?
Solved

Column xxx is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause

Posted on 2011-03-02
8
Medium Priority
?
830 Views
Last Modified: 2012-05-11
Dear All,

I have a very simple query which executed no problems:

select  * from Job

However if I just want to add 1 more stuff, it complains the error in the title:

select MAX(mas_no), * from Job


I know I could use query like:

select top 1 * from Job order by mas_no desc

to replace. But my question is: if I still want to use my query, how to do that? I don't want to list all columns in Job table like:

select MAX(mas_no), * from Job
group by Job.seq, job.type, job.xxx, job.yyy......


because there are too many colums.....




P.S. My final problem is i need to get a count(..) from another table, plus all Job table columns, like:

select count(r.Rec_Seq),Job.* from Job
Left JOIN [References] r on r.Rec_Seq  = Job.Seq
GROUP BY
   r.Rec_Seq,
   (... do not want to list all job's columns...)


Thanks heaps!


0
Comment
Question by:chenyuhao88
[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
  • 5
  • 2
8 Comments
 
LVL 57

Accepted Solution

by:
Raja Jegan R earned 1200 total points
ID: 35023803
>> select MAX(mas_no), * from Job

Few recommended coding practices:

1. Don't have * in your SELECT statement and type in all columns.
2. Don't leave a column without a name. For aggregated/ derived columns you can use alias like

select MAX(mas_no) as Max_mas_no, *
from Job
GROUP BY job.* -- Replace * with all column names present in the table.

in order to avoid issues like this.

>> select count(r.Rec_Seq),Job.* from Job
Left JOIN [References] r on r.Rec_Seq  = Job.Seq
GROUP BY
   r.Rec_Seq,

If you want to avoid all columns in GROUP BY, then you can use like this

select count(r.Rec_Seq) over ( partition by job.primar_key_or_some_other_column) cnt ,Job.*
from Job
Left JOIN [References] r on r.Rec_Seq  = Job.Seq
0
 
LVL 22

Assisted Solution

by:Thomasian
Thomasian earned 800 total points
ID: 35023831
>>P.S. My final problem is i need to get a count(..) from another table, plus all Job table columns, like:

You could avoid using GROUP BY by getting the count in a subquery before joining the tables
select Job.*, r.ReqCount 
from Job Left JOIN
    (SELECT Req_Seq, COUNT(1) ReqCount
       FROM [References]
       GROUP BY Req_Seq
    ) r on r.Rec_Seq = Job.Seq

Open in new window

0
 

Author Comment

by:chenyuhao88
ID: 35023865
Hi rrjegan17, you solution looks quite good. just one issue: the result gives me duplicated Job records when that job has multiple records in References table.

I found another solution like:


select match.cnt, j.* from Job j
left join
(
      select Rec_Seq,COUNT(*) as cnt
      from [References]
      group by Rec_Seq
)  match
on match.Rec_Seq = j.Seq



If my Job table has 200 rows of records, and it gives me exact 200 rows of record, plus 1st column tells me how many matching records in References table.




Is it possible to further change your solution?


Thanks again!


0
Get real performance insights from real users

Key features:
- Total Pages Views and Load times
- Top Pages Viewed and Load Times
- Real Time Site Page Build Performance
- Users’ Browser and Platform Performance
- Geographic User Breakdown
- And more

 

Author Comment

by:chenyuhao88
ID: 35023876
Thomasian, your one is the same as my solution! Thanks mate!
0
 
LVL 57

Expert Comment

by:Raja Jegan R
ID: 35023891
>> just one issue: the result gives me duplicated Job records when that job has multiple records in References table.

Then try this..

select DISTINCT count(r.Rec_Seq) over ( partition by job.Seq) cnt ,Job.*
from Job
Left JOIN [References] r on r.Rec_Seq  = Job.Seq
0
 

Author Comment

by:chenyuhao88
ID: 35024008
Hi rrjegan17,

Here is the error, which I think it's because there are 5 columns in Job table are "Text" type.


Msg 421, Level 16, State 1, Line 2
The text data type cannot be selected as DISTINCT because it is not comparable.
Msg 421, Level 16, State 1, Line 2
The text data type cannot be selected as DISTINCT because it is not comparable.
Msg 421, Level 16, State 1, Line 2
The text data type cannot be selected as DISTINCT because it is not comparable.
Msg 421, Level 16, State 1, Line 2
The text data type cannot be selected as DISTINCT because it is not comparable.
Msg 421, Level 16, State 1, Line 2
The text data type cannot be selected as DISTINCT because it is not comparable.



how could I do that? (assume not change from Text to varchar(max))

Thanks again!!

0
 

Author Comment

by:chenyuhao88
ID: 35024061
Hi rrjegan17,

Don't worry about that. I changed column type to Varchar(Max) and it works perfect!

Thanks again all!!

0
 

Author Closing Comment

by:chenyuhao88
ID: 35024065
2 different solution, and 1st one is perfect!
0

Featured Post

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

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

Slowly Changing Dimension Transformation component in data task flow is very useful for us to manage and control how data changes in SSIS.
The Delta outage: 650 cancelled flights, more than 1200 delayed flights, thousands of frustrated customers, tens of millions of dollars in damages – plus untold reputational damage to one of the world’s most trusted airlines. All due to a catastroph…
Using examples as well as descriptions, and references to Books Online, show the documentation available for date manipulation functions and by using a select few of these functions, show how date based data can be manipulated with these functions.
Via a live example, show how to set up a backup for SQL Server using a Maintenance Plan and how to schedule the job into SQL Server Agent.
Suggested Courses

741 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