How can I do something like this in HIVE:
Table1:
ID Name Friends
1 Tom 5
Table2:
ID Name DOB
1 Jerry 10/10/1999
1 Kate Null
1 Peter 02/11/1983
1 Robert Null
1 Mitchell 09/09/2000
What I want to do is:
For each ID in table 1, find num of not null DOB and then divide by Friends
I wrote a query as:
SELECT t.ID, t.Friends, COUNT(s.DOB)/ t. Friends from Table1 t join Table2 s on (t.ID = s.ID) GROUP BY t.ID
When I do this, I get the error as FRIENDS is not part of the GROUP BY Key
The answer I am expecting is: 3/5
Just add the FRIENDS to your GROUP BY section:
SELECT t.ID, t.FRIENDS, COUNT(s.DOB)/ t. FRIENDS
from Table1 t
join Table2 s
on (t.ID = s.ID)
GROUP BY t.ID. t.FRIENDS
I prefer to write this kind of query like this:
SELECT t.ID, case when t.FIREND>0 then Cnt / t.FRIENDS ELSE 0 END
FROM Table t1
JOIN (Select ID, Count(*) AS Cnt from Table 1 GROUP BY Id) t2
ON t1.ID = t2.ID
if you have declare id,friends in table 1 as integer and id in table2 as integer, then below query will get you intended output
select a.name, concat(cast(b.cnt as string),'/',cast(a.friends as string))
from table1 a
join
(select id, count(DOB) as cnt from table2 where DOB is not null group by id) b
on (a.id = b.id)
Related
I am trying to join a table and get a count but I cannot count an ID twice in the table for the count.
Table 1:
ID animal
-- ------
1 dog
2 dog
3 cat
4 cat
5 dog
Table 2:
ID
--
2
2
3
5
5
I need to get a count of how many of each type of animal are in table 2. I can get it to join and change the ID to the type of animal and then get a count of each.
The issue is that each ID can only get counted once. So the expected output would be.
dog:2
cat:1
Where my output is
dog:4
cat:1
Try like below
select t1.animal, count( distinct t2.ID)
from table1 t1 join table2 t2 on t1.ID=t2.ID
group by t1.animal
You can try below using count distinct id
select b.animal,count(distinct a.id) from table2 a
inner join table1 b on a.id=b.id
group by b.animal
Try this:
SELECT t1.animal AS "Animal", COUNT(DISTINCT t1.ID) AS "No. of Animals"
FROM TABLE2 t2, TABLE1 t1
WHERE t2.ID = t1.ID
GROUP BY t1.animal
You can Try Nested Selects here.
SELECT
t.animal,
COUNT(t.ID) AS Count
FROM
(SELECT DISTINCT a.animal, b.ID FROM table1 a INNER JOIN table2 b ON a.ID = b.ID)t
GROUP BY t.animal
this is tested image
We are having a following sqlite3 table named 'atable'
id student assignment grade
-----------------------------------
1 A 123 9
2 A 456 9
3 A 234 8
4 B 534 7
5 B 654 9
6 C 322 7
id is unique and incremented for each records. We are fetching latest assignment for each user by running query
SELECT student, assignment, grade from atable where id in
(select max(id) from atable group by student) order by id desc
This is working fine. However, we also need to fetch number of assignments for each user where user received a particular grade in the same query, say 9.
Any idea suggestion how to enhance or rewrite above query to return count as well. As mentioned, we are using sqlite3.
Thanks
You can use this correlated query:
SELECT t.student, t.assignment, t.grade,
(SELECT COUNT(*) FROM atable s
WHERE s.student = t.student and s.grade >= 9) as total_above_9
from atable t
where t.id in
(select max(id) from atable group by student)
order by t.id desc
It would be better to join to a derived table that contains an aggregated version of the original table:
select t1.student, t1.assignment, t1.grade, t2.cnt
from mytable as t1
join (
select student, max(id) as id,
count(case when grade = 9 then 1 end) as cnt
from mytable
group by student
) as t2 on t1.id = t2.id
Try this;)
select t1.student, t1.assignment, t1.grade, t2.count
from atable t1
inner join (select max(id) as id, count(if(grade=9, 1, null)) as count from atable group by student) t2
on t1.id = t2.id
order by t1.id desc
Example:
SELECT SUM(SALARY) FROM (SELECT * FROM table1 WHERE id > 10) a LEFT JOIN table2 b on a.person = b.person
I want join table2 records only to (SELECT * FROM table1 WHERE id > 10) records, my example is not correct.
table1 contain 100mln records and I cant join table2 to all records I must use subquery
I'm assuming, you salary is not summing up correctly (you are getting more than you expect). This is because LEFT JOIN will leave NULL for the rowsthat doesn't have match in b.
For this SQL:
SELECT a.*, b.*
FROM (select * from (SELECT 123 AS Salary,
'Tom' AS person
UNION
SELECT 343 AS Salary,
'Bob' AS person
UNION
SELECT 877 AS Salary,
'Tom' AS person) as t where t.Salary > 123) a
LEFT JOIN (SELECT *
FROM (SELECT 'Tom' AS person,
1 AS id
UNION
SELECT 'Bob' AS person,
2 AS id) AS t
WHERE t.id = 1) AS b
ON a.person = b.person
you will have this output:
So INNER JOIN should work for you.
SELECT SUM(SALARY) FROM (SELECT * FROM table1 WHERE id > 10) a
LEFT JOIN table2 b on a.person = b.person
Hopefully this will get you going in the correct direction....
select sum(a.salary)
from table1 a
left join table2 b on a.person = b.person and b.salary_type = "something"
where a.id > 10
;
table1
cid
itemdesc
itemprice
table2
cid
imagename
status
My 1st table is has unique cid (no duplicate) I want it to LEFT JOIN TO table2 but it has multiple rows per cid
cid imagename status
1 image1-of-cid1 test1
1 image2-of-cid1 test2
2 image1-of-cid2 test3
2 image2-of-cid2 test4
2 image3-of-cid2 test5
But I only want the Query to return the the 1st row only of the each record fom table 1
Thanks
I agree with John Woo's answer above. You need a subquery of some kind to actually retrieve the first row of table 2. Something like:
SELECT
t1.[id],
t2.*
FROM table1 AS t1
LEFT JOIN table2 AS t2
ON t2.cid = (SELECT TOP 1 cid FROM table2 WHERE cid = t1.cid)
you need to create an extra subquery that gets one imagename per cid. try this,
SELECT a.*, b.*
FROM table1 a
LEFT JOIN
(
SELECT cid, MIN(imagename) minImage
FROM table2
GROUP BY cid
) c ON a.cid = c.cid
LEFT JOIN table2 b
ON c.cid = b.cid AND
b.imageName = c.minImage
SQLFiddle Demo
Select
distinct a.cid,a.itemdesc,b.imagename,a.itemprice,b.status
from table1 a,
table2 b
where a.cid=b.cid
Try this:
SELECT a.cid, a.itemdesc, a.itemprice, b.imagename, b.status
FROM table1 a
LEFT OUTER JOIN table2 AS b ON a.cid = b.cid
GROUP BY a.cid, a.itemdesc, a.itemprice;
How to select last date in 2 records in Mysql ?
TABLE A
SID NAME Sex
1 Jam M
2 Da F
TABLE B
ID Title SID Date
1 A 1 2012-07-31 09:57:10
2 NULL 1 2012-07-31 09:57:13
3 A 2 2012-07-31 10:10:13
4 NULL 2 2012-07-31 10:57:13
I want to inner join those two tables,
but select only one record only of Table B(distinct of SID) where title not null then show the
biggest Date of title is null field.
Result want the biggest Date of Null title:
ID Name Title SID Date
1 Jam A 1 **2012-07-31 09:57:13**
2 DA A 2 **2012-07-31 10:57:13**
How to do that ?
SELECT b.ID, a.NAME, b.Title, b.SID, b.Datea
from tablea a left outer join
(( SELECT sid, MAX(datea) AS latest
FROM tableb
where title is not null
GROUP BY sid) AS dt
INNER JOIN tableb b ON b.sid= dt.sid and b.datea=dt.latest )
on a.sid=b.sid
I think something like this will work for you:
SELECT b.ID, a.NAME, b.Title, b.SID, b.Date
FROM TABLEA a
INNER JOIN (SELECT SID, IFNULL(Title, "") AS Title,
MAX(IF(Title IS NULL, Date, NULL)) Date
FROM TABLEB GROUP BY SID) b
ON a.SID = b,SID;
Try my code (it works):
SELECT b.id, b.title, b.sid, b.date
FROM table_b as b
JOIN table_a as a ON a.sid = b.sid
WHERE b.id in (SELECT MAX(id) FROM table_b GROUP BY sid)