Let's imagine a table with two columns ex:
| Value | ID |
+-------+----+
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| 1 | 2 |
| 2 | 2 |
| 2 | 2 |
What I am trying to do is to calculate the sum of those with similar id and display them in different table like:
| Sum | ID |
+-----+----+
| 9 | 1 |
| 5 | 2 |
and so on.
I could find a sum of a known id by
SELECT SUM(VALUE) FROM MYTABLE WHERE ID = 1;
However not sure on how to find sum of different id's separately, could you give an idea on how to proceed?
Select SUM(VALUE),ID FROM MYTABLE GROUP BY ID
Use GROUP BY clause:
SELECT SUM(VALUE) Sum, ID FROM MYTABLE GROUP BY ID;
SELECT SUM(VALUE),ID FROM MYTABLE Group By ID
Related
I have the following table:
|-----|-----|
| i d | val |
|-----|-----|
| 1 | 1 |
|-----|-----|
| 2 | 4 |
|-----|-----|
| 3 | 3 |
|-----|-----|
| 4 | 7 |
|-----|-----|
Can I get the following output:
|-----|
| sum |
|-----|
| 1 |
|-----|
| 5 |
|-----|
| 8 |
|-----|
| 1 5 |
|-----|
using a single SQLite3 SELECT-query? I know it could be easily achieved using variables, but SQLite3 lacks those. Maybe some recursive query? Thanks.
No.
In a relational database table rows do not have any order. If you specify an order for the rows, then it's possible to write a query.
Now, you could add an extra column to sort the rows. For example:
| val | sort
|-----|-----
| 1 | 10
| 4 | 20
| 3 | 30
| 7 | 40
The query could be:
select
sum(val) over(order by sort)
from my_table
For the updated question, you can write:
select
sum(val) over(order by id)
from my_table
By using the order of the id column and if you want only the sum column, you can do this:
select (select sum(val) from tablename where id <= t.id) sum
from tablename t
This is my table...
+----+--------+
| id | amount |
+----+--------+
| 1 | 100 |
| 1 | 50 |
| 1 | 0 |
| 2 | 500 |
| 2 | 100 |
| 3 | 300 |
| 3 | -2 |
| 4 | 400 |
| 4 | 200 |
+----+--------+
I would like to choose from it each value of id that does not have a nonpositive (i.e. negative or 0) value associated with it, and the smallest amount associated with that id.
If I use this code...
SELECT DISTINCT id, amount
FROM table t
WHERE amount = (SELECT MIN(amount) FROM table WHERE id= t.id)
... then these results show...
+----+--------+
| id | amount |
+----+--------+
| 1 | 0 |
| 2 | 100 |
| 3 | -2 |
| 4 | 200 |
+----+--------+
But what I want the statement to return is...
+----+--------+
| id | amount |
+----+--------+
| 2 | 100 |
| 4 | 200 |
+----+--------+
Just add amount>0 in your query. You missed out that condition in your query. That should do it.
SELECT DISTINCT id, amount FROM table t
WHERE amount = (SELECT MIN(amount) FROM table WHERE id= t.id)
and amount>0;
If you want to display id, where min(amount) > 0, the use this.
SELECT id, min(amount) as amount
FROM table t
group by id
having min(amount) > 0;
Please try the following...
SELECT id,
MIN( amount )
FROM table
WHERE amount > 0
GROUP BY id
ORDER BY id;
This statement starts by selecting all records WHERE amount is larger than 0.
The records from the resulting dataset are then grouped by each surviving value of id and the smallest value of amount is chosen for that GROUP / id.
The resulting pairs of values are then sorted by ORDER id and returned to the user.
If you have any questions or comments, then please feel free to post a Comment accordingly.
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"...
I have a table like:
| ID | Val |
+-------+-----+
| abc-1 | 10 |
| abc-2 | 30 |
| cde-1 | 10 |
| cde-2 | 10 |
| efg-1 | 20 |
| efg-2 | 11 |
and would like to get the result based on the substring(ID, 1, 3) and minimum value and ist must be only the first in case the Val has duplicates
| ID | Val |
+-------+-----+
| abc-1 | 10 |
| cde-1 | 10 |
| efg-2 | 11 |
the problem is that I am stuck, because I cannot use group by substring(id,1,3), ID since it will then have again 2 rows (each for abc-1 and abc-2)
WITH
sorted
AS
(
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY substring(id,1,3) ORDER BY val, id) AS sequence_id
FROM
yourTable
)
SELECT
*
FROM
sorted
WHERE
sequence_id = 1
SELECT SUBSTRING(id,1,3),MIN(val) FROM Table1 GROUP BY SUBSTRING(id,1,3);
You were grouping the columns using both SUBSTRING(id,1,3),id instead of just SUBSTRING(id,1,3). It works perfectly fine.Check the same example in this below link.
http://sqlfiddle.com/#!3/fd9fc/1