When using Array_agg, it returns the same values in different orders. I tried using distinct in a few places and it didn't work. I tried using an order before and after the array and it would fail or not properly exclude results.
I am trying to find all fields in the field column that share the same time and same ID and put them into an array.
Columns are Fieldname, ID, Time
select b.Field, count(*)
from (select Time, ID, array_agg(fieldname) as Field
from a
group by 1,2
order by 3) b
group by b.field
order by 1 desc
This produces duplicate results
For example I will have:
Field Name Count
Ghost,Mark 1234
Mark,Ghost 1234
I also tried this below where I add a subquery where I first order the fields alphabetically when grouping time and ID but it failed to execute. I think due to array_agg not being the root query?
select a.Field, count(*)
from
(select Time, ID, array_agg(fieldname) as field
from
(select Time, ID, fieldname
from a
group by 1,2
order by 3 desc) a
group by 1,2 ) b
group by 1
order by 2 desc
Related
I'm trying to make an SQL query that returns the greatest number from a column and its respective id.
For more information I have two columns ID and NUMBER. Both of them have 2 entries and I want to get the highest number with the ID next to it. This is what I tried but didn't success.
SELECT ID, MAX(NUMBER) AS MAXNUMB
FROM TABLE1
GROUP BY ID, MAXNUMB;
The problem I'm experiencing is that it just shows ALL the entries and if I add a "where" expression it just shows the same (all entries [ids+numbers]).
Pd.: Yes, I got what I wanted but only with one column (number) if I add another column (ID) to select it "brokes".
Try:
SELECT
ID,
A_NUMBER
FROM TABLE1
WHERE A_NUMBER = (
SELECT MAX(A_NUMBER)
FROM TABLE1);
Presuming you want the IDs* of the row with the highest number (and not, instead, the highest number for each ID -- if IDs were not unique in your table, for example).
* there may be more than one ID returned if there are two or more IDs with equal maximum numbers
you can try this
Select ID,maxNumber
From
(
SELECT
ID,
(Select Max(NUMBER) from Tmp where Id = t.Id) maxNumber
FROM
Tmp t
)T1
Group By ID,maxNumber
The query you posted has an illegal column name (number) and is group by the alias for the max value, which is illegal and also doesn't make sense; and you can't include the unaliased max() within the group-by either. So it's likely you're actually doing something like:
select id, max(numb) as maxnumb
from table1
group by id;
which will give one row per ID, with the maximum numb (which is the new name I've made up for your numeric column) for each ID. Or as you said you get "ALL the entries" you might have group by id, numb, which would show all rows from the table (unless there are duplicate combinations).
To get the maximum numb and the corresponding id you could group by id only, order by descending maxnumb, and then return the first row only:
select id, max(numb) as maxnumb
from table1
group by id
order by maxnumb desc
fetch first 1 row only
If there are two ID with the same maxnumb then you would only get one of them - and which one is indeterminate unless you modify the order by - but in that case you might prefer to use first 1 row with ties to see them all.
You could achieve the same thing with a subquery and analytic function to generating a ranking, and have the outer query return the highest-ranking row(s):
select id, numb as maxnumb
from (
select id, numb, dense_rank() over (order by numb desc) as rnk
from table1
)
where rnk = 1
You could also use keep to get the same result as first 1 row only:
select max(id) keep (dense_rank last order by numb) as id, max(numb) as maxnumb
from table1
fiddle
I have a table in postgres with 2 fields: they are columns of ids of users who have looked at some data, under two conditions:
viewee viewer
------ ------
93024 66994
93156 93151
93163 113671
137340 93161
92992 93161
93161 93135
93156 93024
And I want to group them by both viewee and viewer field, and count the number of occurrences, and return that count
from high to low:
id count
------ -----
93161 3
93156 2
93024 2
137340 1
66994 1
92992 1
93135 1
93151 1
93163 1
I have been running two queries, one for each column, and then combining the results in my JavaScript application code. My query for one field is...
SELECT "viewer",
COUNT("viewer")
FROM "public"."friend_currentfriend"
GROUP BY "viewer"
ORDER BY count DESC;
How would I rewrite this query to handle both fields at once?
You can combine to columns from the table into a single one by using union all then use group by as below:
select id ,count(*) Count from (
select viewee id from vv
union all
select viewer id from vv) t
group by id
order by count(*) desc
Results:
This is a good place to use a lateral join:
select v.viewx, count(*)
from t cross join lateral
(values (t.viewee), (t.viewer)) v(viewx)
group by v.viewx
order by count(*) desc;
You can try this :
SELECT a.ID,
SUM(a.Total) as Total
FROM (SELECT t.Viewee AS ID,
COUNT(t.Viewee) AS Total
FROM #Temp t
GROUP BY t.Viewee
UNION
SELECT t.Viewer AS ID,
COUNT(t.Viewer) AS Total
FROM #Temp t
GROUP BY t.Viewer
) a
GROUP BY a.ID
ORDER BY SUM(a.Total) DESC
Select id, name , max(modify_time)
from customer
group by id, name
but I get all records.
Order by modify_time desc and use row_number to number the row for id,name combination.Then select each combination with row_number = 1
select id,modify_time,name
from (
select id,modify_time,name,row_number() over(partition by id order by modify_time desc) as r_no
from customer
) a
where a.r_no=1
Ids are unique, which means grouping them by the id, will result in the same table.
My suggestion would be, to order the table by "modify_time" descending and limit the result to 1 (Maybe something like the following):
Select id, name modify_time from customer ORDER BY modify_time DESC limit 1
The reason you are getting the whole table as a result is because you are grouping by id AND name. That means every unique combination of id and name is returned. And since all names per id are different, the whole table is returned.
If you want the last modification per id (or name) you should only group by id (or name respectively).
I want to make a select query which groups rows based on a given column and then sorts by size of such groups.
Let's say we have this sample data:
id type
1 c
2 b
3 b
4 a
5 c
6 b
I want to obtain the following by grouping and sorting the column 'type' in a descending way:
id type
2 b
3 b
6 b
1 c
5 c
4 a
As of now I am only able to get the count of each group but that is not exactly what I need:
SELECT *, COUNT(type) AS typecount
FROM sampletable
GROUP BY type
ORDER BY typecount DESC, type ASC
id type count
2 b 3
1 c 2
4 a 1
Can anybody please give me a hand with this query?
Edit:
Made 'b' the biggest group to avoid coming to the same solution by using only SORT BY
You can't use a column alias in your GROUP BY; just repeat the expression:
SELECT type, COUNT(type) AS count
FROM sampletable
GROUP BY type
ORDER BY COUNT(*) DESC, type ASC
Note that I changed the SELECT clause - you can't use * in your SELECT either since expressions in the SELECT need to either be in the GROUP BY clause or an aggregation.
It may not be the best way, but it will give you what you want.
You work out the totals for each group and then join that "virtual" table to your original table by the determined counts.
SELECT *
FROM sampletable s1
INNER JOIN (SELECT count(type) AS iCount,type
FROM sampletable
GROUP BY type) s2 ON s2.type = s1.type
ORDER BY s2.iCount DESC, s1.type ASC
http://sqlfiddle.com/#!9/f6b0c4/6/0
You can't perform GROUP BY operation on COLUMN ALIAS.
The reason why you can't use ALIAS on the GROUP BY clause that is created on the same level of the SELECT statement is because the GROUP BY is executed before the SELECT clause in which the ALIAS is created.
This is the SQL Order of Operation:
FROM clause
WHERE clause
GROUP BY clause
HAVING clause
SELECT clause
ORDER BY clause
Try following query:
SELECT type, COUNT(type) AS count
FROM sampletable
GROUP BY type
ORDER BY COUNT(*) DESC, type ASC;
EDIT:-
SELECT id, type
FROM sampletable
ORDER BY type DESC, id ASC;
The below statement retrieves the top 2 records within each group in SQL Server. It works correctly, however as you can see it doesn't scale at all. I mean that if I wanted to retrieve the top 5 or 10 records instead of just 2, you can see how this query statement would grow very quickly.
How can I convert this query into something that returns the same records, but that I can quickly change it to return the top 5 or 10 records within each group instead, rather than just 2? (i.e. I want to just tell it to return the top 5 within each group, rather than having 5 unions as the below format would require)
Thanks!
WITH tSub
as (SELECT CustomerID,
TransactionTypeID,
Max(EventDate) as EventDate,
Max(TransactionID) as TransactionID
FROM Transactions
WHERE ParentTransactionID is NULL
Group By CustomerID,
TransactionTypeID)
SELECT *
from tSub
UNION
SELECT t.CustomerID,
t.TransactionTypeID,
Max(t.EventDate) as EventDate,
Max(t.TransactionID) as TransactionID
FROM Transactions t
WHERE t.TransactionID NOT IN (SELECT tSub.TransactionID
FROM tSub)
and ParentTransactionID is NULL
Group By CustomerID,
TransactionTypeID
Use Partition by to solve this type problem
select values from
(select values ROW_NUMBER() over (PARTITION by <GroupColumn> order by <OrderColumn>)
as rownum from YourTable) ut where ut.rownum<=5
This will partitioned the result on the column you wanted order by EventDate Column then then select those entry having rownum<=5. Now you can change this value 5 to get the top n recent entry of each group.