I have say a table with as under
Values
--------
12Null345XXX23456
6712356
Expected Output
----------------
23456
123
How can I do this using TSql(prefereable set based) ?
Thanks
So, you want the longest substring of consecutive numeric digits? Maybe something like this (t here is recreating your original table; if it already exists, skip that CTE and use your table)?:
;with ten as (
select i from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10)) x(i)
), subs as (
select SUBSTRING('0123456789',s.i,l.i) as sub, l.i as run
from ten as s
join ten as l on s.i+l.i<=11
), t as (
select s
from (
values ('12Null345XXX23456'),('6712356'))x(s)
)
select x.sub
from t
cross apply (select top 1 subs.sub
from subs
where CHARINDEX(subs.sub,t.s)>0
order by subs.run desc) x
Related
I have two tables Q and T, both containing a column of float numbers.
What I want to do is, for each number in Q, I want to find a number in T that has the smallest distance to it.
For example, for T={1,7,9} and Q={2,6,10}, I want to return Q,T pairs as {(2,1),(6,7),(10,9)}.
How should I express this query with SQL?
In addition, is that possible to accelerate this join by index, e.g. add an operator class which bind "FOR ORDER BY <->" with fabs calculation?
create table t (val_t integer);
create table q (val_q integer);
insert into t values (1),(7),(9);
insert into q values (2),(6),(10);
Start with a query that cross joins the two tables and adds a rank based on the difference:
SELECT val_q, val_t, rank() OVER (PARTITION BY val_q ORDER BY abs(val_t - val_q))
FROM t
JOIN q ON true ;
Use this query in a cte or subquery and filter by rank:
WITH src AS(
SELECT val_q, val_t, rank() OVER (PARTITION BY val_q ORDER BY abs(val_t - val_q))
FROM t
JOIN q ON true )
SELECT val_q, val_t FROM src
WHERE rank = 1;
val_q | val_t
-------+-------
2 | 1
6 | 7
10 | 9
See https://www.postgresql.org/docs/12/tutorial-window.html
Given this schema:
create table t (tn float);
insert into t values (1), (7), (9);
create table q (qn float);
insert into q values (2), (6), (10);
DISTINCT ON is the most straightforward way:
select distinct on (qn) qn, tn
from q
cross join t
order by qn, abs(qn - tn);
Exploiting a numeric range may perform better depending on your data sizes. If performance is an issue, then you can create an actual temp table for the range_tn CTE and put a gist index on it:
with all_tn as (
select tn
from t
union select null
), range_tn as (
select numrange(tn::numeric, (lead(tn) over w)::numeric, '[]') as tr
from all_tn
window w as (order by tn nulls first)
)
select qn,
case
when lower_inf(tr) then upper(tr)
when upper_inf(tr) then lower(tr)
when 2 * qn - lower(tr) - upper(tr) > 0 then upper(tr)
else lower(tr)
end as tn
from q
join range_tn
on qn::numeric <# tr;
Fiddle here
I need to create a DB view with parts of sequential combinations of string parts of a source column. Example:
IN:
tag
--------
A_B_C_D
X_Y_Z
OUT:
subtag
--------
A
A_B
A_B_C
A_B_C_D
X
X_Y
X_Y_Z
The answer seems to be somewhere around WITH RECURSIVE, but I cannot put it all together.
demo:db<>fiddle
SELECT
array_to_string( -- 3
array_agg(t.value) OVER (PARTITION BY tags ORDER BY t.number), --2
'_'
) AS subtag
FROM
tags,
regexp_split_to_table(tag, '_') WITH ORDINALITY as t(value, number) -- 1
Split the string into one row per element. The WITH ORDINALITY adds a row count which can be used to hold the original order of the elements
Using array_agg() window function to aggregate the elements. The ORDER BY makes it cumulative
Reaggregate the array into a string.
You can use a recursive query:
WITH RECURSIVE s AS (
SELECT tag FROM tag
UNION
SELECT regexp_replace(tag, '_[^_]*$', '') FROM s
)
SELECT * FROM s;
tag
---------
A_B_C_D
X_Y_Z
A_B_C
X_Y
A_B
X
A
(7 rows)
The idea is to successively cut off _* at the end.
Thanks a lot #laurenz-albe! There is a problem with your code that it's missing recursion break condition. So I ended up with this:
WITH RECURSIVE s AS (
SELECT tag FROM tag
UNION
SELECT regexp_replace(tag, '_[^_]*$', '')
FROM s
WHERE tag LIKE '%\_%'
)
SELECT * FROM s;
db<>fiddle
Let's say I have the following data that represents taxes:
SELECT trunc(i*i, 3) tax
FROM generate_series(1.17, 5) i;
tax
--------
1.368
4.708
10.048
17.388
(4 rows)
Is there any nice way in PostgreSQL to put mill remainder into next line and if current line is the last it must have all leftovers.
So, I need to make it the following:
tax
--------
1.360
4.710
10.050
17.392
(4 rows)
It could be a query or SQL / PL/pgSQL function.
Next row and last row make sense only when the sort order is defined. I assume that the sort order is defined by tax asc.
The first subquery adds row numbers to the data, while the second one calculates the number of rows. The next part is a recursion based on increasing row numbers:
with recursive data as (
select trunc(i*i, 3) tax, row_number() over (order by i) as rn
from generate_series(1.17, 5) i
),
count as (
select count(*)
from data
),
result as (
select
tax, rn,
floor(tax* 100)/100 as new_tax,
tax- floor(tax* 100)/100 as remainder
from data
where rn = 1
union all
select
d.tax, d.rn,
case d.rn
when count then d.tax+ r.remainder
else floor((d.tax+ r.remainder)* 100)/100 end as new_tax,
d.tax+ r.remainder- floor((d.tax+ r.remainder)* 100)/100 as remainder
from data d
join result r on d.rn = r.rn+ 1
cross join count
)
select new_tax as tax
from result
order by rn;
Live demo in rextester.
In a table I have records with id's 2,4,5,8. How can I receive a list with values 1,3,6,7. I have tried in this way
SELECT t1.id + 1
FROM table t1
WHERE NOT EXISTS (
SELECT *
FROM table t2
WHERE t2.id = t1.id + 1
)
but it's not working correctly. It doesn't bring all available positions.
Is it possible without another table?
You can get all the missing ID's from a recursive CTE, like this:
with recursive numbers as (
select 1 number
from rdb$database
union all
select number+1
from rdb$database
join numbers on numbers.number < 1024
)
select n.number
from numbers n
where not exists (select 1
from table t
where t.id = n.number)
the number < 1024 condition in my example limit the query to the max 1024 recursion depth. After that, the query will end with an error. If you need more than 1024 consecutive ID's you have either run the query multiple times adjusting the interval of numbers generated or think in a different query that produces consecutive numbers without reaching that level of recursion, which is not too difficult to write.
I am wondering if there is some easy way, a function, or other method to return data from a query with the following results.
I have a SQL Express DB 2008 R2, a table that contains numerical data in a given column, say col T.
I am given a value X in code and would like to return up to three records. The record where col T equals my value X, and the record before and after, and nothing else. The sort is done on col T. The record before may be beginning of file and therefore not exist, likewise, if X equals the last record then the record after would be non existent, end of file/table.
The value of X may not exist in the table.
This I think is similar to get a range of results in numerical order.
Any help or direction in solving this would be greatly appreciated.
Thanks again,
It might not be the most optimal solution, but:
SELECT T
FROM theTable
WHERE T = X
UNION ALL
SELECT *
FROM
(
SELECT TOP 1 T
FROM theTable
WHERE T > X
ORDER BY T
) blah
UNION ALL
SELECT *
FROM
(
SELECT TOP 1 T
FROM theTable
WHERE T < X
ORDER BY T DESC
) blah2
DECLARE #x int = 100
;WITH t as
(
select ROW_NUMBER() OVER (ORDER BY T ASC) AS row_nm,*
from YourTable
)
, t1 as
(
select *
from t
WHERE T = #x
)
select *
from t
CROSS APPLY t1
WHERE t.row_nm BETWEEN t1.row_nm -1 and t1.row_nm + 1