I have a single table laid out as such:
id | name | count
1 | John |
2 | Jim |
3 | John |
4 | Tim |
I need to fill out the count column such that the result is the number of times the specific name shows up in the column name.
The result should be:
id | name | count
1 | John | 2
2 | Jim | 1
3 | John | 2
4 | Tim | 1
I can get the count of occurrences of unique names easily using:
SELECT COUNT(name)
FROM table
GROUP BY name
But that doesn't fit into an UPDATE statement due to it returning multiple rows.
I can also get it narrowed down to a single row by doing this:
SELECT COUNT(name)
FROM table
WHERE name = 'John'
GROUP BY name
But that doesn't allow me to fill out the entire column, just the 'John' rows.
you can do that with a common table expression:
with counted as (
select name, count(*) as name_count
from the_table
group by name
)
update the_table
set "count" = c.name_count
from counted c
where c.name = the_table.name;
Another (slower) option would be to use a co-related sub-query:
update the_table
set "count" = (select count(*)
from the_table t2
where t2.name = the_table.name);
But in general it is a bad idea to store values that can easily be calculated on the fly:
select id,
name,
count(*) over (partition by name) as name_count
from the_table;
Another method : Using a derived table
UPDATE tb
SET count = t.count
FROM (
SELECT count(NAME)
,NAME
FROM tb
GROUP BY 2
) t
WHERE t.NAME = tb.NAME
Related
I'm trying to find duplicate rows in a large database (300,000 records). Here's an example of how it looks:
| id | title | thedate |
|----|---------|------------|
| 1 | Title 1 | 2021-01-01 |
| 2 | Title 2 | 2020-12-24 |
| 3 | Title 3 | 2021-02-14 |
| 4 | Title 2 | 2021-05-01 |
| 5 | Title 1 | 2021-01-13 |
I found this excellent (i.e. fast) answer here: Find duplicate rows with PostgreSQL
-- adapted from #MatthewJ answering in https://stackoverflow.com/questions/14471179/find-duplicate-rows-with-postgresql/14471928#14471928
select * from (
SELECT id, title, TO_DATE(thedate,'YYYY-MM-DD'),
ROW_NUMBER() OVER(PARTITION BY title ORDER BY id asc) AS Row
FROM table1
) dups
where
dups.Row > 1
Which I'm trying to use as a base to solve my specific problem: I need to find duplicates according to column values like in the example, but only for records posted within 15 days of each other (the date of record insertion in the column "thedate" in my DB).
I reproduced it in this fiddle http://sqlfiddle.com/#!15/ae109/2, where id 5 (same title as id 1, and posted within 15 days of each other) should be the only acceptable answer.
How would I implement that condition in the query?
With the LAG function you can get the date from the previous row with the same title and then filter based on the time difference.
WITH with_prev AS (
SELECT
*,
LAG(thedate, 1) OVER (PARTITION BY title ORDER BY thedate) AS prev_date
FROM table1
)
SELECT id, title, thedate
FROM with_prev
WHERE thedate::timestamp - prev_date::timestamp < INTERVAL '15 days'
You don't necessarily need window funtions for this, you an use a plain old self-join, like:
select p.id, p.thedate, n.id, n.thedate, p.title
from table1 p
join table1 n on p.title = n.title and p.thedate < n.thedate
where n.thedate::date - p.thedate::date < 15
http://sqlfiddle.com/#!15/a3a73a/7
This has the advantage that it might use some of your indexes on the table, and also, you can decide if you want to use the data (i.e. the ID) of the previous row or the next row from each pair.
If your date column however is not unique, you'll need to be a little more specific in your join condition, like:
select p.id, p.thedate, n.id, n.thedate, p.title
from table1 p
join table1 n on p.title = n.title and p.thedate <= n.thedate and p.id <> n.id
where n.thedate::date - p.thedate::date < 15
This question already has answers here:
Select first row in each GROUP BY group?
(20 answers)
Closed 4 years ago.
I have next data:
id | name | amount | datefrom
---------------------------
3 | a | 8 | 2018-01-01
4 | a | 3 | 2018-01-15 10:00
5 | b | 1 | 2018-02-20
I can group result with the next query:
select name, max(amount) from table group by name
But I need the id of selected row too. Thus I have tried:
select max(id), name, max(amount) from table group by name
And as it was expected it returns:
id | name | amount
-----------
4 | a | 8
5 | b | 1
But I need the id to have 3 for the amount of 8:
id | name | amount
-----------
3 | a | 8
5 | b | 1
Is this possible?
PS. This is required for billing task. At some day 2018-01-15 configuration of a was changed and user consumes some resource 10h with the amount of 8 and rests the day 14h -- 3. I need to count such a day by the maximum value. Thus row with id = 4 is just ignored for 2018-01-15 day. (for next day 2018-01-16 I will bill the amount of 3)
So I take for billing the row:
3 | a | 8 | 2018-01-01
And if something is wrong with it. I must report that row with id == 3 is wrong.
But when I used aggregation function the information about id is lost.
Would be awesome if this is possible:
select current(id), name, max(amount) from table group by name
select aggregated_row(id), name, max(amount) from table group by name
Here agg_row refer to the row which was selected by aggregation function max
UPD
I resolve the task as:
SELECT
(
SELECT id FROM t2
WHERE id = ANY ( ARRAY_AGG( tf.id ) ) AND amount = MAX( tf.amount )
) id,
name,
MAX(amount) ma,
SUM( ratio )
FROM t2 tf
GROUP BY name
UPD
It would be much better to use window functions
There are at least 3 ways, see below:
CREATE TEMP TABLE test (
id integer, name text, amount numeric, datefrom timestamptz
);
COPY test FROM STDIN (FORMAT csv);
3,a,8,2018-01-01
4,a,3,2018-01-15 10:00
5,b,1,2018-02-20
6,b,1,2019-01-01
\.
Method 1. using DISTINCT ON (PostgreSQL-specific)
SELECT DISTINCT ON (name)
id, name, amount
FROM test
ORDER BY name, amount DESC, datefrom ASC;
Method 2. using window functions
SELECT id, name, amount FROM (
SELECT *, row_number() OVER (
PARTITION BY name
ORDER BY amount DESC, datefrom ASC) AS __rn
FROM test) AS x
WHERE x.__rn = 1;
Method 3. using corelated subquery
SELECT id, name, amount FROM test
WHERE id = (
SELECT id FROM test AS t2
WHERE t2.name = test.name
ORDER BY amount DESC, datefrom ASC
LIMIT 1
);
demo: db<>fiddle
You need DISTINCT ON which filters the first row per group.
SELECT DISTINCT ON (name)
*
FROM table
ORDER BY name, amount DESC
You need a nested inner join. Try this -
SELECT id, T2.name, T2.amount
FROM TABLE T
INNER JOIN (SELECT name, MAX(amount) amount
FROM TABLE
GROUP BY name) T2
ON T.amount = T2.amount
In my Postgresql 9.3 database I have a table stock_rotation:
+----+-----------------+---------------------+------------+---------------------+
| id | quantity_change | stock_rotation_type | article_id | date |
+----+-----------------+---------------------+------------+---------------------+
| 1 | 10 | PURCHASE | 1 | 2010-01-01 15:35:01 |
| 2 | -4 | SALE | 1 | 2010-05-06 08:46:02 |
| 3 | 5 | INVENTORY | 1 | 2010-12-20 08:20:35 |
| 4 | 2 | PURCHASE | 1 | 2011-02-05 16:45:50 |
| 5 | -1 | SALE | 1 | 2011-03-01 16:42:53 |
+----+-----------------+---------------------+------------+---------------------+
Types:
SALE has negative quantity_change
PURCHASE has positive quantity_change
INVENTORY resets the actual number in stock to the given value
In this implementation, to get the current value that an article has in stock, you need to sum up all quantity changes since the latest INVENTORY for the specific article (including the inventory value). I do not know why it is implemented this way and unfortunately it would be quite hard to change this now.
My question now is how to do this for more than a single article at once.
My latest attempt was this:
WITH latest_inventory_of_article as (
SELECT MAX(date)
FROM stock_rotation
WHERE stock_rotation_type = 'INVENTORY'
)
SELECT a.id, sum(quantity_change)
FROM stock_rotation sr
INNER JOIN article a ON a.id = sr.article_id
WHERE sr.date >= (COALESCE(
(SELECT date FROM latest_inventory_of_article),
'1970-01-01'
))
GROUP BY a.id
But the date for the latest stock_rotation of type INVENTORY can be different for every article.
I was trying to avoid looping over multiple article ids to find this date.
In this case I would use a different internal query to get the max inventory per article. You are effectively using stock_rotation twice but it should work. If it's too big of a table you can try something else:
SELECT sr.article_id, sum(quantity_change)
FROM stock_rotation sr
LEFT JOIN (
SELECT article_id, MAX(date) AS date
FROM stock_rotation
WHERE stock_rotation_type = 'INVENTORY'
GROUP BY article_id) AS latest_inventory
ON latest_inventory.article_id = sr.article_id
WHERE sr.date >= COALESCE(latest_inventory.date, '1970-01-01')
GROUP BY sr.article_id
You can use DISTINCT ON together with ORDER BY to get the latest INVENTORY row for each article_id in the WITH clause.
Then you can join that with the original table to get all later rows and add the values:
WITH latest_inventory as (
SELECT DISTINCT ON (article_id) id, article_id, date
FROM stock_rotation
WHERE stock_rotation_type = 'INVENTORY'
ORDER BY article_id, date DESC
)
SELECT article_id, sum(sr.quantity_change)
FROM stock_rotation sr
JOIN latest_inventory li USING (article_id)
WHERE sr.date >= li.date
GROUP BY article_id;
Here is my take on it: First, build the list of products at their last inventory state, using a window function. Then, join it back to the entire list, filtering on operations later than the inventory date for the item.
with initial_inventory as
(
select article_id, date, quantity_change from
(select article_id, date, quantity_change, rank() over (partition by article_id order by date desc)
from stockRotation
where type = 'INVENTORY'
) a
where rank = 1
)
select ii.article_id, ii.quantity_change + sum(sr.quantity_change)
from initial_inventory ii
join stockRotation sr on ii.article_id = sr.article_id and sr.date > ii.date
group by ii.article_id, ii.quantity_change
Let's say I have this 3 tables
Countries ProvOrStates MajorCities
-----+------------- -----+----------- -----+-------------
Id | CountryName Id | CId | Name Id | POSId | Name
-----+------------- -----+----------- -----+-------------
1 | USA 1 | 1 | NY 1 | 1 | NYC
How do you get something like
---------------------------------------------
CountryName | ProvinceOrState | MajorCities
| (Count) | (Count)
---------------------------------------------
USA | 50 | 200
---------------------------------------------
Canada | 10 | 57
So far, the way I see it:
Run the first SELECT COUNT (GROUP BY Countries.Id) on Countries JOIN ProvOrStates,
store the result in a table variable,
Run the second SELECT COUNT (GROUP BY Countries.Id) on ProvOrStates JOIN MajorCities,
Update the table variable based on the Countries.Id
Join the table variable with Countries table ON Countries.Id = Id of the table variable.
Is there a possibility to run just one query instead of multiple intermediary queries? I don't know if it's even feasible as I've tried with no luck.
Thanks for helping
Use sub query or derived tables and views
Basically If You You Have 3 Tables
select * from [TableOne] as T1
join
(
select T2.Column, T3.Column
from [TableTwo] as T2
join [TableThree] as T3
on T2.CondtionColumn = T3.CondtionColumn
) AS DerivedTable
on T1.DepName = DerivedTable.DepName
And when you are 100% percent sure it's working you can create a view that contains your three tables join and call it when ever you want
PS: in case of any identical column names or when you get this message
"The column 'ColumnName' was specified multiple times for 'Table'. "
You can use alias to solve this problem
This answer comes from #lotzInSpace.
SELECT ct.[CountryName], COUNT(DISTINCT p.[Id]), COUNT(DISTINCT c.[Id])
FROM dbo.[Countries] ct
LEFT JOIN dbo.[Provinces] p
ON ct.[Id] = p.[CountryId]
LEFT JOIN dbo.[Cities] c
ON p.[Id] = c.[ProvinceId]
GROUP BY ct.[CountryName]
It's working. I'm using LEFT JOIN instead of INNER JOIN because, if a country doesn't have provinces, or a province doesn't have cities, then that country or province doesn't display.
Thanks again #lotzInSpace.
So I have this table:
create table test (
id integer,
rank integer,
image varchar(30)
);
Then some values:
id | rank | image
---+------+-------
1 | 2 | bbb
1 | 3 | ccc
1 | 1 | aaa
2 | 3 | c
2 | 1 | a
2 | 2 | b
I want to group them by id and concatenate the image name in the order given by rank. In mySQL I can do this:
select id,
group_concat( image order by rank asc separator ',' )
from test
group by id;
And the output would be:
1 aaa,bbb,ccc
2 a,b,c
Is there a way I can have this in postgresql?
If I try to use array_agg() the names will not show in the correct order and apparently I was not able to find a way to sort them. (I was using postgres 8.4 )
In PostgreSQL 8.4 you cannot explicitly order array_agg but you can work around it by ordering the rows passed into to the group/aggregate with a subquery:
SELECT id, array_to_string(array_agg(image), ',')
FROM (SELECT * FROM test ORDER BY id, rank) x
GROUP BY id;
In PostgreSQL 9.0 aggregate expressions can have an ORDER BY clause:
SELECT id, array_to_string(array_agg(image ORDER BY rank), ',')
FROM test
GROUP BY id;