How to select last value in multiple rows in MySQL ?
select name,date,value from table1;
Name Date Value
A 01-Jan-2013 3
A 02-Jan-2013 4
B 04-Jan-2013 2
B 05-Jan-2013 8
Result I need :
Name Date Value
A 01-Jan-2013 3
A 02-Jan-2013 4
Last_value 4
B 04-Jan-2013 2
B 05-Jan-2013 8
Last_value 8
How to do that in MySQL store procedure ?
Regards
There's no "last" record. You need to sort them by the column they should be ordered by, then only select the last row.
/* SELECT ... */ ORDER BY sort_column DESC LIMIT 1
This seems better suited for your Presentation Layer, but it is possible to write it using SQL. Something like this should work using UNION ALL to return that last record per group:
SELECT Name, Dt, Value
FROM (
SELECT Name, Dt, Value, Name Name2
FROM YourTable
UNION ALL
SELECT '' Name, 'Last_Value', T.Value, T.Name Name2
FROM YourTable T
JOIN (
SELECT Name, MAX(dt) MaxDt
FROM YourTable
GROUP BY Name
) T2 ON T.Name = T2.Name AND T.dt = T2.MaxDt
ORDER BY Name2, Dt, Value
) t
SQL Fiddle Demo
Resulting in:
NAME DT VALUE
A 2013-01-01 00:00:00 3
A 2013-01-02 00:00:00 4
Last_Value 4
B 2013-01-04 00:00:00 2
B 2013-01-05 00:00:00 8
Last_Value 8
EDIT: To get the Summed Group Total, you have to introduce the use of user defined variables. Here is an example:
SELECT Name, Dt, IF(Name='',summedTotal,Value) Value
FROM (
SELECT #summedTotal:=IF(#prevRow=Name,#summedTotal+Value,Value) summedTotal,
Name, Dt, Value, Name Name2,
#prevRow:=Name
FROM YourTable
JOIN (SELECT #summedTotal:=0) t
UNION ALL
SELECT summedTotal, '' Name, 'Last_Value', T.Value, T.Name Name2, pr
FROM (
SELECT #summedTotal:=IF(#prevRow=Name,#summedTotal+Value,Value) summedTotal,
Name, Dt, Value, Name Name2,
#prevRow:=Name pr
FROM YourTable
JOIN (SELECT #summedTotal:=0) t
) T
JOIN (
SELECT Name, MAX(dt) MaxDt
FROM YourTable
GROUP BY Name
) T2 ON T.Name = T2.Name AND T.dt = T2.MaxDt
ORDER BY Name2, Dt, Value
) t
And more SQL Fiddle
You can select the MAX value using inner join. But you haven't specified your primary key column.
SELECT
m.*
FROM mytable AS m
INNER JOIN (SELECT
MAX(primary_key_column)
FROM mytable
GROUP BY Name) AS l
ON l.primary_key_column = m.primary_key_column
Related
I have a table as follows and what I want is to use get the initial row with least id of each uid group.
The table is as follows
_id uid type
1 a a
2 b bbb #satisfied
3 b ccc
4 b aaa #satisfied
5 a aaa #satisfied
6 b eee
I can already get the initial row using the following correlated subquery
SELECT *
FROM table
WHERE _id IN (
SELECT MIN(_id)
FROM table
WHERE type IN ('aaa','bbb')
GROUP BY uid
);
However, I want the 4th column shown the count of rows satisfied the condition (type IN ('aaa','bbb')), as cnt shown below:
_id uid type cnt
5 a aaa 1
2 b bbb 2
I think I can count this use several joins and then join the result to my code...But this is ugly...Is there any elegant way to achieve this...
You can try this:
SELECT t1.*, t2.cnt
FROM table t1 INNER JOIN (
SELECT MIN(_id) AS id, COUNT(_id) AS cnt
FROM table
WHERE type IN ('aaa','bbb')
GROUP BY uid
) t2 ON t1._id = t2.id
ORDER BY t1.uid
If you are running MySQL 8.0, you can just use window functions for this:
select _id, uid, type, cnt
from (
select
t.*,
count(*) over(partition by uid) cnt,
row_number() over(partition by uid order by _id) rn
from mytable t
where type in ('aaa', 'bbb')
) t
where rn = 1
You can do this without a subquery. In MySQL 8+, you can use this logic:
SELECT DISTINCT MIN(_id) OVER (PARTITION BY uid) as _id,
uid,
FIRST_VALUE(type) OVER (PARTITION BY uid ORDER BY _id) as type,
COUNT(*) OVER (PARTITION BY uid) as cnt
FROM table
WHERE type IN ('aaa', 'bbb');
Unfortunately, MySQL doesn't have a "first" aggregation function, but there is a trick if you like:
SELECT MIN(_id) as _id, uid,
SUBSTRING_INDEX(GROUP_CONCAT(type ORDER BY _id), ',', 1) as type,
COUNT(*) as cnt
FROM table
WHERE type IN ('aaa', 'bbb')
GROUP BY uid;
Sorry for the poor formatting but as part of a larger problem, I have created a query that produces this table:
id id2
4 7
4 6
1 3
1 2
1 1
How would I extract the rows that don't have the highest id2 for each id1.
What I want:
id id2
4 6
1 2
1 1
I can only seem to figure out how to get rid of the max id2 overall but not for each distinct id1. Any help on actually differentiating the max id2 for each id1 would be appreciated.
You can try below way -
select a.id, a.id2
from tablename a
where a.id2 <> (select max(a1.id2) from tablename a1 where a.id=a1.id)
If you are using MySQL 8+, then RANK() provides one option:
WITH cte AS (
SELECT id, id2, RANK() OVER (PARTITION BY id ORDER BY id2 DESC) rnk
FROM yourTable
)
SELECT id, id2
FROM cte
WHERE rnk > 1
ORDER BY id DESC, id2 DESC;
Demo
instead of a correlated subquery in the where, you can LEFT JOIN and apply not in...
select id, id2
from yourTable YT
LEFT JOIN
( select id, max( id2 ) highestID2
from YourTable
group by id ) TopPerID
on YT.ID = TopPerID.ID
AND YT.ID2 != TopPerID.highestID2
where TopPerID.id IS NULL
Since you can have id values with only one id2 value, you need to check for that situation as well, which you can do by comparing the MAX(id2) value with the MIN(id2) value in a JOIN:
SELECT t1.*
FROM Table1 t1
JOIN (SELECT id, MAX(id2) AS max_id2, MIN(id2) AS min_id2
FROM Table1
GROUP BY id) t2 ON t2.id = t1.id
AND (t1.id2 < t2.max_id2 OR t2.min_id2 = t2.max_id2)
If we add a row 2, 5 to your sample data this correctly gives the result as
id id2
4 6
1 2
1 1
2 5
Demo on SQLFiddle
Hello i am having two different table with same field created_date (datetime)
now i want records which counts daywise records with joining table i have done for individual counting as below query :
SELECT DATE(created_date), COUNT(*) FROM table1 GROUP BY DAY(created_date)
SELECT DATE(created_date), COUNT(*) FROM table2 GROUP BY DAY(created_date)
and i am getting results for individuals something like this:
RESULT I NEED :
DATE(created_date) count(table1) count(table2)
2016-12-01 10 3
2016-12-02 1 0
2016-12-05 1 0
2016-11-29 1 0
2016-11-30 4 1
Now i just want to join these result WITH INDIVIDUAL VIEW COUNT ACCORDING TO TABLE can anyone please help me out with this profile....
First take a UNION between your two tables, then use conditional aggregation to determine the counts for each of the two tables. Note that I introduce a field called table_name to keep track of data from each of the two tables.
SELECT t.created_date,
SUM(CASE WHEN t.table_name = 'one' THEN 1 ELSE 0 END) AS count_table_one,
SUM(CASE WHEN t.table_name = 'two' THEN 1 ELSE 0 END) AS count_table_two
FROM
(
SELECT DATE(created_date) AS created_date, 'one' AS table_name
FROM table1
UNION ALL
SELECT DATE(created_date), 'two'
FROM table2
) t
GROUP BY t.created_date
I used DATE consistently everywhere to make the query correct.
Try This:
SELECT created_date, sum(countTable1) countTable1,
sum(countTable2) countTable2
FROM (
SELECT DATE(created_date) created_date, COUNT(*) countTable1, NULL countTable2
FROM table1 GROUP BY DAY(created_date)
UNION ALL
SELECT DATE(created_date) created_date, NULL, COUNT(*) countTable2
FROM table2 GROUP BY DAY(created_date)) t GROUP BY t.created_date
You have a problem in your queries, you are grouping by DAY(date) and showing 'date' so the result will be first date with day(date), yet repeating it to avoid misunderstanding :)
select IFNULL(A.cd, B.cd), A.cnt, B.cnt from
(SELECT DAY(created_date) d, DATE(created_date) cd, COUNT(*) cnt
FROM table1 GROUP BY DAY(created_date)) as A
LEFT JOIN
(SELECT DAY(created_date) d, DATE(created_date) cd , COUNT(*) cnt
FROM table2 GROUP BY DAY(created_date)) B ON B.d = A.d
Its not too hard just use union if no need to allow duplicate row else use union all for all(means allow duplicate as well).
SELECT DATE(created_date), COUNT(*) FROM table1 GROUP BY DAY(created_date)
UNION
SELECT DATE(created_date), COUNT(*) FROM table2 GROUP BY DAY(created_date)
SELECT T.create_date,ISNULL(T.count,0)AS Counttable1,ISNULL(X.count,0)AS Counttable2 FROM(SELECT DATE(created_date) AS create_date,COUNT(*) as count FROM table1 GROUP BY DAY(created_date)) AS T LEFTJOIN(SELECT DATE(created_date) AS create_date, COUNT(*) as count FROM table2 GROUP BY DAY(created_date))AS X ON T.create_date=X.create_date
You actually need a SQL UNION. JOIN natuarually eliminate counts becuase the maytch fields. I.e. if you had 2016-12-01 in both table1 andtable2 then a JOIN on created_date would give you a count of 1 instead of a count of 2.
SELECT DATE(total.created_date), COUNT(*)
FROM (
SELECT created_date FROM table1
UNION ALL
SELECT created_date FROM table2) as total
GROUP BY total.created_date
HERE you simply union the two tables since they have a matching column name. Then you get back every date from both tables. That is in the inner query. The outer query then does the counting.
Hope that makes sense.
I have a table:
ID CLUSTERID
1 56
1 24
1 24
1 35
2 13
2 24
Now, i want to get the following:
I want to count per id, which cluster id repeats most of the time.
For example, in ID=1, CLUSTERID=24 repeats most of the time
In ID=2 i have 2 CLUSTER IDs that repeats the same.
So in the output i will have:
ID CLUSTERID
1 24
2 13
2 24
The answer that i wrote (and works)
TT is my original table that have 2 columns: ID and CLUSTER ID
SELECT t3.ID,t3.ClusterID,t3.ListingAmount
FROM
(SELECT ID, ClusterID, COUNT() AS ListingAmount
FROM tt
GROUP BY ID, ClusterID) AS t3 LEFT JOIN
(SELECT ID, MAX(ListingAmount) AS amount
FROM
(SELECT ID, ClusterID, COUNT() AS ListingAmount
FROM tt
GROUP BY ID, ClusterID) AS t2
GROUP BY ID) AS BB ON BB.id=t3.id
WHERE BB.amount=t3.ListingAmount
Can't think of a more elegant solution right now (I'm sure there is), but it seems to do the job:
select t1.id,
t1.clusterid,
t1.cnt
from (
select id,
clusterid,
count(*) as cnt
from foo
group by id, clusterid
) t1
join (select id,
max(cnt) as max_count
from (
select id,
clusterid,
count(*) as cnt
from foo
group by id, clusterid
) tm
group by id
) t2 on t1.id = t2.id
and t1.cnt = t2.max_count
order by t1.id, t1.cnt;
SQLFiddle example: http://sqlfiddle.com/#!2/2cacc/3
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;