I have mysql table like this
I want to get row that has minimum 2 or more than 2 (multiple) row only from this table, so the result would be like this
What do i do?
thank you
Use GROUP BY and HAVING clauses
SELECT t.* FROM my_table t
JOIN (
SELECT cust_id, MIN(transaction_no) AS transaction_no
FROM my_table
GROUP BY cust_id
HAVING COUNT(cust_id) > 1
) agg ON t.transaction_no = agg.transaction_no
Related
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
I have a table having id and no field, what I really want is the result raw will be repeated no filed times, if the no field is 2 then that raw must be repeated twice in result.
this is my sample table structure:
id no
1 3
2 2
3 1
now I need to get a result like:
1 3
1 3
1 3
2 2
2 2
3 1
I tried to write mysql query to get the result like above, but failed.
You need a table of numbers to accomplish this. For just three values, this is easy:
select t.id, t.no
from t join
(select 1 as n union all select 2 union all select 3
) n
on t.no <= n.no;
This query must do what you want to achieve:
select t.id, t.no from test t cross join test y where t.id>=y.id
not completely solve your problem, but this one can help
set #i=0;
select
test_table.*
from
test_table
join
(select
#i:=#i+1 as i
from
any_table_with_number_of_rows_greater_than_max_no_of_test_table
where
#i < (select max(no) from test_table)) tmp on no >= i
order by
id desc
EDIT :
This is on SQL Server. I checked online and see that CTEs work on MySQL too. Just couldn't get them to work on SQLFiddle
Try this, remove unwanted columns
create table #temp (id int, no int)
insert into #temp values (1, 2),(2, 3),(3, 5)
select * from #temp
;with cte as
(
select id, no, no-1 nom from #temp
union all
select c.id, c.no, c.nom-1 from cte c inner join #temp t on t.id = c.id and c.nom < t.no and c.nom > 0
)
select * from cte order by 1
drop table #temp
Let's say I have two columns: id and date.
I want to give it an id and it'll find all the duplicates of the value date of the column id.
Example:
id |date
1 |2013-09-16
2 |2013-09-16
3 |2013-09-23
4 |2013-09-23
I want to give it id 1 (without giving anything about date) and it'll give me a table of 2 columns listing the duplicates of id 1's date
Thanks in advance!
select * from your_table
where `date` in
(
select `date`
from your_table
where id = 1
)
or if you like to use a join
select t.*
from your_table t
inner join
(
select `date`
from your_table
where id = 1
) x on x.date = t.date
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;
We read values from a set of sensors, occasionally a reading or two is lost for a particular sensor , so now and again I run a query to see if all sensors have the same record count.
GROUP BY sensor_id HAVING COUNT(*) != xxx;
So I run a query once to visually get a value of xxx and then run it again to see if any vary.
But is there any clever way of doing this automatically in a single query?
You could do:
HAVING COUNT(*) != (SELECT MAX(count) FROM (
SELECT COUNT(*) AS count FROM my_table GROUP BY sensor_id
) t)
Or else group again by the count in each group (and ignore the first result):
SELECT count, GROUP_CONCAT(sensor_id) AS sensors
FROM (
SELECT sensor_id, COUNT(*) AS count FROM my_table GROUP BY sensor_id
) t
GROUP BY count
ORDER BY count DESC
LIMIT 1, 18446744073709551615
SELECT sensor_id,COUNT(*) AS count
FROM table
GROUP BY sensor_id
ORDER BY count
Will show a list of the sensor_id along with a count of all the records it has, you can then manually check to see if any vary.
SELECT * FROM (
SELECT sensor_id,COUNT(*) AS count
FROM table
GROUP BY sensor_id
) AS t1
GROUP BY count
Will show all the counts that vary, but the group by will lose information about which sensor_ids have which counts.
---EDIT---
Taken a bit from both mine and eggyal's answer and created this, for the count that is most frequent I call the id default, and then for any values that stand out I have given them separate rows. This way you maintain the readability of a table if you have many results Multi Row, but also have a simple one row column if all counts are the same One Row. If however you are happy with the concocted strings then go with eggyal's answer.
Might be a bit over the top but here goes:
select 'default' as id,t5.c1 as count from(
select id,count(*) as c1 from your_table group by id having count(*)=
(select t4.count from
(
select max(t3.count2) as max,t3.count as count from
(
select count(*) as count2,t2.count from
(
SELECT id,COUNT(*) AS count
FROM your_table
GROUP BY id
) as t2
GROUP BY count
) as t3
) as t4)) as t5 group by count
union all
select t5.id as id,t5.c1 as count from(
select id,count(*) as c1 from your_table group by id having count(*)<>
(select t4.count from
(
select max(t3.count2) as max,t3.count as count from
(
select count(*) as count2,t2.count from
(
SELECT id,COUNT(*) AS count
FROM your_table
GROUP BY id
) as t2
GROUP BY count
) as t3
) as t4)) as t5