Advertisement

11.29.2005 at 02:35PM PST, ID: 21647154
[x]
Attachment Details
[x]
The Solution Rating System

With so many solutions, how can you tell which solutions are most likely to help you and which ones are not? To provide you with a tool to use, we rate our solutions based on various elements that most accurately determine if a solution is a quality solution. To explain what factors affect the solution rating, here are the elements we take into consideration when formulating our solution rating.

  • The Grade of the Solution
  • The Zone Rank of the Expert Providing the Solution
  • The Number of Author and Expert Comments
  • The Number of Experts Contributing
  • The Feedback of the Community

Your Input Matters
Because of the way the system is set up, the most important variable in this equation is you. As a member of Experts Exchange, you are able to cast your vote on the quality of the solutions in regard to how complete, accurate, helpful and easy to understand each solution is. When you provide your feedback, each rating is adjusted accordingly. So, if you see a solution that has a poor rating that you think is a good solution, let us know by rating it. As you do, the rating will be adjusted and will become more accurate for other members of our site.

If you have any suggestions that you would like to make for our rating system, please ask a question in the Suggestions Zone of Community Support.

Thank you!

8.4

Need to SUM the sizes of DISTINCT files, so we don't count the sizes more than once

Asked by BigDavy in MS SQL Server

Hi, I need help with the SUM command.  I have a table called "today" that lists ads worked on today; some ads were worked on more than once.  I have another table called "sizes" with the sizes of all the ads in my database (not just today).   I want to subtotal the count and size of the ads that were worked on today, but I only want to count each ad once when adding up the sizes.

Table "today":
User   Ad
Joe    ABC
Mary  DEF
Mike   ABC
Joe     GHI

Table "sizes":
Ad      Inches
ABC    2
DEF    3
GHI     2
JKL      4

Desired result (2+3+2=7 inches):
Adcount   Inches
3             7

The following doesn't work right:

select count(distinct today.ad) as Adcount,sum(sizes.inches) as Inches from today,sizes where today.ad=sizes.ad

which produces the following result (2+3+2+2=9):
Adcount   Inches
3             9
which is wrong because the size of ad ABC was counted twice.

The following is also bad:

select count(distinct today.ad) as Adcount,sum(distinct sizes.inches) as Inches from today,sizes where today.ad=sizes.ad

which produces the following result (2+3=5):
Adcount   Inches
3             5
which is wrong because the three 2-inch ads are only counted as one 2-inch ad.

I'm a newbie, looking for a simple approach if there is one! Thanks!Start Free Trial
[+][-]11.29.2005 at 02:45PM PST, ID: 15384064

At Experts Exchange, members can ask their questions to thousands of technology professionals, also known as Experts. Experts compete and collaborate to answer those questions by leaving comments like this one.

Start your 7-day free trial to view this Expert Comment or ask the Experts your question.

 
[+][-]11.29.2005 at 02:51PM PST, ID: 15384105

At Experts Exchange, members can ask their questions to thousands of technology professionals, also known as Experts. Experts compete and collaborate to answer those questions by leaving comments like this one.

Start your 7-day free trial to view this Expert Comment or ask the Experts your question.

 
[+][-]11.29.2005 at 03:02PM PST, ID: 15384185

At Experts Exchange, members can ask their questions to thousands of technology professionals, also known as Experts. Experts compete and collaborate to answer those questions by leaving comments like this one.

Start your 7-day free trial to view this Expert Comment or ask the Experts your question.

 
[+][-]11.29.2005 at 04:30PM PST, ID: 15384584

Often, when Experts are collaborating with members who have asked questions, they will request additional information about the problem. Askers respond with an author comment like this one.

Start your 7-day free trial to view this Author Comment or ask the Experts your question.

 
[+][-]11.29.2005 at 07:16PM PST, ID: 15385206

View this solution now by starting your 7-day free trial. Setting up your free trial is quick, easy, and secure. We will return you to this solution, unlocked, when you're done.

 

About this solution

Zone: MS SQL Server
Sign Up Now!
Solution Provided By: acperkins
Participating Experts: 3
Solution Grade: A
 
 
[+][-]11.30.2005 at 01:00AM PST, ID: 15386543

Assisted solutions are selected by the member who asked the question as a comment that contributed to their question's solution.

Start your 7-day free trial to view this Assisted Solution or ask the Experts your question.

 
[+][-]11.30.2005 at 01:02AM PST, ID: 15386551

At Experts Exchange, members can ask their questions to thousands of technology professionals, also known as Experts. Experts compete and collaborate to answer those questions by leaving comments like this one.

Start your 7-day free trial to view this Expert Comment or ask the Experts your question.

 
[+][-]11.30.2005 at 02:19PM PST, ID: 15392237

Often, when Experts are collaborating with members who have asked questions, they will request additional information about the problem. Askers respond with an author comment like this one.

Start your 7-day free trial to view this Author Comment or ask the Experts your question.

 
[+][-]12.02.2005 at 02:29PM PST, ID: 15408640

Often, when Experts are collaborating with members who have asked questions, they will request additional information about the problem. Askers respond with an author comment like this one.

Start your 7-day free trial to view this Author Comment or ask the Experts your question.

 
[+][-]12.02.2005 at 05:52PM PST, ID: 15409512

Often, when Experts are collaborating with members who have asked questions, they will request additional information about the problem. Askers respond with an author comment like this one.

Start your 7-day free trial to view this Author Comment or ask the Experts your question.

 
 
Loading Advertisement...
20081112-EE-VQP-42