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

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!


chenyuhao88Asked:
Who is Participating?
 
Raja Jegan RSQL Server DBA & Architect, EE Solution GuideCommented:
>> 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
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ThomasianCommented:
>>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

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chenyuhao88Author Commented:
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!


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chenyuhao88Author Commented:
Thomasian, your one is the same as my solution! Thanks mate!
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Raja Jegan RSQL Server DBA & Architect, EE Solution GuideCommented:
>> 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
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chenyuhao88Author Commented:
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!!

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chenyuhao88Author Commented:
Hi rrjegan17,

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

Thanks again all!!

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chenyuhao88Author Commented:
2 different solution, and 1st one is perfect!
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