I would like to get average of product=A that a client have. Say inner select return 1,2,1,4,4,4 for 6 clients
I would like to see result as 4 which means the avg product count a client can have is 4
Can somebody please confirm the following.
E.g
Select avg(count)
From (
Select count(*) as count
From Table1
Where product = A
Group by client)
as counts
Having sample data is important to getting assistance. It's still difficult to determine how your data looks. Let's assume it looks like this:
create table table1 (
client varchar(10),
product varchar(10)
);
insert into table1 values
('xxx', 'A'),
('bbb', 'A'),
('bbb', 'A'),
('ccc', 'A'),
('ddd', 'A'),
('ddd', 'A'),
('ddd', 'A'),
('ddd', 'A'),
('tt', 'A'),
('tt', 'A'),
('tt', 'A'),
('tt', 'A'),
('bdad', 'A'),
('bdad', 'A'),
('bdad', 'A'),
('bdad', 'A');
I don't have access to a DB2 database, but this query works for most dbms types. You may need to tweak to fit DB2.
select purchased as most_common_value
from (
select client, count(*) as purchased
from table1
where product = 'A'
group by client
)z
group by purchased
order by count(client) desc
limit 1
Output of query is:
most_common_value
4
Related
I'm new to sql and I'm wondering how to extract the relevant data from the sites and plugins table using the sites_plugins table. The data I am interested in is sites.description, plugins.fullName, plugins.currentVersion and sites_plugins.syncedAt
Below are the sql tables
INSERT INTO sites (id, name, description) VALUES
(1, "facebook", "Facebook"),
(2, "amazon", "Amazon"),
(3, "google", "Google");
INSERT INTO plugins (id, name, fullName, currentVersion) VALUES
(1, "yoast", "Yoast SEO", "16.8"),
(2, "jetpack", "Jetpack", "9.9.1"),
(3, "akismet", "Akismet", "4.1.10"),
(4, "wordfence", "Wordfence Security", "7.5.4"),
(5, "contact-form", "Contact Form 7", "7.5.4.2");
INSERT INTO sites_plugins (siteId, pluginId, version, syncedAt) VALUES
(1, 1, "16.8", NULL),
(1, 3, "3.8", '2021-07-01 10:00:00'),
(2, 3, "4.1.10", NULL),
(2, 5, "7.0", NULL),
(2, 4, "7.5.3", '2021-06-15 12:00:00');
Ultimately, I would very much like to achieve a data format like the following
{["Amazon", "Jetpack", "16.8", "NULL"]}
Thanks for any advice, Best Kacper
You must join all tables by thier linking column and pick for the SELECT the wanted columns, which you can freely pick from all tables
SELECT
s.description
,p.fullName
,sp.version
, sp.syncedAt
FROM sites_plugins sp
INNER JOIN sites s ON s.id = sp.siteId
INNER JOIN plugins p = p.id = sp.pluginId
But your wanted result is not possible with the data you provided
I have a table customer which contains 2 columns, 1 is a customer_id column, and the other one is a date column named order_date that records what dates did the customers purchased a product. Now I want to count for how many days each customer went in and made a purchase. I tried to do the following but only got an error message saying sum(date) doesn't exist.
select customer_id, sum(order_date)
from customer;
How can I do this correctly?
---- Edit, adding the query to create table:
CREATE TABLE sales (
"customer_id" VARCHAR(1),
"order_date" DATE
);
INSERT INTO sales
("customer_id", "order_date")
VALUES
('A', '2021-01-01'),
('A', '2021-01-01'),
('A', '2021-01-07'),
('A', '2021-01-10'),
('A', '2021-01-11'),
('A', '2021-01-11'),
('B', '2021-01-01'),
('B', '2021-01-02'),
('B', '2021-01-04'),
('B', '2021-01-11'),
('B', '2021-01-16'),
('B', '2021-02-01'),
('C', '2021-01-01'),
('C', '2021-01-01'),
('C', '2021-01-07');
You'll want just this:
SELECT
customer_id,
COUNT( DISTINCT "order_date" ) AS count_days_they_bought_something
FROM
sales
GROUP BY
customer_id
I have below data
CREATE TABLE #EmployeeData
(
EmpID INT,
Designation VARCHAR(100),
Grade CHAR(1)
)
INSERT INTO #EmployeeData (EmpID, Designation, Grade)
VALUES (1, 'TeamLead', 'A'),
(2, 'Manager', 'B'),
(3, 'TeamLead', 'B'),
(4, 'SeniorTeamLead', 'A'),
(5, 'TeamLead', 'C'),
(6, 'Manager', 'C'),
(7, 'TeamLead', 'D'),
(8, 'SeniorTeamLead', 'B')
SELECT Designation,CASE WHEN COUNT(DISTINCT GRADE)>1 THEN 'MultiGrade' ELSE Grade END FROM
#EmployeeData
GROUP BY Designation
Desired result:
Designation Grade
--------------------------
Manager MultiGrade
TeamLead MultiGrade
SeniorTeamLead A
Note:
If designation has more than one grade then it is multigrade
If single grade is there then the particular grade
In case there is a combination with A and B then it should be A only
I tried with a query using case but I get this error:
Column '#EmployeeData.Grade' is invalid in the select list because it is not contained in either` an aggregate function or the GROUP BY clause.
Can anyone suggest the query to fetch the desired result?
As the error says, you need to aggregate the columns you are not grouping by. So use MAX and MIN (as Jeroen commented).
SELECT Designation
, CASE WHEN MAX(Grade) = 'B' AND MIN(Grade) = 'A' THEN 'A' WHEN MAX(Grade) <> MIN(Grade) THEN 'MultiGrade' ELSE MIN(Grade) END Grade
FROM #EmployeeData
GROUP BY Designation
ORDER BY Designation;
Your real world situation might be more complex, but the same principle applies.
That seems to be quite simple, but there is no solution right now. I would like to determine from a table the person who has the least entries. If there are several, I just want to limit it to TOP 1.
In this example it would be Person 2 (or Person 5) with the least entries.
Id, Person
1, Person 1
2, Person 3
3, Person 4
4, Person 1
5, Person 1
6, Person 3
7, Person 2
8, Person 5
9, Person 6
Use GROUP BY and ORDER BY:
select top (1) person
from t
group by person
order by count(*);
The question specifically asks for one row in the result set. If you want all of them, then use top (1) with ties`.
Use aggregation and TOP:
SELECT TOP 1 person
FROM mytable
GROUP BY person
ORDER BY COUNT(*), person
This will give you a unique record with the name of the person that has fewest entries. If there are ties, the person with the alphabetically first name will show up.
Demo on DB Fiddle with your sample data:
| person |
| :------- |
| Person 2 |
You can also use
SELECT TOP 1 Person -- You can add WITH TIES too
FROM
(
SELECT Person, COUNT(Person) Cnt
FROM
(
VALUES
(1, 'Person 1'),
(2, 'Person 3'),
(3, 'Person 4'),
(4, 'Person 1'),
(5, 'Person 1'),
(6, 'Person 3'),
(7, 'Person 2'),
(8, 'Person 5'),
(9, 'Person 6')
) T(ID, Person)
GROUP BY Person
) TT
ORDER BY Cnt;
Demo
You could also use OFFSET and FETCHif you have 2012+ version as
SELECT Person
FROM
(
SELECT Person, COUNT(Person) Cnt
FROM
(
VALUES
(1, 'Person 1'),
(2, 'Person 3'),
(3, 'Person 4'),
(4, 'Person 1'),
(5, 'Person 1'),
(6, 'Person 3'),
(7, 'Person 2'),
(8, 'Person 5'),
(9, 'Person 6')
) T(ID, Person)
GROUP BY Person
) TT
ORDER BY Cnt
OFFSET 0 ROWS
FETCH NEXT 1 ROWS ONLY;
I have a table like this
sub_id reference
1 A
1 A
1 A
1 A
1 A
1 A
1 C
2 B
2 B
3 D
3 D
I want to make sure all the references in each group have the same reference.
Meaning, for example, all references in:
group 1 should be A
group 2 should be B
group 3 should be D
If they are not, then I would like to have returned a list of sub_id's.
So for the table above my result would be: 1
Ideally, with these conditions reference would be in a separate table with sub_id as PK, but I need to fix first for a massive dataset before I can move on restructuring the database.
You could use the following method:
select t.sub_id
from YourTable t
group by t.sub_id
having max(t.reference) <> min(t.reference)
Change YourTable to suit.
Are you looking for simple aggregation ?
select sub_id
from table t
group by sub_id
having count(distinct reference) > 1;
The query you want:
SELECT sub_id
FROM test_sub
GROUP BY sub_id HAVING count(DISTINCT reference) > 1
;
Here is what I used to test it:
CREATE TABLE `test_sub` (
sub_id int(11) NOT NULL,
reference varchar(45) DEFAULT NULL
);
INSERT INTO test_sub (sub_id, reference) VALUES
(1, 'A'),
(1, 'A'),
(1, 'A'),
(1, 'A'),
(1, 'C'),
(2, 'B'),
(2, 'B'),
(3, 'D'),
(3, 'D'),
(3, 'D'),
(4, 'E'),
(4, 'E'),
(4, 'E'),
(5, 'F'),
(5, 'G')
;