I have table :
+------+-------+-----------------+
| id | name | code | desc |
+------+-------+-----------------+
| 1 | aa | 032016 | grape |
| 1 | aa | 012016 | apple |
| 1 | aa | 032016 | grape |
| 1 | aa | 022016 | orange |
| 1 | aa | 012016 | apple |
| 1 | aa | 032016 | grape |
+------+-------+-----------------+
i tried with query:
SELECT id, name, code, desc, COUNT(code) as view
FROM mytable
GROUP BY id, name, code, desc
and the result is :
+------+-------+------------------------+
| id | name | code | desc | view |
+------+-------+------------------------+
| 1 | aa | 012016 | apple | 2 |
| 1 | aa | 022016 | orange | 1 |
| 1 | aa | 032016 | grape | 3 |
+------+-------+------------------------+
what i expected is like this :
+------+-------+----------------------------------------------------+
| id | name | code | desc | view |
+------+-------+----------------------------------------------------+
| 1 | aa | 012016,022016,032016 | apple,orange,grape | 2,1,3 |
+------+-------+----------------------------------------------------+
can anyone help me how to aggregate the result?
thanks in advance
Your table design has me a bit worried. Is it coincidence that one fruit always has the same code in the table? Then why store it redundantly? There should be a fruit table holding each fruit and its code only once. You know why this is called a relational database system, don't you?
However, with your query you are almost where you wanted to get. You have the counts per id, name, code, and desc. Now you want to aggregate even further. So in the next step group by id and name, because you want one result row per id and name it seems. Use LISTAGG to concatenate the strings in the group:
SELECT
id,
name,
listagg(code, ',') within group(order by code) as codes,
listagg(desc, ',') within group(order by code) as descs,
listagg(view, ',') within group(order by code) as views
FROM
(
SELECT id, name, code, desc, COUNT(*) as view
FROM mytable
GROUP BY id, name, code, desc
)
GROUP BY id, name
ORDER BY id, name;
Related
I have a (mssql) table like this:
+----+----------+---------+--------+--------+
| id | username | date | scoreA | scoreB |
+----+----------+---------+--------+--------+
| 1 | jim | 01/2020 | 100 | 0 |
| 2 | max | 01/2020 | 0 | 200 |
| 3 | jim | 01/2020 | 0 | 150 |
| 4 | max | 02/2020 | 150 | 0 |
| 5 | jim | 02/2020 | 0 | 300 |
| 6 | lee | 02/2020 | 100 | 0 |
| 7 | max | 02/2020 | 0 | 200 |
+----+----------+---------+--------+--------+
What I need is to get the best "combined" score per date. (With "combined" score I mean the best scores per user and per date summarized)
The result should look like this:
+----------+---------+--------------------------------------------+
| username | date | combined_score (max(scoreA) + max(scoreB)) |
+----------+---------+--------------------------------------------+
| jim | 01/2020 | 250 |
| max | 02/2020 | 350 |
+----------+---------+--------------------------------------------+
I came this far:
I can group the scores by user like this:
SELECT
username, (max(scoreA) + max(scoreB)) AS combined_score,
FROM score_table
GROUP BY username
ORDER BY combined_score DESC
And I can get the best score per date with PARTITION BY like this:
SELECT *
FROM
(SELECT t.*, row_number() OVER (PARTITION BY date ORDER BY scoreA DESC) rn
FROM score_table t) as tmp
WHERE tmp.rn = 1
ORDER BY date
Is there a proper way to combine these statements and get the result I need? Thank you!
Btw. Don't care about possible ties!
You can combine window functions and aggregation functions like this:
SELECT s.*
FROM (SELECT username, date, (max(scoreA) + max(scoreB)) AS combined_score,
ROW_NUMBER() OVER (PARTITION BY date ORDER BY max(scoreA) + max(scoreB) DESC) as seqnum
FROM score_table
GROUP BY username, date
) s
ORDER BY combined_score DESC;
Note that date needs to be part of the aggregation.
Question
Say I have a table with such rows:
id | country | place | last_action | second_to_last_action
----------------------------------------------------------
1 | US | 2 | reply |
1 | US | 2 | | comment
4 | DE | 5 | reply |
4 | | | | comment
What I want to do is to combine these by id, country and place so that the last_action and second_to_last_action would be on the same row
id | country | place | last_action | second_to_last_action
----------------------------------------------------------
1 | US | 2 | reply | comment
4 | DE | 5 | reply | comment
How would I approach this? I guess I would need an aggregate here but my mind is hitting completely blank on which one should I use.
It can be expected that there will always be a matching pair.
Background:
Note: this table has been derived from something like this:
id | country | place | action | time
----------------------------------------------------------
1 | US | 2 | reply | 16:15
1 | US | 2 | comment | 15:16
1 | US | 2 | view | 13:16
4 | DE | 5 | reply | 17:15
4 | DE | 5 | comment | 16:16
4 | DE | 5 | view | 14:12
Code used to partition was:
row_number() over (partition by id order by time desc) as event_no
And then I got the last and second_to_last action by getting event_no 1 & 2. So if there's more efficient way to get the last two actions in two distinct columns I would be happy to hear that.
You can fix your first data by using aggregation:
select id, country, place, max(last_action), max(second_to_last_action)
from derived
group by id, country, place;
You can do this from the original table using conditional aggregation:
select id, country, place,
max(case when seqnum = 1 then action end) as last_action,
max(case when seqnum = 2 then action end) as second_to_last_action
from (select t.*,
row_number() over (partition by id order by time desc) as seqnum
from t
) t
group by id, country, place;
I have the following table
+---------+----------+------------+------------+----------+----------+
| PN | code | date | F2 | PO | Supplier |
+---------+----------+------------+------------+----------+----------+
| CDS0055 | NBR00008 | 16.06.2017 | 19.06.2017 | 9872786 | S00021XC |
| CDS0055 | NBR00008 | 16.06.2017 | 03.07.2017 | | S00021XC |
| CDS0055 | NBR00008 | 16.06.2017 | 04.07.2017 | | S0000000 |
+---------+----------+------------+------------+----------+----------+
and I want to get this result (group min (f2)):
+---------+----------+------------+------------+---------+----------+
| PN | code | date | F2 | PO | Supplier |
+---------+----------+------------+------------+---------+----------+
| CDS0055 | NBR00008 | 16.06.2017 | 19.06.2017 | 9872786 | S00021XC |
+---------+----------+------------+------------+---------+----------+
Query:
select distinct PN, min(f2), date, PO, siplier from Order
group by pn, po, date, supplier
You can get the minimum value for a column within each group (and all the other corresponding values) using the ROW_NUMBER() analytic function:
SELECT *
FROM (
SELECT t.*,
ROW_NUMBER() OVER (
PARTITION BY pn,po,"date",supplier
ORDER BY f2
) AS rn
FROM your_table t
)
WHERE rn = 1
I would like to filter my table by MIN() function but still keep columns which cant be grouped.
I have table:
+----+----------+----------------------+
| ID | distance | geom |
+----+----------+----------------------+
| 1 | 2 | DSDGSAsd23423DSFF |
| 2 | 11.2 | SXSADVERG678BNDVS4 |
| 2 | 2 | XCZFETEFD567687SDF |
| 3 | 24 | SADASDSVG3423FD |
| 3 | 10 | SDFSDFSDF343DFDGF |
| 4 | 34 | SFDHGHJ546GHJHJHJ |
| 5 | 22 | SDFSGTHHGHGFHUKJYU45 |
| 6 | 78 | SDFDGDHKIKUI45 |
| 6 | 15 | DSGDHHJGHJKHGKHJKJ65 |
+----+----------+----------------------+
This is what I would like to achieve:
+----+----------+----------------------+
| ID | distance | geom |
+----+----------+----------------------+
| 1 | 2 | DSDGSAsd23423DSFF |
| 2 | 2 | XCZFETEFD567687SDF |
| 3 | 10 | SDFSDFSDF343DFDGF |
| 4 | 34 | SFDHGHJ546GHJHJHJ |
| 5 | 22 | SDFSGTHHGHGFHUKJYU45 |
| 6 | 15 | DSGDHHJGHJKHGKHJKJ65 |
+----+----------+----------------------+
it is possible when I use MIN() on distance column and grouping by ID but then I loose my geom which is essential.
The query looks like this:
SELECT "ID", MIN(distance) AS distance FROM somefile GROUP BY "ID"
the result is:
+----+----------+
| ID | distance |
+----+----------+
| 1 | 2 |
| 2 | 2 |
| 3 | 10 |
| 4 | 34 |
| 5 | 22 |
| 6 | 15 |
+----+----------+
but this is not what I want.
Any suggestions?
One common approach to this is to find the minimum values in a derived table that you join with:
SELECT somefile."ID", somefile.distance, somefile.geom
FROM somefile
JOIN (
SELECT "ID", MIN(distance) AS distance FROM somefile GROUP BY "ID"
) t ON t.distance = somefile.distance AND t.ID = somefile.ID;
Sample SQL Fiddle
You need a window function to do this:
SELECT "ID", distance, geom
FROM (
SELECT "ID", distance, geom, rank() OVER (PARTITION BY "ID" ORDER BY distance) AS rnk
FROM somefile) sub
WHERE rnk = 1;
This effectively orders the entire set of rows first by the "ID" value, then by the distance and returns the record for each "ID" where the distance is minimal - no need to do a GROUP BY.
select a.*,b.geom from
(SELECT ID, MIN(distance) AS distance FROM somefile GROUP BY ID) as a
inner join somefile as b on a.id=b.id and a.distance=b.distance
You can use "distinct on" clause of the PostgreSQL.
select distinct on(id) id, distance, geom
from table_name
order by distance;
I think this is what you are exactly looking for.
For more details on how "distinct on" works, refer the documentation and the example.
But, remember, using "distinct on" does not comply to SQL standards.
I have a table, and I'd like to select rows with the highest value. For example:
----------------
| user | index |
----------------
| 1 | 1 |
| 2 | 1 |
| 2 | 2 |
| 3 | 4 |
| 3 | 7 |
| 4 | 1 |
| 5 | 1 |
----------------
Expected result:
----------------
| user | index |
----------------
| 1 | 1 |
| 2 | 2 |
| 3 | 7 |
| 4 | 1 |
| 5 | 1 |
----------------
How may I do so? I assume it can be done by some oracle function I am not aware of?
Thanks in advance :-)
You can use MAX() function for that with grouping user column like this:
SELECT "user"
,MAX("index") AS "index"
FROM Table1
GROUP BY "user"
ORDER BY "user";
Result:
| USER | INDEX |
----------------
| 1 | 1 |
| 2 | 2 |
| 3 | 7 |
| 4 | 1 |
| 5 | 1 |
See this SQLFiddle
if you have more than one column
select user , index
from (
select u.* , row_number() over (partition by user order by index desc) as rnk
from some_table u)
where rnk = 1
user is a reserved word - you should use a different name for the column.
select user,max(index) index from tbl
group by user;
Alternatively, you can use analytic functions:
select user,index, max(index) over (partition by user order by 1 ) highest from YOURTABLE
Note: Try NOT to use words like user, index, date etc.. as your column names, as they are reserved words for Oracle. If you will use, then use them with quotation marks, eg. "index", "date"...