I have multiples tables with Names and Ages like this:
| Name | Age |
------------------
| Carlos | 25 |
| Mauricio | 28 |
| Cesar | 19 |
| Hernan | 7 |
And I need to retrieve all the names that are above the average Age.
I tried
select Name from Table1 where Age > avg(Age)
but I found that the where clause does not work with aggregate functions, so I tried
select Name from Table 1 having Age > avg(Age)
But it does not work either.
You can do it with following query:
select Name from Table1 where Age > (select avg(Age) from Table1)
Related
I am trying to learn SQL queries and have this scenario where I have this table:
Table1
ID | Name | Hour
----------------
1 | Mark | 2
2 | ken | 1.5
3 | jake | 3
1 | Mark | 1.8
2 | ken | 1
Expected result
ID | Name | Hour
----------------
1 | Mark | 3.8
2 | ken | 2.5
3 | jake | 3
I have tried to use the sum() function but I get an error.
My query:
Select ID, Name, Sum(Hour)
From Table1
Where ID = ID
Response:
Kindly use Group by clause whenever the Aggregate functions (min(),max(),sum(),count(),...etc.,) and columns are used together.
Non aggregated columns present in SELECT columns should be used in GROUP BY clause.
For using aggregate function you need to use Group By like this:
Select ID, Name , Sum(Hour) AS Hour From Table1
Group By ID, Name
Order By ID
I am trying to identify the count of each distinct value in one column (name) in a table called brgy.
---------------------
| ID | name |
---------------------
| 1 | Alfonso |
| 2 | Arakan |
| 3 | Poblacion |
| 4 | Ilaya |
| 5 | Poblacion |
----------------------
I tried using this code but it keeps giving the COUNT as 1 despite Poblacion appearing twice in the name column:
SELECT name,COUNT(name) AS distinct_name
FROM (
SELECT DISTINCT name
FROM brgy
GROUP BY name
)
GROUP BY name;
The intended output should eliminate duplicate names but sum up the number of times the distinct name appears in the name column:
Expected Output is as below,
-----------------------------
| name | distinct_name |
-----------------------------
| Alfonso | 1 |
| Arakan | 1 |
| Ilaya | 1 |
| Poblacion | 2 |
-----------------------------
A simple GROUP BY name:
SELECT name, COUNT(name) AS distinct_name
FROM brgy
GROUP BY name;
you don't need the subquery:
SELECT DISTINCT name FROM brgy GROUP BY name
because GROUP BY name takes care of it.
Just removed the subquery because it will give you unique entry :
select name, count(*) as distinct_name
from brgy
group by name
order by distinct_name, name;
Lets say I have following table:
id | name | no
--------------
1 | A | 10
1 | A | 20
1 | A | 40
2 | B | 20
2 | B | 20
And I want to perform a select query in SQL server which sums the value of "no" field which have same id.
Result should look like this,
id | name | no
--------------
1 | A | 70
2 | B | 40
Simple GROUP BY and SUM should work.
SELECT ID, NAME, SUM([NO])
FROM Your_TableName
GROUP BY ID, NAME;
Use SUM and GROUP BY
SELECT ID,NAME, SUM(NO) AS TOTAL_NO FROM TBL_NAME GROUP BY ID, NAME
SELECT *, SUM(no) AS no From TABLE_NAME GROUP BY name
This will return the same table by summing up the no column of the same name column.
I am working on Terradata SQL. I would like to get the duplicate fields with their count and other variables as well. I can only find ways to get the count, but not exactly the variables as well.
Available input
+---------+----------+----------------------+
| id | name | Date |
+---------+----------+----------------------+
| 1 | abc | 21.03.2015 |
| 1 | def | 22.04.2015 |
| 2 | ajk | 22.03.2015 |
| 3 | ghi | 23.03.2015 |
| 3 | ghi | 23.03.2015 |
Expected output :
+---------+----------+----------------------+
| id | name | count | // Other fields
+---------+----------+----------------------+
| 1 | abc | 2 |
| 1 | def | 2 |
| 2 | ajk | 1 |
| 3 | ghi | 2 |
| 3 | ghi | 2 |
What am I looking for :
I am looking for all duplicate rows, where duplication is decided by ID and to retrieve the duplicate rows as well.
All I have till now is :
SELECT
id, name, other-variables, COUNT(*)
FROM
Table_NAME
GROUP BY
id, name
HAVING
COUNT(*) > 1
This is not showing correct data. Thank you.
You could use a window aggregate function, like this:
SELECT *
FROM (
SELECT id, name, other-variables,
COUNT(*) OVER (PARTITION BY id) AS duplicates
FROM users
) AS sub
WHERE duplicates > 1
Using a teradata extension to ISO SQL syntax, you can simplify the above to:
SELECT id, name, other-variables,
COUNT(*) OVER (PARTITION BY id) AS duplicates
FROM users
QUALIFY duplicates > 1
As an alternative to the accepted and perfectly correct answer, you can use:
SELECT {all your required 'variables' (they are not variables, but attributes)}
, cnt.Count_Dups
FROM Table_NAME TN
INNER JOIN (
SELECT id
, COUNT(1) Count_Dups
GROUP BY id
HAVING COUNT(1) > 1 -- If you want only duplicates
) cnt
ON cnt.id = TN.id
edit: According to your edit, duplicates are on id only. Edited my query accordingly.
try this,
SELECT
id, COUNT(id)
FROM
Table_NAME
GROUP BY
id
HAVING
COUNT(id) > 1
I have an SQL SELECT query that also uses a GROUP BY,
I want to count all the records after the GROUP BY clause filtered the resultset.
Is there any way to do this directly with SQL? For example, if I have the table users and want to select the different towns and the total number of users:
SELECT `town`, COUNT(*)
FROM `user`
GROUP BY `town`;
I want to have a column with all the towns and another with the number of users in all rows.
An example of the result for having 3 towns and 58 users in total is:
Town
Count
Copenhagen
58
New York
58
Athens
58
This will do what you want (list of towns, with the number of users in each):
SELECT `town`, COUNT(`town`)
FROM `user`
GROUP BY `town`;
You can use most aggregate functions when using a GROUP BY statement
(COUNT, MAX, COUNT DISTINCT etc.)
Update:
You can declare a variable for the number of users and save the result there, and then SELECT the value of the variable:
DECLARE #numOfUsers INT
SET #numOfUsers = SELECT COUNT(*) FROM `user`;
SELECT DISTINCT `town`, #numOfUsers FROM `user`;
You can use COUNT(DISTINCT ...) :
SELECT COUNT(DISTINCT town)
FROM user
The other way is:
/* Number of rows in a derived table called d1. */
select count(*) from
(
/* Number of times each town appears in user. */
select town, count(*)
from user
group by town
) d1
Ten non-deleted answers; most do not do what the user asked for. Most Answers mis-read the question as thinking that there are 58 users in each town instead of 58 in total. Even the few that are correct are not optimal.
mysql> flush status;
Query OK, 0 rows affected (0.00 sec)
SELECT province, total_cities
FROM ( SELECT DISTINCT province FROM canada ) AS provinces
CROSS JOIN ( SELECT COUNT(*) total_cities FROM canada ) AS tot;
+---------------------------+--------------+
| province | total_cities |
+---------------------------+--------------+
| Alberta | 5484 |
| British Columbia | 5484 |
| Manitoba | 5484 |
| New Brunswick | 5484 |
| Newfoundland and Labrador | 5484 |
| Northwest Territories | 5484 |
| Nova Scotia | 5484 |
| Nunavut | 5484 |
| Ontario | 5484 |
| Prince Edward Island | 5484 |
| Quebec | 5484 |
| Saskatchewan | 5484 |
| Yukon | 5484 |
+---------------------------+--------------+
13 rows in set (0.01 sec)
SHOW session status LIKE 'Handler%';
+----------------------------+-------+
| Variable_name | Value |
+----------------------------+-------+
| Handler_commit | 1 |
| Handler_delete | 0 |
| Handler_discover | 0 |
| Handler_external_lock | 4 |
| Handler_mrr_init | 0 |
| Handler_prepare | 0 |
| Handler_read_first | 3 |
| Handler_read_key | 16 |
| Handler_read_last | 1 |
| Handler_read_next | 5484 | -- One table scan to get COUNT(*)
| Handler_read_prev | 0 |
| Handler_read_rnd | 0 |
| Handler_read_rnd_next | 15 |
| Handler_rollback | 0 |
| Handler_savepoint | 0 |
| Handler_savepoint_rollback | 0 |
| Handler_update | 0 |
| Handler_write | 14 | -- leapfrog through index to find provinces
+----------------------------+-------+
In the OP's context:
SELECT town, total_users
FROM ( SELECT DISTINCT town FROM canada ) AS towns
CROSS JOIN ( SELECT COUNT(*) total_users FROM canada ) AS tot;
Since there is only one row from tot, the CROSS JOIN is not as voluminous as it might otherwise be.
The usual pattern is COUNT(*) instead of COUNT(town). The latter implies checking town for being not null, which is unnecessary in this context.
With Oracle you could use analytic functions:
select town, count(town), sum(count(town)) over () total_count from user
group by town
Your other options is to use a subquery:
select town, count(town), (select count(town) from user) as total_count from user
group by town
If you want to order by count (sound simple but i can`t found an answer on stack of how to do that) you can do:
SELECT town, count(town) as total FROM user
GROUP BY town ORDER BY total DESC
You can use DISTINCT inside the COUNT like what milkovsky said
in my case:
select COUNT(distinct user_id) from answers_votes where answer_id in (694,695);
This will pull the count of answer votes considered the same user_id as one count
I know this is an old post, in SQL Server:
select isnull(town,'TOTAL') Town, count(*) cnt
from user
group by town WITH ROLLUP
Town cnt
Copenhagen 58
NewYork 58
Athens 58
TOTAL 174
If you want to select town and total user count, you can use this query below:
SELECT Town, (SELECT Count(*) FROM User) `Count` FROM user GROUP BY Town;
if You Want to use Select All Query With Count Option, try this...
select a.*, (Select count(b.name) from table_name as b where Condition) as totCount from table_name as a where where Condition
Try the following code:
select ccode, count(empno)
from company_details
group by ccode;