Solved

Join where LIKE?

Posted on 2007-11-16
3
149 Views
Last Modified: 2010-03-19
See tables below. The table PipeFull contains varchars up to 4,000 characters long that may or may not contain an acct number from table la_missing_nate.

The following SQL is running, but is this proper?

      select a.*, b.*
      from dbo.la_missing_nate a
      join dbo.PipeFull b
      on b.test like a.acct


CREATE TABLE [dbo].[PipeFull](
      [test] [varchar](max))

CREATE TABLE [dbo].[la_missing_nate](
      [acct] [nvarchar](50))
0
Comment
Question by:donnatronious
[X]
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
3 Comments
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20302658
why the like word?

select a.*, b.*
      from dbo.la_missing_nate a
      join dbo.PipeFull b
      on b.test = a.acct
0
 
LVL 27

Accepted Solution

by:
MikeToole earned 250 total points
ID: 20303771
For a LIKE comparison you need to specify wildcard characters. A % stands for any number of characters.

      select a.*, b.*
      from dbo.la_missing_nate a
      join dbo.PipeFull b
      on b.test like '%' + a.acct +'%'
0
 
LVL 25

Assisted Solution

by:imitchie
imitchie earned 250 total points
ID: 20303849
like Mike said, if you had 'Account X1234 has been archived' in PipeFull and 'X1234' in la_missing_nate
b.test like a.acct   means it has to math 'X1234' exactly, whereas
'%' + a.acct + '%' turns it into '%X1234%' which means

it can have any number of characters either side of a.acct
HOWEVER, there's a catch. if any of your account codes contain '%', your search expression becomes funny. The attached code far below shows you the what happens.  This code is more robust by escaping the %'s in the search string:

      select a.*, b.*
      from dbo.la_missing_nate a
      join dbo.PipeFull b on b.test like '%' + replace(a.acct, '%', '\%') + '%' escape '\'
create table ab ( a varchar(max) )
insert into ab values ('test account X123 is closed')
insert into ab values ('test account X1 test 23 is closed')
create table ac ( a varchar(10) )
insert into ac values ('X123')
insert into ac values ('X1%23')
 
select * from ab join ac on ab.a like '%' + ac.a + '%' escape '\'
select * from ab join ac on ab.a like '%' + replace(ac.a, '%', '\%') + '%' escape '\'

Open in new window

0

Featured Post

On Demand Webinar - Networking for the Cloud Era

This webinar discusses:
-Common barriers companies experience when moving to the cloud
-How SD-WAN changes the way we look at networks
-Best practices customers should employ moving forward with cloud migration
-What happens behind the scenes of SteelConnect’s one-click button

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 Update query with connected table data 3 63
SQL Query help 3 24
removing unwanted rows from an sql server ranked table 13 36
Need multiple Group By's 8 28
Occasionally there is a need to clean table columns, especially if you have inherited legacy data. There are obviously many ways to accomplish that, including elaborate UPDATE queries with anywhere from one to numerous REPLACE functions (even within…
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.
The goal of the tutorial is to teach the user how to use functions in C++. The video will cover how to define functions, how to call functions and how to create functions prototypes. Microsoft Visual C++ 2010 Express will be used as a text editor an…
The viewer will be introduced to the technique of using vectors in C++. The video will cover how to define a vector, store values in the vector and retrieve data from the values stored in the vector.

733 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