I have a question: it's possible to create an count in count in sql:
my code is:
SELECT COUNT( DISTINCT p.id_participant ) as number
FROM participation p
INNER JOIN message m ON m.id_participation=p.id
AND p.id_event = 4
I want to add in first count another count from table winners with count (id_winner)
Help me please, Exist a solution?
You need to use the aggregate function SUM.
For example,
SQL> SELECT SUM(val)
2 FROM (SELECT Count(*) VAL
3 FROM emp
4 UNION
5 SELECT Count(*) VAL
6 FROM dept);
SUM(VAL)
----------
18
Related
I have a table event_tblwith columns id,add_date,name,events;
and I want count all rows and the latest add_date count, how can I get that in one sql statement?
select count(1) as total
from event_tbl
union
select count(id) as latestCount from event_tbl where add_date=(select max(add_date)
from event_tbl);
but the result is :
------
total|
------
5499|
-----
611 |
what is correct way with efficiency?
I would suggest writing this as:
select count(*) as total, sum(add_date = maxad) as lastestCount
from event_tbl e cross join
(select max(add_date) as maxad from event_tbl) ee;
Could you help me with simple table SUM and COUNT calculating?
I've simple table 'test'
id name value
1 a 4
2 a 5
3 b 3
4 b 7
5 b 1
I need calculate SUM and Count for "a" and "b". I try this sql request:
SELECT name, SUM( value ) AS val, COUNT( * ) AS count FROM `test`
result:
name val count
a 20 5
But should be
name val count
a 9 2
b 11 3
Could you help me with correct sql request?
Add GROUP BY. That will cause the query to return a count and sum per group you defined (in this case, per name).
Without GROUP BY you just get the totals and any of the names (in your case 'a', but if could just as well have been 'b').
SELECT name, SUM( value ) AS val, COUNT( * ) AS count
FROM `test`
GROUP BY name
You need group by
select
name,
sum(value) as value,
count(*) as `count`
from test group by name ;
I have a simple MySQL statement:
SELECT q1, COUNT(q1) FROM results WHERE q1 IN ('1','2','3');
Currently there are only results for 1 and 3 - results are:
1 = 6
3 = 7
But what I need is for MySQL to bring back a result for 1,2 and 3 even though 2 has no data, as this:
1 = 6
2 = 0
3 = 7
Any ideas?
This is tricky because no rows match your value (2), they cannot be counted.
I would solve this by creating a temp table containing the list of values I want counts for:
CREATE TEMPORARY TABLE q ( q1 INT PRIMARY KEY );
INSERT INTO q (q1) VALUES (1), (2), (3);
Then do an OUTER JOIN to your results table:
SELECT q.q1, COALESCE(COUNT(*), 0) AS count
FROM q LEFT OUTER JOIN results USING (q1)
GROUP BY q.q1;
This way each value will be part of the final result set, even if it has no matching rows.
Re comment from #Mike Christensen:
MySQL doesn't support CTE's, in spite of it being requested as far back as 2006: http://bugs.mysql.com/bug.php?id=16244
You could do the same thing with a derived table:
SELECT q.q1, COALESCE(COUNT(*), 0) AS count
FROM (SELECT 1 AS q1 UNION ALL SELECT 2 UNION ALL SELECT 3) AS q
LEFT OUTER JOIN results USING (q1)
GROUP BY q.q1;
But this creates a temp table anyway.
A SQL query doesn't really have a way to refer to the values in your IN clause. I think you'd have to break this down into one query for each value. Something like:
SELECT 1 as q1, COUNT(1) FROM results WHERE q1 = '1'
UNION ALL
SELECT 2 as q1, COUNT(1) FROM results WHERE q1 = '2'
UNION ALL
SELECT 3 as q1, COUNT(1) FROM results WHERE q1 = '3'
Fiddle
Note: If there are a lot of values in your IN clause, you might be better off to write your code in a way where missing values are assumed to have zero.
In general, you cannot query something that does not exists. So, you must create data for it. Use union to add those missing data values.
select q1, COUNT(*)
from results
where q1 in ('1','2','3')
group by q1
union
select q1, 0
from (
select '1' as q1
union
select '2'
union
select '3'
) as q
where q1 not in (
select q1
from results
)
I have table with, folowing structure.
tbl
id name
1 AAA
2 BBB
3 BBB
4 BBB
5 AAA
6 CCC
select count(name) c from tbl
group by name having c >1
The query returning this result:
AAA(2) duplicate
BBB(3) duplicate
CCC(1) not duplicate
The names who are duplicates as AAA and BBB. The final result, who I want is count of this duplicate records.
Result should be like this:
Total duplicate products (2)
The approach is to have a nested query that has one line per duplicate, and an outer query returning just the count of the results of the inner query.
SELECT count(*) AS duplicate_count
FROM (
SELECT name FROM tbl
GROUP BY name HAVING COUNT(name) > 1
) AS t
Use IF statement to get your desired output:
SELECT name, COUNT(*) AS times, IF (COUNT(*)>1,"duplicated", "not duplicated") AS duplicated FROM <MY_TABLE> GROUP BY name
Output:
AAA 2 duplicated
BBB 3 duplicated
CCC 1 not duplicated
For List:
SELECT COUNT(`name`) AS adet, name
FROM `tbl` WHERE `status`=1 GROUP BY `name`
ORDER BY `adet` DESC
For Total Count:
SELECT COUNT(*) AS Total
FROM (SELECT COUNT(name) AS cou FROM tbl GROUP BY name HAVING cou>1 ) AS virtual_tbl
// Total: 5
why not just wrap this in a sub-query:
SELECT Count(*) TotalDups
FROM
(
select Name, Count(*)
from yourTable
group by name
having Count(*) > 1
) x
See SQL Fiddle with Demo
The accepted answer counts the number of rows that have duplicates, not the amount of duplicates. If you want to count the actual number of duplicates, use this:
SELECT COALESCE(SUM(rows) - count(1), 0) as dupes FROM(
SELECT COUNT(1) as rows
FROM `yourtable`
GROUP BY `name`
HAVING rows > 1
) x
What this does is total the duplicates in the group by, but then subtracts the amount of records that have duplicates. The reason is the group by total is not all duplicates, one record of each of those groupings is the unique row.
Fiddle: http://sqlfiddle.com/#!2/29639a/3
SQL code is:
SELECT VERSION_ID, PROJECT_ID, VERSION_NO, COUNT(VERSION_NO) AS dup_cnt
FROM MOVEMENTS
GROUP BY VERSION_NO
HAVING (dup_cnt > 1 && PROJECT_ID = 11660)
I'm using this query for my own table in PHP, but it only gives me one result whereas I'd like to the amount of duplicate per username, is that possible?
SELECT count(*) AS duplicate_count
FROM (
SELECT username FROM login_history
GROUP BY username HAVING COUNT(time) > 1
) AS t;
I already read (this), but couldn't figure out a way to implement it to my specific problem. I know SUM() is an aggregate function and it doesn't make sense not to use it as such, but in this specific case, I have to SUM() all of the results while maintaining every single row.
Here's the table:
--ID-- --amount--
1 23
2 11
3 8
4 7
I need to SUM() the amount, but keep every record, so the output should be like:
--ID-- --amount--
1 49
2 49
3 49
4 49
I had this query, but it only sums each row, not all results together:
SELECT
a.id,
SUM(b.amount)
FROM table1 as a
JOIN table1 as b ON a.id = b.id
GROUP BY id
Without the SUM() it would only return one single row, but I need to maintain all ID's...
Note: Yes this is a pretty basic example and I could use php to do this here,but obviously the table is bigger and has more rows and columns, but that's not the point.
SELECT a.id, b.amount
FROM table1 a
CROSS JOIN
(
SELECT SUM(amount) amount FROM table1
) b
You need to perform a cartesian join of the value of the sum of every row in the table to each id. Since there is only one result of the subselect (49), it basically just gets tacked onto each id.
With MS SQL you can use OVER()
select id, SUM(amount) OVER()
from table1;
select id, SUM(amount) OVER()
from (
select 1 as id, 23 as amount
union all
select 2 as id, 11 as amount
union all
select 3 as id, 8 as amount
union all
select 4 as id, 7 as amount
) A
--- OVER PARTITION ID
PARTITION BY which is very useful when you want to do SUM() per MONTH for example or do quarterly reports sales or yearly...
(Note needs distinct it is doing for all rows)
select distinct id, SUM(amount) OVER(PARTITION BY id) as [SUM_forPARTITION]
from (
select 1 as id, 23 as amount
union all
select 1 as id, 23 as amount
union all
select 2 as id, 11 as amount
union all
select 2 as id, 11 as amount
union all
select 3 as id, 8 as amount
union all
select 4 as id, 7 as amount
) OverPARTITIONID
Join the original table to the sum with a subquery:
SELECT * FROM table1, (SELECT SUM(amount) FROM table1 AS amount) t
This does just one sum() query, so it should perform OK:
SELECT a.id, b.amount
FROM table1 a
cross join (SELECT SUM(amount) as amount FROM table1 AS amount) b
in case someone else has the same problem and without joining we can do the following
select *
,totcalaccepted=(select sum(s.acceptedamount) from cteresult s)
, totalpay=(select sum(s.payvalue) from cteresult s)
from cteresult t
end
Using Full Join -
case when you need sum of amount field from tableB and all data from tableA on behalf of id match.
SELECT a.amount, b.* FROM tableB b
full join (
select id ,SUM(amount) as amount FROM tableA
where id = '1' group by id
) a
on a.id = b.id where a.id ='1' or b.id = '1' limit 1;