Solved

SUM up DISTINCT GROUP BY results MySQL

Posted on 2008-10-01
8
800 Views
Last Modified: 2010-04-21
I'm using a query:

SELECT DISTINCT COUNT(o.organizationId) as rowcount
FROM CatalogListings cl, Organization o
WHERE cl.active = '1'
GROUP BY alphabetizeUnderLetter


This returns 2 results lets say, since its grouping by...

___Rowcount__
  3
  2

I want to get the numbers merged into "5" at the MySQL level through the query.
Thanks guys!
0
Comment
Question by:MattKenefick
  • 4
  • 2
  • 2
8 Comments
 
LVL 142

Accepted Solution

by:
Guy Hengel [angelIII / a3] earned 500 total points
ID: 22614482
this should do it:
SELECT SUM(rowcount)
FROM (
SELECT DISTINCT COUNT(o.organizationId) as rowcount 
FROM CatalogListings cl, Organization o 
WHERE cl.active = '1'
GROUP BY alphabetizeUnderLetter
)

Open in new window

0
 
LVL 6

Expert Comment

by:RemcovC
ID: 22614495
I'd say remove the GROUP BY clause
0
 
LVL 4

Author Comment

by:MattKenefick
ID: 22614596
For that query above.. I get:

#1248 - Every derived table must have its own alias


Remcov : Can't remove the GROUP BY. There is more to the query (searching) that makes it necessary.
0
PRTG Network Monitor: Intuitive Network Monitoring

Network Monitoring is essential to ensure that computer systems and network devices are running. Use PRTG to monitor LANs, servers, websites, applications and devices, bandwidth, virtual environments, remote systems, IoT, and many more. PRTG is easy to set up & use.

 
LVL 4

Author Comment

by:MattKenefick
ID: 22614676
Nevermind.. if I use an "as X" identifier at the end, it works.

Thanks !
0
 
LVL 4

Author Closing Comment

by:MattKenefick
ID: 31501977
make sure to add an "AS tablename" or you'll catch errors
0
 
LVL 6

Expert Comment

by:RemcovC
ID: 22615235
MattKanefick

I thought so, but a coworker of mine was absolutely sure that it was the solution and persuaded me to post it.

Sorry for the inconvenience
0
 
LVL 142

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 22615308
>I thought so, but a coworker of mine was absolutely sure that it was the solution and persuaded me to post it.

the problem with just removing the GROUP BY is if 2 values of alphabetizeUnderLetter have the same value of organizationId, that suggested method will return 1, and it should return 2 (1 for each Letter value)...

which leads to a "simpler" version:
SELECT DISTINCT COUNT( CONCAT( o.organizationId,alphabetizeUnderLetter)  ) as rowcount 
FROM CatalogListings cl, Organization o 
WHERE cl.active = '1'

Open in new window

0
 
LVL 4

Author Comment

by:MattKenefick
ID: 22615354
Remcov: No problem. All possibly solutions are welcome. Who knows, maybe a wrong answer today will be the right answer tomorrow.
0

Featured Post

Microsoft Certification Exam 74-409

Veeam® is happy to provide the Microsoft community with a study guide prepared by MVP and MCT, Orin Thomas. This guide will take you through each of the exam objectives, helping you to prepare for and pass the examination.

Question has a verified solution.

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

Suggested Solutions

Title # Comments Views Activity
Query Optimization 14 44
Restore of mysql database from .SQL file - using Coldfusion 5 37
SQL Insert parts by customer 12 34
syntax sql error 2 12
I have a large data set and a SSIS package. How can I load this file in multi threading?
Introduction This article is intended for those who are new to PHP error handling (https://www.experts-exchange.com/articles/11769/And-by-the-way-I-am-New-to-PHP.html).  It addresses one of the most common problems that plague beginning PHP develop…
Familiarize people with the process of retrieving data from SQL Server using an Access pass-thru query. Microsoft Access is a very powerful client/server development tool. One of the ways that you can retrieve data from a SQL Server is by using a pa…
This video shows how to set up a shell script to accept a positional parameter when called, pass that to a SQL script, accept the output from the statement back and then manipulate it in the Shell.

777 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