you can test my code here :
https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=61a67764f626bfadb1e9594e1ae08229
code to print the first n terms of the fibonacci sequel:
with fibo(s2,s1,n) as(
select 1 ,1 ,1 from dual
union all
select s1+s2,s2,n+1 from fibo where n<12
)
select s2 from fibo;
It works but it use probably twice as much memory as needed. Each line contains the nth and n-1th terms of the sequel before the selection
Therefore i tried with Lag function
with fibo(s,n) as(
select 1,1 from dual
union all
select LAG(s, 1, 0) OVER ( ORDER BY s) +LAG(s, 2, 0) OVER ( ORDER BY s),n+1 from fibo where n<12
)
Select * from fibo
But i get only a sequel of 1. (same things with lead function)
I have tried to understand what happens with this :
with test(s,d1,d2,n) as(
select 1,0,0,1 from dual
union all
select
2*s,LAG(s, 1, 0) OVER (a ORDER BY s) ,
LAG(s, 2, 0) OVER ( ORDER BY s),n+1
from test where n<12
)
select * from test
It seems that lag returns always 0. Is it impossible to use lag and lead in recursive query? Or do i do something false?
"Recursive" queries are not recursive, they are iterative.
The iteration starts with the anchor (the part(s) of the UNION ALL that doesn't refer to the CTE).
For each following iteration, the result set of the previous iteration (aliased by the CTE) is used as the input.
The iteration stops when the result set is empty.
In your specific attempt the anchor returns a single record and so is every following query.
Obviously LAG(s, 2, 0) will always return 0
Related
I tried to create a simple SQL to track query_history usage, but got into trouble when creating my timeslots using the table and generator functions (the CTE named x below).
I got no results at all when limiting the query_history using my timeslots, so after a while I hardcoded an SQL to give the same result (the CTE named y below) and this works fine.
Why does not x work? As far as I can see x and y produce identical result?
To test the example first run the code as it is, this produces no result.
Then comment the line x as timeslots and un-comment the line y as timeslots, this will give the desired result.
with
x as (
select
dateadd('min',seq4()*10,dateadd('min',-60,current_timestamp())) f,
dateadd('min',(seq4()+1)*10,dateadd('min',-60,current_timestamp())) t
from table(generator(rowcount => 6))
),
y as (
select
dateadd('min',n*10,dateadd('min',-60,current_timestamp())) f,
dateadd('min',(n+1)*10,dateadd('min',-60,current_timestamp())) t
from (select 0 n union all select 1 n union all select 2 union all select 3
union all select 4 union all select 5)
)
--select * from x;
--select * from y;
select distinct
user_name,
timeslots.f
from snowflake.account_usage.query_history,
x as timeslots
--y as timeslots
where start_time >= timeslots.f
and start_time < timeslots.t
order by timeslots.f desc;
(I know the code is not optimal, this is only meant to illustrate the problem)
SEQ:
Returns a sequence of monotonically increasing integers, with wrap-around. Wrap-around occurs after the largest representable integer of the integer width (1, 2, 4, or 8 byte).
If a fully ordered, gap-free sequence is required, consider using the ROW_NUMBER window function.
For:
with x as (
select
dateadd('min',seq4()*10,dateadd('min',-60,current_timestamp())) f,
dateadd('min',(seq4()+1)*10,dateadd('min',-60,current_timestamp())) t
from table(generator(rowcount => 6))
)
SELECT * FROM x;
Should be:
with x as (
select
(ROW_NUMBER() OVER(ORDER BY seq4())) - 1 AS n,
dateadd('min',n*10,dateadd('min',-60,current_timestamp())) f,
dateadd('min',(n+1)*10,dateadd('min',-60,current_timestamp())) t
from table(generator(rowcount => 6))
)
SELECT * FROM x;
I'm trying to find a tidier way of doing the below query so that I'm not duplicating my code.
SELECT CASE WHEN <COMPLICATED CODE THAT RETURNS A SINGLE INT> = 0
THEN 1
ELSE <COMPLICATED CODE THAT RETURNS A SINGLE INT> END
Ideally, I would like something like this using an existing function rather than creating my own:
SELECT ISVALUE(COMPLICATED CODE THAT RETURNS A SINGLE INT,0,1)
You can use apply:
SELECT (CASE WHEN v.val = 0 THEN 1 ELSE v.val END)
FROM . . . CROSS APPLY
(VALUES (<COMPLICATED CODE THAT RETURNS A SINGLE INT>)) v(val);
You could also do a series of functions:
select coalesce(nullif(<COMPLICATED CODE THAT RETURNS A SINGLE INT>, 0), 1)
However, I think apply is clearer. In addition the above will turn NULL values into 1 as well as 0.
You can use a CTE (or a subquery) as
WITH CTE AS
(
SELECT <COMPLICATED CODE THAT RETURNS A SINGLE INT> AS Value
FROM ...
)
SELECT CASE WHEN Value = 0 THEN 1 ELSE Value END
FROM CTE
This way you write the complicated code just once, and then use just the Value column.
You can use IIF:
SELECT IIF(1 = 1, 'true', 'false')
https://learn.microsoft.com/en-us/sql/t-sql/functions/logical-functions-iif-transact-sql
use a sub query, then you can figure out a mathematical formula that acts to give the values you desire, that way you can eliminate actual boolean logic and replace with mathematical functions
an example is
SELECT *,(1 - ceiling(cos(atan(abs(cast(x as float)))) -
floor(cos(atan(abs(cast(x as float))))))) +
x * ceiling(cos(atan(abs(cast(x as float)))) -
floor(cos(atan(abs(cast(x as float)))))) as Computed
FROM
( select 0 as x union SELECT 1 X UNION SELECT -.123 UNION SELECT 9.1 UNION SELECT 67000 union select -1) OQ
I have one of the column in oracle table which has below value :
select csv_val from my_table where date='09-OCT-18';
output
==================
50,100,25,5000,1000
I want this values to be in ascending order with select query, output would looks like :
output
==================
25,50,100,1000,5000
I tried this link, but looks like it has some restriction on number of digits.
Here, I made you a modified version of the answer you linked to that can handle an arbitrary (hardcoded) number of commas. It's pretty heavy on CTEs. As with most LISTAGG answers, it'll have a 4000-char limit. I also changed your regexp to be able to handle null list entries, based on this answer.
WITH
T (N) AS --TEST DATA
(SELECT '50,100,25,5000,1000' FROM DUAL
UNION
SELECT '25464,89453,15686' FROM DUAL
UNION
SELECT '21561,68547,51612' FROM DUAL
),
nums (x) as -- arbitrary limit of 20, can be changed
(select level from dual connect by level <= 20),
splitstr (N, x, substring) as
(select N, x, regexp_substr(N, '(.*?)(,|$)', 1, x, NULL, 1)
from T
inner join nums on x <= 1 + regexp_count(N, ',')
order by N, x)
select N, listagg(substring, ',') within group (order by to_number(substring)) as sorted_N
from splitstr
group by N
;
Probably it can be improved, but eh...
Based on sample data you posted, relatively simple query would work (you need lines 3 - 7). If data doesn't really look like that, query might need adjustment.
SQL> with my_table (csv_val) as
2 (select '50,100,25,5000,1000' from dual)
3 select listagg(token, ',') within group (order by to_number(token)) result
4 from (select regexp_substr(csv_val, '[^,]+', 1, level) token
5 from my_table
6 connect by level <= regexp_count(csv_val, ',') + 1
7 );
RESULT
-------------------------
25,50,100,1000,5000
SQL>
I have noticed strange behaviour in some SQL code used for address matching at the company I work for & have created some test SQL to illustrate the issue.
; WITH Temp (Id, Diff) AS (
SELECT 9218, 0
UNION
SELECT 9219, 0
UNION
SELECT 9220, 0
)
SELECT TOP 1 * FROM Temp ORDER BY Diff DESC
Returns 9218 but
; WITH Temp (Id, Name) AS (
SELECT 9218, 'Sonnedal'
UNION
SELECT 9219, 'Lammermoor'
UNION
SELECT 9220, 'Honeydew'
)
SELECT TOP 1 *, DIFFERENCE(Name, '') FROM Temp ORDER BY DIFFERENCE(Name, '') DESC
returns 9219 even though the Difference() is 0 for all records as you can see here:
; WITH Temp (Id, Name) AS (
SELECT 9218, 'Sonnedal'
UNION
SELECT 9219, 'Lammermoor'
UNION
SELECT 9220, 'Honeydew'
)
SELECT *, DIFFERENCE(Name, '') FROM Temp ORDER BY DIFFERENCE(Name, '') DESC
which returns
9218 Sonnedal 0
9219 Lammermoor 0
9220 Honeydew 0
Does anyone know why this happens? I am writing C# to replace existing SQL & need to return the same results so I can test that my code produces the same results. But I can't see why the actual SQL used returns 9219 rather than 9218 & it doesn't seem to make sense. It seems it's down to the Difference() function but it returns 0 for all the record in question.
When you call:
SELECT TOP 1 *, DIFFERENCE(Name, '')
FROM Temp l
ORDER BY DIFFERENCE(Name, '') DESC
All three records have a DIFFERENCE value of zero, and hence SQL Server is free to choose from any of the three records for ordering. That is to say, there is no guarantee which order you will get. The same is true for your second query. Actually, it is possible that the ordering for the same query could even change over time. In practice, if you expect a certain ordering, you should provide exact logic for it, e.g.
SELECT TOP 1 *
FROM Temp
ORDER BY Id;
I'm not new in sql and t-sql, but at past I've never used recursive query - all problems were solved with WHILE or CURSOR. I just got 1 question - how to organaze recursion query for following problem: I want to manipulate with last row of data in certain partition. Can't understand how to stop my recursion at last level of partition.
CREATE TABLE #temp
(i int
, s int
, v int);
INSERT INTO #temp
SELECT 1, 1, 10
UNION
SELECT 1, 2, 20
UNION
SELECT 2, 1, 5
UNION
SELECT 2, 2, 5
UNION
SELECT 2, 3, 2
WITH CTE AS
(
SELECT i
, s
, v
FROM #temp
WHERE s=1
UNION ALL
SELECT t.i
, t.s
, t.v + cte.v as new_v
FROM #temp t
INNER JOIN cte
ON (cte.i=t.i)
WHERE t.s>1
)
SELECT *
FROM cte
OPTION(MAXRECURSION 0)
I want to get 5 rows as result:
result
I know that it could be solved with OUTER APPLY, JOINS, WHILE or CURSOR methods. Could you please share any features for my to understand how to get same result with recurcive cte query? SUM function there is just for example - for that problem recurcive query is best way cause I will use many scalar functions in big CASE which will use value from last row in partition and value of current row partition.
Thanks.
Sorry for my bad english level.
Will it be correctly if I'll try same problem with following example? I guess that need to correctly say in which order way recursive query gonna do any data manipulating. So below code which will help you understand what did I want to solve:
CREATE TABLE #temp
(i_key int
, step int
, step_h int
, value int);
INSERT INTO #temp
SELECT 1, 1, NULL, 20
UNION
SELECT 1, 2, 1, 20
UNION
SELECT 2, 1, NULL, 10
UNION
SELECT 2, 2, 1, 10
UNION
SELECT 2, 3, 2, 5
WITH CTE AS
(
SELECT i_key
, step
, value
FROM #temp
WHERE step=1
--AND i_key=2
UNION ALL
SELECT t.i_key
, t.step
, CASE
WHEN cte.value - t.value <=0 THEN 0
ELSE cte.value - t.value
END as value
FROM #temp t
INNER JOIN cte
ON (cte.i_key=t.i_key
AND cte.step=t.step_h)
--WHERE t.step>1
)
SELECT *
FROM CTE
OPTION(MAXRECURSION 0)
Is parent-child structure always need for solving this problems?
So i guess it could be done with another join (without column of parent-child).
AND cte.step=t.step-1
For your particular example, recursion is unnecessary. All you need is SQL Server 2012 or later version:
select t.*,
sum(t.v) over(partition by t.i order by t.s) as [RT]
from #temp t
order by t.i, t.s;
If you need to access previos / next row, there are lag() / lead() ranking functions that were introduced in the same aforementioned version of SQL Server.
EDIT: Ah, I see. You simply want to know how to write recursive CTEs properly. Here is a (seemingly) correct code for your second example:
with cte as (
select t.i_key, t.step, t.value
from #temp t
where t.step_h is null
union all
select c.i_key, t.step, case
when c.value < t.value then 0
else c.value - t.value
end as [Value]
from #temp t
inner join cte c on c.step = t.step_h
and c.i_key = t.i_key
)
select *
from cte c
order by c.i_key, c.step;
In the end, it stops by itself when an iteration does not produce any new rows.