Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people, just like you, are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
Solved

How to remove char(13) + char(10)  from the END of a string

Posted on 2014-07-18
9
464 Views
Last Modified: 2014-07-21
I have a sql variable that holds appended record set values that include char(13) + char(10) .

At the very end of the string (not within the string) i will need to remove the char(13) + char(10) .

For example i could have a string variable with the following:

Address: char(13) + char(10) 689 Will Street char(13) + char(10) Columbus char(13) + char(10) GA

But the actual result i would be looking to be removed is bolded below:
Address: char(13) + char(10) 689 Will Street char(13) + char(10) Columbus char(13) + char(10) GA


Thanks,
0
Comment
Question by:MoreThanDoubled
  • 3
  • 2
  • 2
  • +1
9 Comments
 
LVL 47

Expert Comment

by:Dale Fye (Access MVP)
ID: 40205600
Didn't we just build this string the other day?

That isn't actually "at the very end of the string" is it, since the state characters follow those values.

You might start by using the InstrRev( ) function to look for the last position where those to characters are found together in the sting.

To check your table, you could try:

SELECT [yourField], Left(yourField, instrrev([yourField], chr$(13) & chr$(10))-1)
FROM yourTable
WHERE InstrRev(yourString, chr$(13) & chr$(10)) <> 0
0
 
LVL 69

Expert Comment

by:Scott Pletcher
ID: 40205672
Do you just want to SELECT the rows or do you want to UPDATE the values?

Here's the SELECT to do that; if you press Ctrl-T before running, you can see the affects of the CR+LF in the output results:

SELECT
    address_string AS address_string,
    STUFF(address_string, LEN(address_string) - CHARINDEX(char(10) + char(13), REVERSE(address_string)), 2, ' ') AS address_string_new

FROM (
    select char(13) + char(10) + '689 Will Street' + char(13) + char(10) + 'Columbus' + char(13) + char(10) + 'GA' as address_string
) AS test_data
0
 
LVL 32

Expert Comment

by:awking00
ID: 40206374
with t as
(select reverse(address) as address,charindex(char(10) + char(13), reverse(address) - 1) as pos
 from yourtable)
select reverse(left(address,pos) + ' ' + right(address,len(address) - pos - 2)) as newaddress
from t;
0
Optimizing Cloud Backup for Low Bandwidth

With cloud storage prices going down a growing number of SMBs start to use it for backup storage. Unfortunately, business data volume rarely fits the average Internet speed. This article provides an overview of main Internet speed challenges and reveals backup best practices.

 
LVL 47

Expert Comment

by:Dale Fye (Access MVP)
ID: 40206435
Sorry, i missed that this was posted to the SQL Server topic area
0
 
LVL 30

Accepted Solution

by:
hnasr earned 500 total points
ID: 40207125
For the mentioned  pattern, compare with:

declare @p varchar(100)
set @p='char(13) + char(10) 689 Will Street char(13) + char(10) Columbus char(13) + char(10) GA';
select left(@p,len(@p)-23)+right(@p,3)

Open in new window

0
 
LVL 32

Expert Comment

by:awking00
ID: 40208882
How well does that work if the address line was? -
Address: + char(13) + char(10) +  1 Main St. + char(13) + char(10) +  Ames + char(13) + char(10) +  IA
0
 
LVL 30

Expert Comment

by:hnasr
ID: 40209053
awking00,

For the mentioned  pattern, compare with:
set @p='+ char(13) + char(10) +  1 Main St. + char(13) + char(10) +  Ames + char(13) + char(10) +  IA '
set @p= rtrim(replace(@p,'  ',' '));
select left(@p,len(@p)-26)+right(@p,3)

Open in new window

0
 
LVL 32

Expert Comment

by:awking00
ID: 40209109
hnasr,
I agree that what you have posted will return what is wanted in these two cases, but I think what is really desired is code that will work for any combination of addresses.
0
 
LVL 69

Expert Comment

by:Scott Pletcher
ID: 40209132
Apparently that's not what is really desired, since I posted such code two days before hnasr's code.
0

Featured Post

Master Your Team's Linux and Cloud Stack!

The average business loses $13.5M per year to ineffective training (per 1,000 employees). Keep ahead of the competition and combine in-person quality with online cost and flexibility by training with Linux Academy.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
MS SQL BCP Extra Lines Between Records 2 24
Currency in SQL? 2 30
SQL Syntax 14 34
Clone table from one server.database to another server.database 24 35
This article explains how to reset the password of the sa account on a Microsoft SQL Server.  The steps in this article work in SQL 2005, 2008, 2008 R2, 2012, 2014 and 2016.
This article describes how to use the timestamp of existing data in a database to allow Tableau to calculate the prior work day instead of relying on case statements or if statements to calculate the days of the week.
Nobody understands Phishing better than an anti-spam company. That’s why we are providing Phishing Awareness Training to our customers. According to a report by Verizon, only 3% of targeted users report malicious emails to management. With compan…

860 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question