Solved

SQL Query Assistance (Disinct Count)

Posted on 2014-12-30
25
128 Views
Last Modified: 2015-01-20
Hello, I am looking for assistance creating a query.  I need to count the number of distinct records based on the name of the record and rated by category.  See sample below.

SELECT DISTINCT COUNT(*) AS Count, [Category]
FROM [table name]
WHERE NAME LIKE [Name]
GROUP BY [Category]

I want to list the following columns
Count| Name| Category

Thank you
0
Comment
Question by:blackcatkempo
[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
  • 12
  • 12
25 Comments
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40524313
It might help if you give some sample data and sample expected output,
0
 

Author Comment

by:blackcatkempo
ID: 40524342
Attached is sample data.  I need to sort through thousands of records similar.  Key columns Category and Name.  I want a count of a distinct name based on category.
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40524402
Nothing attached.
0
Best Practices: Disaster Recovery Testing

Besides backup, any IT division should have a disaster recovery plan. You will find a few tips below relating to the development of such a plan and to what issues one should pay special attention in the course of backup planning.

 
LVL 69

Expert Comment

by:Scott Pletcher
ID: 40524421
COUNT can use DISTINCT inside it, which is really useful:

SELECT COUNT(DISTINCT Name) AS Count_Distinct_Names, [Category]
FROM [table name]
WHERE NAME LIKE [Name] --??
GROUP BY [Category]
0
 

Author Comment

by:blackcatkempo
ID: 40524638
0
 

Author Comment

by:blackcatkempo
ID: 40524813
I think I need to clarify.  Using the sample-data.csv.  I would like to report the total number of distinct records (name) that meet a specific criteria (category).  

SELECT DISTINCT COUNT(*) AS Count, Risk
FROM [tablename]
WHERE NAME LIKE 'Microsoft XML Parser%'
GROUP BY Category
ORDER BY Category ASC

Results in (columns in pipes) ---- I want another column with the name, which is related/assigned a category.
Count|Category
32| High

Need to result with following columns.

Count of Distinct Name | Name | Category

Please let me know if I need to provide further detail.  Thank you.
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40525030
Without a sample data set leading to a sample result, I cannot help you.

Your attached sample data does not lead to a result of 32, High, and therefore I would still be guessing as to what you want.
0
 

Author Comment

by:blackcatkempo
ID: 40525441
Attached is a sanitized version for sample data.  Using this file, I would like to see the following columns in a query displaying the number of distinct occurrences of the name record.

Count of Distinct Name | Name | Risk

Count of distinct name - the number of instances of the name
Name - the name of the risk
Risk - the risk level

Does this help?
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40525532
Select [Name], Count(Distinct [Name]) as CountOfDistinctName, Risk
From [tablename]
Group by [Name], Risk

From what you have described, the only thing wrong with your previous example was the placement of the word "Distinct".
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40525535
Note that it is very similar to Scott's solution.
0
 

Author Comment

by:blackcatkempo
ID: 40525624
Thank you for the query. I would like to see the number of distinct occurrences (risks) displayed for each name .  The script your provided only displays a "1" for each row (name).  I apologize if i am not explaining properly.
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40525939
Then

Select [Name], Count(Distinct [Risk]) as CountOfDistinctName, Risk
From [tablename]
Group by [Name], Risk
0
 

Author Comment

by:blackcatkempo
ID: 40525960
Sorry that query provided the same result - only displays a "1".  I have confirmed there are more then one instance per name using a separate individual query.
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40525975
Select [Name],
    (select Count(Distinct [Risk])
    From [tablename] as U
    Where T.[Name] = U.[Name]) as CountOfDistinctName,
Risk
From [tablename] as T
Group by [Name], Risk
0
 

Author Comment

by:blackcatkempo
ID: 40525995
Same result.
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40527653
Try this:

Select [Name], 
    (select Count([Risk]) 
    From [tablename] as U
    Where T.[Name] = U.[Name]) as CountOfDistinctName,
Risk
From [tablename] as T
Group by [Name], Risk

Open in new window

0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40527654
This gives this result:

Adobe Reader Detection      1      None
Adobe Reader Enabled in Browser (Internet Explorer)      1      None
Antivirus Software Check      1      None
BIOS Version (WMI)      1      None
Citrix Online Plug-in Installed      1      None
Citrix Receiver / Online Plug-in Remote Code Execution (CTX134681)      1      High
Common Platform Enumeration (CPE)      4      None
Computer Manufacturer Information (WMI)      1      None
DCE Services Enumeration      4      None
0
 

Author Comment

by:blackcatkempo
ID: 40527673
Thanks Phillip, can I also use Count(Distinct Host)?  The code you provided works great, however the data table contains multiple risks for the same host.  I just want to count the single risk by host.
0
 
LVL 24

Accepted Solution

by:
Phillip Burton earned 500 total points
ID: 40527678
Yes, you can - note the difference in DCE Services Enumeration:

Adobe Reader Detection      1      None
Adobe Reader Enabled in Browser (Internet Explorer)      1      None
Antivirus Software Check      1      None
BIOS Version (WMI)      1      None
Citrix Online Plug-in Installed      1      None
Citrix Receiver / Online Plug-in Remote Code Execution (CTX134681)      1      High
Common Platform Enumeration (CPE)      4      None
Computer Manufacturer Information (WMI)      1      None
DCE Services Enumeration      1      None
0
 

Author Closing Comment

by:blackcatkempo
ID: 40544906
Thank you!
0
 

Author Comment

by:blackcatkempo
ID: 40559117
Hi Phillip,

One more follow up question regarding this query you crafted for me.  For some reason I can not use the WHERE IN clause for the column risk. I am specifically looking for the number of Critical, High, etc.  The results provide the number of all rows with a risk.  See query below. Thank you.

Select Distinct [Host], Risk,
    (select Count (Distinct [Name])
    From [tablename] as U
    Where T.[Host] = U.[Host]) as CountOfDistinctRisk
From [tablename] as T
WHERE Risk IN ('Critical','High')
Group by [Host], Name, Risk
ORDER by [CountofDistinctRisk] DESC
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40559373
Shouldn't it be inside the brackets?

Select Distinct [Host], Risk,
    (select Count (Distinct [Name])
    From [tablename] as U
    Where T.[Host] = U.[Host] AND Risk IN ('Critical','High')) as CountOfDistinctRisk
From [tablename] as T
Group by [Host], Name, Risk
ORDER by [CountofDistinctRisk] DESC

Open in new window

0
 

Author Comment

by:blackcatkempo
ID: 40559524
Having the AND statement inside of the sub query provides different results that are not accurate.  I actually do no know what it is reporting/resulting, essentially the countofdistinctrisk column produces a number that is not made up the total risks aggregate or something else.  We need the count of total number of critical, high, per hostname.
0
 
LVL 24

Expert Comment

by:Phillip Burton
ID: 40559579
Sounds like this would work, if all you want is a filtered list:

Select [Name], 
    (select Count([Risk]) 
    From [tablename] as U
    Where T.[Name] = U.[Name]) as CountOfDistinctName,
Risk
From [tablename] as T
WHERE Risk IN ('Critical','High')
Group by [Name], Risk

Open in new window


I wouldn't understand why it wouldn't work.
0

Featured Post

What Is Transaction Monitoring and who needs it?

Synthetic Transaction Monitoring that you need for the day to day, which ensures your business website keeps running optimally, and that there is no downtime to impact your customer experience.

Question has a verified solution.

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

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…
In this article I will describe the Copy Database Wizard method as one possible migration process and I will add the extra tasks needed for an upgrade when and where is applied so it will cover all.
Monitoring a network: how to monitor network services and why? Michael Kulchisky, MCSE, MCSA, MCP, VTSP, VSP, CCSP outlines the philosophy behind service monitoring and why a handshake validation is critical in network monitoring. Software utilized …
This is my first video review of Microsoft Bookings, I will be doing a part two with a bit more information, but wanted to get this out to you folks.

717 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