I have two table table1 and table2
table1:
id amount
1 394
2 897
2 345
3 123
table2:
id amount
2 876
3 890
3 876
4 908
I need to generate table3. join two table and SUM id wise and show top 30 amount holder.
table3:
id amount
2 2118
3 1889
4 908
1 394
You can get data in followng way:
select table3.id, SUM (table3.amount) AS amount
FROM
(SELECT * FROM table1 UNION ALL SELECT * FROM table2) table3
GROUP BY table3.id
If you want to create table with above data, you can do like this:
CREATE TABLE new_table select table3.id, SUM (table3.amount) AS amount
FROM
(SELECT * FROM table1 UNION ALL SELECT * FROM table2) table3
GROUP BY table3.id ;
Try using this query
SELECT a.id, SUM(a.amount)
FROM (
SELECT * FROM table1
UNION
SELECT * FROM table2) AS a
GROUP BY a.id
ORDER BY SUM(a.amount)
LIMIT 30 DESC;
Above query will generate results, now to create new table use following
CREATE TABLE table3(
SELECT a.id, SUM(a.amount)
FROM (
SELECT * FROM table1
UNION
SELECT * FROM table2) AS a
GROUP BY a.id
ORDER BY SUM(a.amount)
LIMIT 30 DESC);
Hope this helps...
Related
I am trying to display all records from table1 even if the catid not existing in table2 (all employee in table2 should have all catid from table1 with 0 days if not exising in table2) with the following sql query but getting an error
Error Code: 1054. Unknown column 'catid' in 'group statement'
select empid,days from table2 union select catid from
table1 group by empid, catid;
table1:
catid
1
2
3
table2:
empid catid days (computed column count(*))
1000 1 8
1000 3 10
expected result:
empid catid days
1000 1 8
1000 2 0 <---catid 2 and days 0 if catid not existing in table2 for empid 1000
1000 3 10
That is not the function of the union statement. Union statement does a set like capability which merging two sets. What you are looking for a is a join with the table 1 where you do a count and group by catid. Your data model to achieve this output itself is grievously wrong ;)
select employeeid, catid, sum(days) from table1, table2 group by employeeid, catid;
You just need a LEFT JOIN:
Select tab2.empid, tab2.catid, ifnull(tab2.days, 0)
from tab2
left join tab1 on tab2.catid = tab1.catid
Please note : While doing a UNION the number and type of the columns present in the first select should be the same as the next Selects.
So you need to first make the select columns in sync first.
can you check this and add empid similarly.
SELECT TABLE1.CATID, IFNULL(TABLE2.DAYS,0) FROM table1 LEFT OUTER JOIN
table2 ON table1.catid = table2.catid
Please use LEFT JOIN with IFNULL.
Select table2.empid, table1.catid, IFNULL(table2.days, 0) from table2
LEFT JOIN table1 ON table2.catid = table1.catid;
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;
table1
row_id row_one row_two
1 1 5
2 1 5
3 2 5
4 2 5
5 2 6
table2
row2_id row2_one row2_two
1 1 somevalue
2 2 somevalue2
"select distinct row_one from table1 where row_two=5"
result
row_one
1
2
after that i want select
select * from table2 where row2_one=1
select * from table2 where row2_one=2
i want select with one query.
i am trying this query
select * from table2 where row2_one in (select distinct row_one from table1 where
row_two where row_two=5)
but it took 8s
Showing rows 0 - 14 ( 15 total, Query took 8.3255 sec)
why is it so slow. i want select faster.
please help me!
You don't need the DISTINCT in there. You can just do:
SELECT *
FROM table2
WHERE row2_one IN (SELECT row_one FROM table1 WHERE row_two=5)
And using EXISTS might be faster:
SELECT *
FROM table2 A
WHERE EXISTS (SELECT * FROM table1 WHERE row_two=5 AND row_one = A.row2_one)
Let me add to the that was said bellow - use GROUP BY or EXISTS instead of DISTINCT it can really improve your performance.
Assuming that this is your query:
select *
from table2 where row2_one in (select distinct row_one from table1 where row_two=5)
Then this is well-formed. One thing, you don't need the distinct in the subquery.
If you add an index to the table1 on the column row_two, you should get better performance. An index on row2_one in table2 would also speed it up.
Select distinct table2.*
from table1 t1, table2 t2
where t1.row_two =5 and t1.row2_one = t2.row2_one
I have two tables
The first with only 5 rows
The second with 800 rows
I'm using this query:
SELECT *
FROM table1 t1
JOIN (SELECT * FROM table2 ORDER BY RAND() LIMIT 5) t2
But I'm getting 5 rows from the first table for each result of the second table.
I don't need a condition when joining, I just want 5 random results from the second table to join the 5 results from the first.
Example:
--------------------------------------------------------
|table1 (always with same order)| table2(random order) |
--------------------------------------------------------
item1 | item4
item2 | item2
item3 | item5
item4 | item1
item5 | item3
Do you mean UNION ?
SELECT * FROM table1
UNION SELECT * FROM table2 ORDER BY RAND() LIMIT 5;
Update: revised answer after modification of your question:
SELECT field1 FROM table1
UNION SELECT field2 FROM table2 ORDER BY RAND() LIMIT 5;
To my understanding, you just need one field from each table. If you need several ones, you can list them: field2, field2, ... as long as the number of fields is the same in both SELECTs.
Update 2: ok, I think I see what you mean now. Here is a (dirty) way to do it, I'm quite confident someone can come with a more elegant solution though:
SET #num1=0, #num2=0;
SELECT t1.field1, t2.field2
FROM (
SELECT field1, #num1:=#num1+1 AS num
FROM table1
) AS t1
INNER JOIN (
SELECT field2, #num2:=#num2+1 AS num
FROM (
SELECT field2
FROM table2
ORDER BY RAND()
LIMIT 5
) AS t
) AS t2
ON t1.num = t2.num;
Try use subquery in select. The subquery part pick an id for each row of table1.
SELECT
id AS table1_id,
(
SELECT id FROM table2 ORDER BY RAND() LIMIT 1
) AS table2_id
FROM table1
The query result would be like this:
table1_id
table2_id
1
24
2
13
3
36
4
68
5
5
You may join with table2 to select other table2 column:
SELECT table1.*, table2.*
FROM (
SELECT
id AS table1_id,
(
SELECT id FROM table2 ORDER BY RAND() LIMIT 1
) AS table2_id
FROM table1
) t
JOIN table1 on t.table1_id = table1.id
JOIN table2 on t.table2_id = table2.id