I have table with descriptions of smth. For example:
My_Table
id description
================
1 ABC
2 ABB
3 OPAC
4 APEЧ
I need to get all unique symbols from all "description" columns.
Result should look like that:
symbol
================
A
B
C
O
P
E
Ч
And it shoud work for all languages, so, as I see, regular expressions cant help.
Please help me. Thanks.
with cte (c,description_suffix) as
(
select substr(description,1,1)
,substr(description,2)
from mytable
where description is not null
union all
select substr(description_suffix,1,1)
,substr(description_suffix,2)
from cte
where description_suffix is not null
)
select c
,count(*) as cnt
from cte
group by c
order by c
or
with cte(n) as
(
select level
from dual
connect by level <= (select max(length(description)) from mytable)
)
select substr(t.description,c.n,1) as c
,count(*) as cnt
from mytable t
join cte c
on c.n <= length(description)
group by substr(t.description,c.n,1)
order by c
+---+-----+
| C | CNT |
+---+-----+
| A | 4 |
| B | 3 |
| C | 2 |
| E | 1 |
| O | 1 |
| P | 2 |
| Ч | 1 |
+---+-----+
Create a numbers table and populate it with all the relevant ids you'd need (in this case 1..maxlength of string)
SELECT DISTINCT
locate(your_table.description, numbers.id) AS symbol
FROM
your_table
INNER JOIN
numbers
ON numbers.id >= 1
AND numbers.id <= CHAR_LENGTH(your_table.description)
SELECT DISTINCT(SUBSTR(ll,LEVEL,1)) OP --Here DISTINCT(SUBSTR(ll,LEVEL,1)) is used to get all distinct character/numeric in vertical as per requirment
FROM
(
SELECT LISTAGG(DES,'')
WITHIN GROUP (ORDER BY ID) ll
FROM My_Table --Here listagg is used to convert all values under description(des) column into a single value and there is no space in between
)
CONNECT BY LEVEL <= LENGTH(ll);
Related
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3
I'd like to get a count of all of the Ids that have have the same value (Drops) as other Ids. For instance, the illustration below shows you that ID 1 and 3 have A drops so the query would count them. Similarly, ID 7 & 18 have B drops so that's another two IDs that the query would count totalling in 4 Ids that share the same values so that's what my query would return.
+------+-------+
| ID | Drops |
+------+-------+
| 1 | A |
| 2 | C |
| 3 | A |
| 7 | B |
| 18 | B |
+------+-------+
I've tried the several approaches but the following query was my last attempt.
With cte1 (Id1, D1) as
(
select Id, Drops
from Posts
),
cte2 (Id2, D2) as
(
select Id, Drops
from Posts
)
Select count(distinct c1.Id1) newcnt, c1.D1
from cte1 c1
left outer join cte2 c2 on c1.D1 = c2.D2
group by c1.D1
The result if written out in full would be a single value output but the records that the query should be choosing should look as follows:
+------+-------+
| ID | Drops |
+------+-------+
| 1 | A |
| 3 | A |
| 7 | B |
| 18 | B |
+------+-------+
Any advice would be great. Thanks
You can use a CTE to generate a list of Drops values that have more than one corresponding ID value, and then JOIN that to Posts to find all rows which have a Drops value that has more than one Post:
WITH CTE AS (
SELECT Drops
FROM Posts
GROUP BY Drops
HAVING COUNT(*) > 1
)
SELECT P.*
FROM Posts P
JOIN CTE ON P.Drops = CTE.Drops
Output:
ID Drops
1 A
3 A
7 B
18 B
If desired you can then count those posts in total (or grouped by Drops value):
WITH CTE AS (
SELECT Drops
FROM Posts
GROUP BY Drops
HAVING COUNT(*) > 1
)
SELECT COUNT(*) AS newcnt
FROM Posts P
JOIN CTE ON P.Drops = CTE.Drops
Output
newcnt
4
Demo on SQLFiddle
You may use dense_rank() to resolve your problem. if drops has the same ID then dense_rank() will provide the same rank.
Here is the demo.
with cte as
(
select
drops,
count(distinct rnk) as newCnt
from
( select
*,
dense_rank() over (partition by drops order by id) as rnk
from myTable
) t
group by
drops
having count(distinct rnk) > 1
)
select
sum(newCnt) as newCnt
from cte
Output:
|newcnt |
|------ |
| 4 |
First group the count of the ids for your drops and then sum the values greater than 1.
select sum(countdrops) as total from
(select drops , count(id) as countdrops from yourtable group by drops) as temp
where countdrops > 1;
Is there SQL statement to match up multiple records to an exact match of multiple records in another table?
Lets say I have table A
ID | List# | Item
1 | 5 | A
2 | 5 | C
3 | 5 | B
4 | 6 | A
5 | 6 | D
*I purposely made Items 'ABC' out of order as the order of the records I receive may be out of order.
Table B
ID | Group | Item
1 | AAA | A
2 | AAA | B
3 | AAA | C
4 | AAA | D
5 | BBB | A
6 | BBB | B
7 | BBB | C
8 | DDD | A
If looking at the first table, I would want List# 5 to return a match only for group 'BBB', as all (and only) three records match.
The simplest way is to aggregate into a string or array and join. Standard SQL supports listagg(), so you can do:
select a.list, b.list, a.items
from (select a.list, listagg(item, ',') within group (order by item) as items
from a
group by a.list
) a join
(select b.list, listagg(item, ',') within group (order by item) as items
from b
group by b.list
) b
on a.items = b.items;
Not all databases support listagg(). Many -- but not all -- have similar functionality. This is simpler than the "standard" SQL approach.
You can simulate database division. It's a little bit cumbersome but here it is:
with
x as (
select
from a
where a.list = 5
),
y as (
select grp, count(*) as cnt
from b
join x on x.item = b.item
group by grp
)
select grp
from y
where cnt = (select count(*) from x)
How can we create all the combinations of any length for the values in one column and return the distinct count of another column for that combination?
Table:
+------+--------+
| Type | Name |
+------+--------+
| A | Tom |
| A | Ben |
| B | Ben |
| B | Justin |
| C | Ben |
+------+--------+
Output Table:
+-------------+-------+
| Combination | Count |
+-------------+-------+
| A | 2 |
| B | 2 |
| C | 1 |
| AB | 3 |
| BC | 2 |
| AC | 2 |
| ABC | 3 |
+-------------+-------+
When the combination is only A, there are Tom and Ben so it's 2.
When the combination is only B, 2 distinct names so it's 2.
When the combination is A and B, 3 distinct names: Tom, Ben, Justin so it's 3.
I'm working in Amazon Redshift. Thank you!
NOTE: This answers the original version of the question which was tagged Postgres.
You can generate all combinations with this code
with recursive td as (
select distinct type
from t
),
cte as (
select td.type, td.type as lasttype, 1 as len
from td
union all
select cte.type || t.type, t.type as lasttype, cte.len + 1
from cte join
t
on 1=1 and t.type > cte.lasttype
)
You can then use this in a join:
with recursive t as (
select *
from (values ('a'), ('b'), ('c'), ('d')) v(c)
),
cte as (
select t.c, t.c as lastc, 1 as len
from t
union all
select cte.type || t.type, t.type as lasttype, cte.len + 1
from cte join
t
on 1=1 and t.type > cte.lasttype
)
select type, count(*)
from (select name, cte.type, count(*)
from cte join
t
on cte.type like '%' || t.type || '%'
group by name, cte.type
having count(*) = length(cte.type)
) x
group by type
order by type;
There is no way to generate all possible combinations (A, B, C, AB, AC, BC, etc) in Amazon Redshift.
(Well, you could select each unique value, smoosh them into one string, send it to a User-Defined Function, extract the result into multiple rows and then join it against a big query, but that really isn't something you'd like to attempt.)
One approach would be to create a table containing all possible combinations — you'd need to write a little program to do that (eg using itertools in Python). Then, you could join the data against that reasonably easy to get the desired result (eg IF 'ABC' CONTAINS '%A%').
Here's what I'm trying to do. Let's say I have this table t:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
2 | 18 | 2012-05-19 | y
3 | 18 | 2012-08-09 | z
4 | 19 | 2009-06-01 | a
5 | 19 | 2011-04-03 | b
6 | 19 | 2011-10-25 | c
7 | 19 | 2012-08-09 | d
For each id, I want to select the row containing the minimum record_date. So I'd get:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
The only solutions I've seen to this problem assume that all record_date entries are distinct, but that is not this case in my data. Using a subquery and an inner join with two conditions would give me duplicate rows for some ids, which I don't want:
key_id | id | record_date | other_cols
1 | 18 | 2011-04-03 | x
5 | 19 | 2011-04-03 | b
4 | 19 | 2009-06-01 | a
How about something like:
SELECT mt.*
FROM MyTable mt INNER JOIN
(
SELECT id, MIN(record_date) AS MinDate
FROM MyTable
GROUP BY id
) t ON mt.id = t.id AND mt.record_date = t.MinDate
This gets the minimum date per ID, and then gets the values based on those values. The only time you would have duplicates is if there are duplicate minimum record_dates for the same ID.
I could get to your expected result just by doing this in mysql:
SELECT id, min(record_date), other_cols
FROM mytable
GROUP BY id
Does this work for you?
To get the cheapest product in each category, you use the MIN() function in a correlated subquery as follows:
SELECT categoryid,
productid,
productName,
unitprice
FROM products a WHERE unitprice = (
SELECT MIN(unitprice)
FROM products b
WHERE b.categoryid = a.categoryid)
The outer query scans all rows in the products table and returns the products that have unit prices match with the lowest price in each category returned by the correlated subquery.
I would like to add to some of the other answers here, if you don't need the first item but say the second number for example you can use rownumber in a subquery and base your result set off of that.
SELECT * FROM
(
SELECT
ROW_NUM() OVER (PARTITION BY Id ORDER BY record_date, other_cols) as rownum,
*
FROM products P
) INNER
WHERE rownum = 2
This also allows you to order off multiple columns in the subquery which may help if two record_dates have identical values. You can also partition off of multiple columns if needed by delimiting them with a comma
This does it simply:
select t2.id,t2.record_date,t2.other_cols
from (select ROW_NUMBER() over(partition by id order by record_date)as rownum,id,record_date,other_cols from MyTable)t2
where t2.rownum = 1
If record_date has no duplicates within a group:
think of it as of filtering. Simpliy get (WHERE) one (MIN(record_date)) row from the current group:
SELECT * FROM t t1 WHERE record_date = (
select MIN(record_date)
from t t2 where t2.group_id = t1.group_id)
If there could be 2+ min record_date within a group:
filter out non-min rows (see above)
then (AND) pick only one from the 2+ min record_date rows, within the given group_id. E.g. pick the one with the min unique key:
AND key_id = (select MIN(key_id)
from t t3 where t3.record_date = t1.record_date
and t3.group_id = t1.group_id)
so
key_id | group_id | record_date | other_cols
1 | 18 | 2011-04-03 | x
4 | 19 | 2009-06-01 | a
8 | 19 | 2009-06-01 | e
will select key_ids: #1 and #4
SELECT p.* FROM tbl p
INNER JOIN(
SELECT t.id, MIN(record_date) AS MinDate
FROM tbl t
GROUP BY t.id
) t ON p.id = t.id AND p.record_date = t.MinDate
GROUP BY p.id
This code eliminates duplicate record_date in case there are same ids with same record_date.
If you want duplicates, remove the last line GROUP BY p.id.
This a old question, but this can useful for someone
In my case i can't using a sub query because i have a big query and i need using min() on my result, if i use sub query the db need reexecute my big query. i'm using Mysql
select t.*
from (select m.*, #g := 0
from MyTable m --here i have a big query
order by id, record_date) t
where (1 = case when #g = 0 or #g <> id then 1 else 0 end )
and (#g := id) IS NOT NULL
Basically I ordered the result and then put a variable in order to get only the first record in each group.
The below query takes the first date for each work order (in a table of showing all status changes):
SELECT
WORKORDERNUM,
MIN(DATE)
FROM
WORKORDERS
WHERE
DATE >= to_date('2015-01-01','YYYY-MM-DD')
GROUP BY
WORKORDERNUM
select
department,
min_salary,
(select s1.last_name from staff s1 where s1.salary=s3.min_salary ) lastname
from
(select department, min (salary) min_salary from staff s2 group by s2.department) s3