I have a table like this
I want to check the all rows in Column A with column B and get the count of duplicates.
For example, I want to get the
count of 12 as 3(2 times in A+1 time in B)
count of 11 as 2(2 times in A+0 time in B)
count of 13 as 2(1 time in A+0 time in B)
How can I acheive it?
You can calculate the total occurrences from a union all. A where clause can show only the values that occur in the A column:
select nr
, count(*)
from (
select A as nr
from YourTable
union all
select B
from YourTable
) sub
where nr in -- only values that occur at least once in the A column
(
select A
from YourTable
)
group by
nr
having count(*) > 1 -- show only duplicates
You can combine all values in A and B then do the group by.
Then only select those values found in column A.
Select A, count(A) as cnt
From (
Select A
from yourTable
Union All
Select B
from yourTable) t
Where t.A in
(select distinct A from yourTable)
Group by t.A
Order by t.A;
Result:
A cnt
11 2
12 3
13 1
See demo: http://sqlfiddle.com/#!9/9fcfe9/3
Related
I am facing a problem with MySQL query which is a variant of "Id for row with max value". I am either getting error or incorrect result for all my trials.
Here is the table structure
Row_id
Group_id
Grp_col1
Grp_col2
Field_for_aggregate_func
Another_field_for_row
For all rows with a particular group_id, I want to group by fields Grp_col1, Grp_col2 then get max value of Field_for_aggregate_func and then corresponding value of Another_field_for_row.
Query I have tried is like below
SELECT c.*
FROM mytable as c left outer join mytable as c1
on (
c.group_id=c1.group_id and
c.Grp_col1 = c1.Grp_col1 and
c.Grp_col2 = c1.Grp_col2 and
c.Field_for_aggregate_func > c1.Field_for_aggregate_func
)
where c.group_id=2
Among alternative solutions for this problem I want a high performance solution as this will be used for large set of data.
EDIT: Here is the sample set of row and expected answer
Group_ID Grp_col1 Grp_col2 Field_for_aggregate_func Another_field_for_row
2 -- N 12/31/2015 35
2 -- N 1/31/2016 15 select 15 from group for max value 1/31/2016
2 -- Y 12/31/2015 5
2 -- Y 1/1/2016 15
2 -- Y 1/2/2016 25
2 -- Y 1/3/2016 30 select 30 from group for max value 1/3/2016
You can use a sub-query to find the maximums, then join that with the original table, along the lines of:
select m1.group_id, m1.grp_col1, m1.grp_col2, m1.another_field_for_row, max_value
from mytable m1, (
select group_id, grp_col1, grp_col2, max(field_for_aggregate_func) as max_value
from mytable
group by group_id, grp_col1, grp_col2) as m2
where m1.group_id=m2.group_id
and m1.grp_col1=m2.grp_col1
and m1.grp_col2=m2.grp_col2
and m1.field_for_aggregate_func=m2.max_value;
Watch out for when there is more than one max_value for the given grouping. You'll get multiple rows for that grouping. Fiddle here.
Try this.
See Fiddle demo here
http://sqlfiddle.com/#!9/9a3c26/8
Select t1.* from table1 t1 inner join
(
Select a.group_id,a.grp_col2,
A.Field_for_aggregate_func,
count(*) as rnum from table1 a
Inner join table1 b
On a.group_id=b.group_id
And a.grp_col2=b.grp_col2
And a.Field_for_aggregate_func
<=b.Field_for_aggregate_func
Group by a.group_id,
a.grp_col2,
a.Field_for_aggregate_func) t2
On t1.group_id=t2.group_id
And t1.grp_col2=t2.grp_col2
And t1.Field_for_aggregate_func
=t2.Field_for_aggregate_func
And t2.rnum=1
Here first I am assigning a rownumber in descending order based on date. The selecting all the records for that date.
I have a table with the following structure:
id name
1 X
1 X
1 Y
2 A
2 A
2 B
Basically what I am trying to do is to write a query that returns X for 1 because X has repeated more than Y (2 times) and returns A for 2. So if a value occurs more than the other one my query should return that. Sorry if the title is confusing but I could not find a better explanation. This is what I have tried so far:
SELECT MAX(counted) FROM(
SELECT COUNT(B) AS counted
FROM table
GROUP BY A
) AS counts;
The problem is that my query should return the actual value other than the count of it.
Thanks
This should work:
SELECT count(B) as occurrence, A, B
FROM table
GROUP BY B
ORDER BY occurrence DESC
LIMIT 1;
Please check: http://sqlfiddle.com/#!9/dfa09/3
You can try like this using a GROUP BY clause. See a Demo Here
select *, max(occurence) as Maximum_Occurence from
(
select B, count(B) as occurence
from table1
group by B
) xxx
This is how I finally handled my problem. Not the most efficient way but get the job done:
select A,B from
(select A,B, max(cnt) from
(select A ,B ,count(B) as cnt
from myTable
group by A,B
order by cnt desc
) as x group by A
) as xx
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;
Columns a, b and c contain some values of the same nature. I need to select all the unique values. If I had just one column I'd use something like
SELECT DISTINCT a FROM mytable ORDER BY a;
but I need to treat a, b and c columns as one and gett all the unique values ever occurring among them.
As an example, let this be a CSV representation of mytable, the first row naming the columns:
a, b, c
1, 2, 3
1, 3, 4
5, 7, 1
The result of the query is to be:
1
2
3
4
5
7
UPDATE: I don't understand why do all of you suggest wrapping it in an extra SELECT? It seems to me that the answer is
(SELECT `a` AS `result` FROM `mytable`)
UNION (SELECT `b` FROM `mytable`)
UNION (SELECT `c` FROM `mytable`)
ORDER BY `result`;
isn't it?
So you want one column all with unique values from a, b and c? Try this:
(select a as yourField from d1)
union
(select b from d2)
union
(select c from d3)
order by yourField desc
limit 5
Working example
Edited after requirements changed... There you have the order by and limit you requested. Of course, you'll get only 5 records in this example
sorry i miss understood your question. here is updated query.
select a from my table
UNION
select b from my table
UNION
select c from my table
SELECT tmp.a
FROM
(SELECT column_1 AS a
FROM table
UNION
SELECT column_2 AS a
FROM table
UNION
SELECT column_3 AS a
FROM table) AS tmp
ORDER BY `tmp`.`a` ASC
try this:
SELECT b.iResult
FROM
(SELECT a as iResult FROM tableName
UNION
SELECT b as iResult FROM tableName
UNION
SELECT c as iResult FROM tableName) b
ORDER BY b.iResult
LIMIT BY 10 -- or put any number you want to limit.