE is the way to go. 
Prior to V5R3 this UPDATE would take care of single instance of 'SD' in 
the 
field: 

UPDATE test SET f1 = SUBSTR(f1, 1, POSITION('SD' IN f1) - 1) || 
'SC' || 
SUBSTR(f1, POSITION('SD' IN f1) + 2) 
WHERE f1 LIKE '%SD%' 

You could run it multiple times to get all instances taken care of. You'd 
know you're done when you get "Rows not found for update" message (SQLCODE 
100). 

Elvis 

-----Original Message----- 
Subject: AW: Use SQL to search for a substring anywhere in row, and then 
update that substring? 

Hi, 

if you are on release V5R3 or higher, you may use the SQL Scalar function 
REPLACE 

update test 
set MyText = Replace(MyText, 'SC', 'SD'); 

Birgitta 

-----Ursprüngliche Nachricht----- 
Betreff: Use SQL to search for a substring anywhere in row, and then 
update 
that substring? 

I'm almost certain this can't be done, but I've thought that before on 
other 
SQL questions and have been consistently proven wrong. 

Can I use SQL to find a substring of 2 characters anywhere in a given 
field, 
then replace those 2 characters with 2 new characters? 

For example: 
Combination Notice - SC - 42 
EFT Schedule - SC - 19 
Courtesy Letter - SC - 41 

I would like to search for 'SC' in these rows, and replace any occurrences 
of it with 'SD'. The search string "should" exist no more than once in 
each 
row. 

TIA, 
Dan 



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