This question already has answers here:
ORA-00979 not a group by expression
(10 answers)
Closed 7 years ago.
I am trying show all the different companies for which students work. However only companies where more than four students are employed should be displayed.
This is what I have so far:
SELECT EMPLOYER, COUNT (STUDENT_ID)
FROM STUDENT
GROUP BY STUDENT_ID
HAVING COUNT (STUDENT_ID) >4;
I keep getting this message:
ERROR at line 1:
ORA-00979: not a GROUP BY expression
I don't get it. I also tried this earlier:
SELECT STUDENT.EMPLOYER, COUNT (STUDENT.STUDENT_ID)
FROM STUDENT
GROUP BY STUDENT.STUDENT_ID
HAVING COUNT (STUDENT.STUDENT_ID) >4;
but nothing seems to work. Any help is appreciated. I am on SQL*Plus if that helps.
Try:
SELECT EMPLOYER, COUNT (STUDENT_ID)
FROM STUDENT
GROUP BY EMPLOYER
HAVING COUNT (STUDENT_ID) >4;
- this will return a list of all employers with more than 4 students.
When grouping or including aggregated fields, your select statement should only include fields that are either aggregated or included in the group by clause - in your existing select, you are including EMPLOYER in your select clause, but not grouping by it or aggregating it.
Related
This question already has answers here:
Fetch the rows which have the Max value for a column for each distinct value of another column
(35 answers)
Closed 1 year ago.
Write a query to display the student names and the maximum mark scored by them in any subject, ordered by name in ascending order. Give an alias to the maximum mark as MAX_MARK.
I am not able to find the logic for this. Kindly help me with it. Do it in oracle SQL I am at beginner level in SQL.
SELECT MAX(M.VALUE), S2.SUBJECT_ID,M.STUDENT_ID, S2.SUBJECT_NAME,S2.SUBJECT_CODE
from Mark M INNER JOIN SUBJECT S2
ON M.SUBJECT_ID=S2.SUBJECT_ID group BY S2.SUBJECT_ID,
S2.SUBJECT_CODE, S2.SUBJECT_NAME;
I am getting error with this query if I get this student id with the help of the above query then I can easily solve this question using subquery concept.
You don't need subject there. Question asks Max mark per student, regardless of subject:
SELECT s.Student_Name, MAX(M.VALUE) as MAX_MARK
from Student s
inner Join Mark M on m.student_id = s.student_id
group by s.student_id, s.student_name
order by s.student_name;
This question already has answers here:
"You tried to execute a query that does not include the specified aggregate function"
(3 answers)
Closed 6 years ago.
select Customers.cust_id, count(Orders.cust_id)
from Customers left outer join Orders
on Customers.cust_id=Orders.cust_id
group by Customers.cust_id
This correctly displays everything.
select Customers.cust_id, ***Customers.cust_name***, count(Orders.cust_id)
from Customers left outer join Orders
on Customers.cust_id=Orders.cust_id
group by Customers.cust_id
,,Your query does not include the specified expression 'cust_name' as port of an aggregate function."
Why is that? Each cust_id in Customers has a name in cust_name. Why do I get this error message?
When you use an aggregate function count() all other fields (that aren't used with an aggregate function) must appear in the Group By clause.
Here is my explanation as to why:
Aggregate functions operate across groups.
(That is, unless no groups or other fields are specified, in which case they operate across the whole recordset by default. For example, SELECT Sum(Salary) FROM Staff works.)
If you group by cust_id then it knows what to output, a count for each cust_id. But what would it do with the cust_name's? Which cust_name would it, or should it, display for each cust_id output? What if there are several cust_name's for a cust_id? It will only display one row for each cust_id, so what name should it display alongside it? It won't make the assumption that there is exactly one cust_name to correspond to one cust_id.
If there is one cust_name per cust_id then grouping by both will produce the same number of rows (as for cust_id alone) and provide consistent, and reliable, behaviour.
select Customers.cust_id, Customers.cust_name, count(Orders.cust_id)
from Customers left outer join Orders
on Customers.cust_id=Orders.cust_id
group by Customers.cust_id, Customers.cust_name
This question already has answers here:
ORA-00979 not a group by expression
(10 answers)
Closed 8 years ago.
Can anyone please help me figure this out,
I have 3 tables: Customer, Products and Products_ordered and I am trying to find customers who have ordered more than 1 product. Here is my query:
SELECT customer_id, product_id
FROM product_ordered
GROUP BY customer_id
HAVING COUNT (customer_id)>1;
I am getting this error:
Error report:
SQL Error: ORA-00979: not a GROUP BY expression
00979. 00000 - "not a GROUP BY expression"
Thanks for helping
try
select customer_id, product_id from product_ordered group by customer_id,product_id having count (customer_id)>1;
Try:
SELECT customer_id
FROM product_ordered
GROUP BY customer_id
HAVING COUNT (customer_id)>1;
The issue is that product_id is not part of the group by. If you do a group by, you can only select columns in the group by or use an aggregate function. That query will return the customer_id's that occur more then once. I don't know your table structure, but if you want more data then just the id let us know what sql version you are using, SQL Sever, MYSQL, or Oracle, and I can try to write something with windowing functions.
Don't you really want to select Customers who ordered more than one product?
More than one order line, or more than one product, or more than one unique product?
If you run as an inline query
(select customer_id from product_ordered group by customer_id having count (customer_id) > 1)
You will see all customers who placed more than one order line. But there could be multiple lines in an order, or multiple orders of one line, yada yada...
Try select customer_id from product_ordered group by customer_id having count(distinct product_id)>1 which will let you actually see customers who bought more than one unique product.
This question already has an answer here:
PGError: ERROR: column "p.name" must appear in the GROUP BY clause or be used in an aggregate function
(1 answer)
Closed 8 years ago.
I am using postgreSQL version
PostgreSQL 9.1.9 on x86_64-unknown-linux-gnu, compiled by gcc (Ubuntu/Linaro 4.7.2-22ubuntu5) 4.7.2, 64-bit,my question is am joining two tables,Let name it as temp1 and temp2 ,here i need to join this two table
Table structure is
marks_map
marks int
stud_id int
student
stud_id int
class_id int
here my query
select class_id,stud_id,count(marks)
from student as s
inner join marks_map as m on (s.stud_id=m.stud_id) group by stud_id
Here i get error as
ERROR: column "s.class_id" must appear in the GROUP BY clause or be used in an aggregate function
Why does this error happen? If I use class_id in group by it's running successfully.
You have to add the class_id attribute to your group by clause because in your select part of the statement there is no aggregation function over this attribue.
In GROUP BY statments you have to add all the attributes over which you haven't aggregated after the GROUP BY clause.
For example:
SELECT
non-aggregating-attr-1, non-aggregating-attr2, non-aggregating-attr3, sum(attr4)
FROM
table
GROUP BY
non-aggregating-attr-1, non-aggregating-attr2, non-aggregating-attr3
That's the way group by work.
You can check your data like
select
array_agg(class_id) as arr_class_id,
stud_id, count(marks)
from student as s
inner join marks_map as m on (s.stud_id=m.stud_id)
group by stud_id
and see how much class_id you have for each group. Sometimes your class_id is dependant from stud_id (you have only one elemnet in array for each group), so you can use dummy aggregate like:
select
max(class_id) as class_id,
stud_id, count(marks)
from student as s
inner join marks_map as m on (s.stud_id=m.stud_id)
group by stud_id
You should be able to understand the problem on a simplified case that doesn't even involve a JOIN.
The query SELECT x,[other columns] GROUP BY x expresses the fact that for every distinct value of x, the [other columns] must be output with only one row for every x.
Now looking at a simplified example where the student table has two entries:
stud_id=1, class_id=1
stud_id=1, class_id=2
And we ask for SELECT stud_id,class_id FROM student GROUP BY class_id.
There is only one distinct value of stud_id, which is 1.
So we're telling the SQL engine, give me one row with stud_id=1 and the value of class_id that comes with it. And the problem is that there is not one, but two such values, 1 and 2. So which one to choose? Instead of choosing randomly, the SQL engine yields an error saying the question is conceptually bogus in the first place, because there's no rule that says each distinct value of stud_id has its own corresponding class_id.
On the other hand, if the non-GROUP'ed output columns are aggregate functions that transform a series of values into just one, like min, max, or count, then they provide the missing rules that say how to get only one value from several. That's why the SQL engine is OK with, for instance: SELECT stud_id,count(class_id) FROM student GROUP BY stud_id;.
Also, when faced with the error column "somecolumn" must appear in the GROUP BY clause, you don't want to just add columns to the GROUP BY until the error goes away, as if it was purely a syntax problem. It's a semantic problem, and each column added to the GROUP BY changes the sense of the question submitted to the SQL engine.
That is, GROUP BY x,y means for each distinct value of the (x,y) couple. It does not mean GROUP BY x, and hey, since it leads to an error, let's throw in the y as well!
This question already has answers here:
ORA-00979 not a group by expression
(10 answers)
Closed 8 years ago.
I'm relatively new to databases. I am using Oracle and I'm trying to implement this query to find the number of personal training sessions the member has had.
The tables are;
MEMBERS
MEMBERS_ID(NUMBER),
MEMBERSHIP_TYPE_CODE(VARCHAR),
ADDRESS_ID(NUMBER), CLUB_ID(NUMBER)
MEMBER_NAME(VARCHAR),
MEMBER_PHONE(VARCHAR),
MEMBER_EMAIL(VARCHAR)
PERSONAL_TRAINING_SESSIONS
SESSION_ID(VARHCAR),
MEMBER_ID (NUMBER),
STAFF_ID(VARCHAR),
SESSION_DATETIME(DATE)
My query is returing this error:
ORA-00979: not a GROUP BY expression
00979. 00000 - "not a GROUP BY expression"
*Cause:
*Action: Error at Line: 1 Column: 8
SELECT MEMBERS.MEMBER_ID,MEMBERS.MEMBER_NAME, COUNT(personal_training_sessions.session_id)
FROM MEMBERS JOIN personal_training_sessions
ON personal_training_sessions.member_id=members.member_id
GROUP BY personal_training_sessions.session_id;
Can anyone point me in the right direction? I have looked around do I need to separate the count query?
The error says it all, you're not grouping by MEMBERS.MEMBER_ID and MEMBERS.MEMBER_NAME.
SELECT MEMBERS.MEMBER_ID, MEMBERS.MEMBER_NAME
, COUNT(personal_training_sessions.session_id)
FROM MEMBERS
JOIN personal_training_sessions
ON personal_training_sessions.member_id = members.member_id
GROUP BY MEMBERS.MEMBER_ID, MEMBERS.MEMBER_NAME
You want the count of personal sessions per member, so you need to group by the member information.
The basic (of course it can get a lot more complex) GROUP BY, SELECT query is:
SELECT <column 1>, <column n>
, <aggregate function 1>, <aggregate function n>
FROM <table_name>
GROUP BY <column 1>, <column n>
An aggregate function being, as Ken White says, something like MIN(), MAX(), COUNT() etc. You GROUP BY all the columns that are not aggregated.
This will only work as intended if your MEMBERS table is unique on MEMBER_ID, but based on your query I suspect it is. To clarify what I mean, if your table is not unique on MEMBER_ID then you're not counting the number of sessions per MEMBER_ID but the number of sessions per MEMBER_ID and per MEMBER_NAME. If they're in a 1:1 relationship then it's effectively the same thing but if you can have multiple MEMBER_NAMEs per MEMBER_ID then it's not.
SELECT MEMBERS.MEMBER_ID,
MEMBERS.MEMBER_NAME,
COUNT(personal_training_sessions.session_id)
FROM MEMBERS JOIN personal_training_sessions
ON personal_training_sessions.member_id=members.member_id
GROUP BY personal_training_sessions.session_id;
You are using a COUNT function, thus the other columns, MEMBER_ID & MEMBER_NAME, must be included in the group by clause.