Question

SQL SUM and Join

Asked by: flow79

Hello all!

I am working to create a small report from our data to determine the profit per job.  In doing so, I have two tables customer_costs and actual_costs.  The profit can be found by adding the total_price fields per job_number in each table, and then subtracting these sums.  You will see the SQL syntax below.

My issue is this: Whenever I run this query, the answer is double the actual number desired.  I will display an example of each table to illustrate the correct numbers below, and then show you what is actually output by the query.

table: customer_costs
------------------------------------------------------------------
ID  |   job_number  |     total_price
------------------------------------------------------------------
1   |         46           |       135.00
------------------------------------------------------------------
2   |         46           |       247.50
------------------------------------------------------------------

table: actual_costs
------------------------------------------------------------------
ID  |   job_number  |     total_price
------------------------------------------------------------------
1   |         46           |       101.25
------------------------------------------------------------------
2   |         46           |       180.00
------------------------------------------------------------------

As you can see the total customer_cost for job_number 46 is: 382.50.  And the total actual_cost is: 281.25

Thus the profit should be: 382.50 - 281.25 = 101.25

The SQL statement below gives me: 202.50 (double the real amount).  Anyone know why the sum is doubled?  is it due to how many tables are in the join?  If so, how do I work around this (other than simply sub-querying each)?

SELECT SUM(cc.total_price - ac.total_price) FROM customer_costs cc, actual_costs ac WHERE cc.job_number = ac.job_number AND cc.job_number = '46'

Thanks!

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Asked On
2007-04-20 at 07:24:38ID22523881
Tags

sql

,

sum

,

join

Topic

PHP Scripting Language

Participating Experts
3
Points
50
Comments
5

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Answers

 

by: hernst42Posted on 2007-04-20 at 07:34:41ID: 18946699

The duplication is caused by the join. So if have multiple rows in each table with the same id you will need to use subqueries for each part the result you get for your join is:
The resulting table rows are for your example:
46 - 135 - 46 - 101.25
46 - 135 - 46 - 180
46 - 247.50 - 46 - 101.25
46 - 247.50 - 46 - 180

 

by: DanielWilsonPosted on 2007-04-20 at 07:37:53ID: 18946724

it's b/c there are 2 rows in each of the tables.  If you have an example w/ 3 or 4 rows in one or both of the tables, things will get worse.

What DBMS are you on?

This syntax should work in MS SQL ... MySQL would probably be somewhere close.

Select CC.Total_Price - AC.Total_Price From
 (Select sum(total_Price) as Total_Price From customer_cost where job_number = 46) CC,
 (Select sum(total_Price) as Total_Price From Actual_cost where job_number = 46) AC

What this does is summarizes before it joins so there is only 1 record on each side to be joined.

 

by: ZSoderquistPosted on 2007-04-20 at 07:39:59ID: 18946746

The reason you are getting the double amount is because your Join is doubling the records..

For each record in cc, you join to EACH matching record in ac.

So what you end up with is    

135 - 101.25
135 - 180.00
247.50 - 101.25
247.50 - 180.00
_______________+
202.50

 

by: flow79Posted on 2007-04-20 at 08:37:49ID: 18947247

So there is no way to do this without subqueries?  Currently I am using:

SELECT (SELECT SUM(total_price) FROM customer_costs WHERE job_number = '46') - (SELECT SUM(total_price) FROM actual_costs WHERE job_number = '46') as profit

Is this more or less efficient than the similar query written above by DanielWilson?

Thanks for any more info you can give me.

 

by: DanielWilsonPosted on 2007-04-20 at 09:04:15ID: 18947442

>>Is this more or less efficient than the similar query written above by DanielWilson?

I'd have to run it both ways to be sure, but I suspect no difference.

20120131-EE-VQP-002

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