redshift regexp_substr all occurrences - sql

I am trying to return all results that match a specific regex in Redshift database.
Sample query
WITH aa AS
(SELECT DISTINCT
id,
record,
regexp_substr(record, '(#{2})([A-Z]{2,3})',1,1),
REGEXP_COUNT(record, '(#{2})([A-Z]{2,3})')
FROM table_a)
SELECT
*
FROM aa
The above query returns the first occurrence and the total count of the occurrences.
Is there a way to return all occurrences? Maybe an iteration using a variable that is limited by the count?
regexp_substr(record, '(#{2})([A-Z]{2,3})',1,n)
There is a solution for Oracle with the CONNECT BY LEVEL, but that option seems to be unavailable for Redshift.

If I understand correctly, you can use split_part() instead:
split_part(record, '##', <n>)
Then you can attempt to generate numbers and get the results:
with n as (
select row_number() over () as n
from t
limit 10
)
select t.*, split_part(record, '##', n.n)
from t join
n
on split_part(record, '##', n.n) <> ''

Related

How to unnest BigQuery nested records into multiple columns

I am trying to unnest the below table .
Using the below unnest query to flatten the table
SELECT
id,
name ,keyword
FROM `project_id.dataset_id.table_id`
,unnest (`groups` ) as `groups`
where id = 204358
Problem is , this duplicates the rows (except name) as is the case with flattening the table.
How can I modify the query to put the names in two different columns rather than rows.
Expected output below -
That's because the comma is a cross join - in combination with an unnested array it is a lateral cross join. You repeat the parent row for every row in the array.
One problem with pivoting arrays is that arrays can have a variable amount of rows, but a table must have a fixed amount of columns.
So you need a way to decide for a certain row that becomes a certain column.
E.g. with
SELECT
id,
name,
groups[ordinal(1)] as firstArrayEntry,
groups[ordinal(2)] as secondArrayEntry,
keyword
FROM `project_id.dataset_id.table_id`
unnest(groups)
where id = 204358
If your array had a key-value pair you could decide using the key. E.g.
SELECT
id,
name,
(select value from unnest(groups) where key='key1') as key1,
keyword
FROM `project_id.dataset_id.table_id`
unnest(groups)
where id = 204358
But that doesn't seem to be the case with your table ...
A third option could be PIVOT in combination with your cross-join solution but this one has restrictions too: and I'm not sure how computation-heavy this is.
Consider below simple solution
select * from (
select id, name, keyword, offset
from `project_id.dataset_id.table_id`,
unnest(`groups`) with offset
) pivot (max(name) name for offset + 1 in (1, 2))
if applied to sample data in your question - output is
Note , when you apply to your real case - you just need to know how many such name_NNN columns to expect and extend respectively list - for example for offset + 1 in (1, 2, 3, 4, 5)) if you expect 5 such columns
In case if for whatever reason you want improve this - use below where everything is built dynamically for you so you don't need to know in advance how many columns it will be in the output
execute immediate (select '''
select * from (
select id, name, keyword, offset
from `project_id.dataset_id.table_id`,
unnest(`groups`) with offset
) pivot (max(name) name for offset + 1 in (''' || string_agg('' || pos, ', ') || '''))
'''
from (select pos from (
select max(array_length(`groups`)) cnt
from `project_id.dataset_id.table_id`
), unnest(generate_array(1, cnt)) pos
))
Your question is a little unclear, because it does not specify what to do with other keywords or other columns. If you specifically want the first two values in the array for keyword "OVG", you can unnest the array and pull out the appropriate names:
SELECT id,
(SELECT g.name
FROM UNNEST(t.groups) g WITH OFFSET n
WHERE key = 'OVG'
ORDER BY n
LIMIT 1
) as name_1,
(SELECT g.name
FROM UNNEST(t.groups) g WITH OFFSET n
WHERE key = 'OVG'
ORDER BY n
LIMIT 1 OFFSET 1
) as name_2,
'OVG' as keyword
FROM `project_id.dataset_id.table_id` t
WHERE id = 204358;

tricky SQL with substrings

I have a table (postgres) with a varchar field that has content structured like:
".. John;Smith;uuid=7c32e9e1-e29e-4211-b11e-e20b2cb78da9 .."
The uuid can occur in more than one record. But it must not occur for more than one combination of [givenname];[surname], according to a business rule.
That is, if the John Smith example above is present in the table, then if uuid 7c32e9e1.. occurs in any other record, the field in that record most also contain ".. John;Smith; .."
The problem is, this business rule has been violated due to some bug. And I would like to know how many rows in the table contains a uuid such that it occurs in more than one place with different combinations of [givenname];[surname].
I'd appreciate if someone could help me out with the SQL to accomplish this.
Use regular expressions to extract the UUID and the name from the string. Then aggregate per UUID and either count distinct names or compare minimum and maximum name:
select
substring(col, 'uuid=([[:alnum:]]+)') as uuid,
string_agg(distinct substring(col, '([[:alnum:]]+;[[:alnum:]]+);uuid'), ' | ') as names
from mytable
group by substring(col, 'uuid=([[:alnum:]]+)')
having count(distinct substring(col, '([[:alnum:]]+;[[:alnum:]]+);uuid')) > 1;
Demo: https://dbfiddle.uk/?rdbms=postgres_12&fiddle=907a283a754eb7427d4ffbf50c6f0028
If you only want to count:
select
count(*) as cnt_uuids,
sum(num_names) as cnt_names,
sum(num_rows) as cnt_rows
from
(
select
count(*) as num_rows,
count(distinct substring(col, '([[:alnum:]]+;[[:alnum:]]+);uuid')) as num_names
from mytable
group by substring(col, 'uuid=([[:alnum:]]+)')
having count(distinct substring(col, '([[:alnum:]]+;[[:alnum:]]+);uuid')) > 1
) flaws;
But as has been mentioned already: This is not how a database should be used.
I assume you know all the reasons why this is a bad data format, but you are stuck with it. Here is my approach:
select v.user_id, array_agg(distinct names)
from (select v.id,
max(el) filter (where n = un) as user_id,
array_agg(el order by el) filter (where n in (un - 2, un - 1)) as names
from (select v.id, u.*,
max(u.n) filter (where el like 'uuid=%') over (partition by v.id) as un
from (values (1 , 'junkgoeshere;John;Smith;uuid=7c32e9e1-e29e-4211-b11e-e20b2cb78da9; ..'),
(2 , 'junkgoeshere;John;Smith;uuid=7c32e9e1-e29e-4211-b11e-e20b2cb78da9; ..'),
(3 , 'junkgoeshere;John;Smith;uuid=new_7c32e9e1-e29e-4211-b11e-e20b2cb78da9; ..'),
(4 , 'junkgoeshere;John;Jay;uuid=new_7c32e9e1-e29e-4211-b11e-e20b2cb78da9; ..')
) v(id, str) cross join lateral
unnest(regexp_split_to_array(v.str, ';')) with ordinality u(el, n)
) v
where n between un - 2 and un
group by v.id
) v
group by user_id
having min(names) <> max(names);
Here is a db<>fiddle.
This assumes that the fields are separated by semicolons. Your data format is just awful, not just as a string but because the names are not identified. So, I am assuming they are the two fields before the user_id field.
So, this implements the following logic:
Breaks up the string by semicolons, with an identifying number.
Finds the number for the user_id.
Extracts the previous two fields together and the user_id column.
Then uses aggregation to find cases where there are multiple matches.

Substring in a column

I have a column that has several items in which I need to count the times it is called, my column table looks something like this:
Table Example
Id_TR Triggered
-------------- ------------------
A1_6547 R1:23;R2:0;R4:9000
A2_1235 R2:0;R2:100;R3:-100
A3_5436 R1:23;R2:100;R4:9000
A4_1245 R2:0;R5:150
And I would like the result to be like this:
Expected Results
Triggered Count(1)
--------------- --------
R1:23 2
R2:0 3
R2:100 2
R3:-100 1
R4:9000 2
R5:150 1
I've tried to do some substring, but cant seem to find how to solve this problem. Can anyone help?
This solution is X3 times faster than the CONNECT BY solution
performance: 15K records per second
with cte (token,suffix)
as
(
select substr(triggered||';',1,instr(triggered,';')-1) as token
,substr(triggered||';',instr(triggered,';')+1) as suffix
from t
union all
select substr(suffix,1,instr(suffix,';')-1) as token
,substr(suffix,instr(suffix,';')+1) as suffix
from cte
where suffix is not null
)
select token,count(*)
from cte
group by token
;
with x as (
select listagg(Triggered, ';') within group (order by Id_TR) str from table
)
select regexp_substr(str,'[^;]+',1,level) element, count(*)
from x
connect by level <= length(regexp_replace(str,'[^;]+')) + 1
group by regexp_substr(str,'[^;]+',1,level);
First concatenate all values of triggered into one list using listagg then parse it and do group by.
Another methods of parsing list you can find here or here
This is a fair solution.
performance: 5K records per second
select triggered
,count(*) as cnt
from (select id_tr
,regexp_substr(triggered,'[^;]+',1,level) as triggered
from t
connect by id_tr = prior id_tr
and level <= regexp_count(triggered,';')+1
and prior sys_guid() is not null
) t
group by triggered
;
This is just for learning purposes.
Check my other solutions.
performance: 1K records per second
select x.triggered
,count(*)
from t
,xmltable
(
'/r/x'
passing xmltype('<r><x>' || replace(triggered,';', '</x><x>') || '</x></r>')
columns triggered varchar(100) path '.'
) x
group by x.triggered
;

Returning the lowest integer not in a list in SQL

Supposed you have a table T(A) with only positive integers allowed, like:
1,1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18
In the above example, the result is 10. We always can use ORDER BY and DISTINCT to sort and remove duplicates. However, to find the lowest integer not in the list, I came up with the following SQL query:
select list.x + 1
from (select x from (select distinct a as x from T order by a)) as list, T
where list.x + 1 not in T limit 1;
My idea is start a counter and 1, check if that counter is in list: if it is, return it, otherwise increment and look again. However, I have to start that counter as 1, and then increment. That query works most of the cases, by there are some corner cases like in 1. How can I accomplish that in SQL or should I go about a completely different direction to solve this problem?
Because SQL works on sets, the intermediate SELECT DISTINCT a AS x FROM t ORDER BY a is redundant.
The basic technique of looking for a gap in a column of integers is to find where the current entry plus 1 does not exist. This requires a self-join of some sort.
Your query is not far off, but I think it can be simplified to:
SELECT MIN(a) + 1
FROM t
WHERE a + 1 NOT IN (SELECT a FROM t)
The NOT IN acts as a sort of self-join. This won't produce anything from an empty table, but should be OK otherwise.
SQL Fiddle
select min(y.a) as a
from
t x
right join
(
select a + 1 as a from t
union
select 1
) y on y.a = x.a
where x.a is null
It will work even in an empty table
SELECT min(t.a) - 1
FROM t
LEFT JOIN t t1 ON t1.a = t.a - 1
WHERE t1.a IS NULL
AND t.a > 1; -- exclude 0
This finds the smallest number greater than 1, where the next-smaller number is not in the same table. That missing number is returned.
This works even for a missing 1. There are multiple answers checking in the opposite direction. All of them would fail with a missing 1.
SQL Fiddle.
You can do the following, although you may also want to define a range - in which case you might need a couple of UNIONs
SELECT x.id+1
FROM my_table x
LEFT
JOIN my_table y
ON x.id+1 = y.id
WHERE y.id IS NULL
ORDER
BY x.id LIMIT 1;
You can always create a table with all of the numbers from 1 to X and then join that table with the table you are comparing. Then just find the TOP value in your SELECT statement that isn't present in the table you are comparing
SELECT TOP 1 table_with_all_numbers.number, table_with_missing_numbers.number
FROM table_with_all_numbers
LEFT JOIN table_with_missing_numbers
ON table_with_missing_numbers.number = table_with_all_numbers.number
WHERE table_with_missing_numbers.number IS NULL
ORDER BY table_with_all_numbers.number ASC;
In SQLite 3.8.3 or later, you can use a recursive common table expression to create a counter.
Here, we stop counting when we find a value not in the table:
WITH RECURSIVE counter(c) AS (
SELECT 1
UNION ALL
SELECT c + 1 FROM counter WHERE c IN t)
SELECT max(c) FROM counter;
(This works for an empty table or a missing 1.)
This query ranks (starting from rank 1) each distinct number in ascending order and selects the lowest rank that's less than its number. If no rank is lower than its number (i.e. there are no gaps in the table) the query returns the max number + 1.
select coalesce(min(number),1) from (
select min(cnt) number
from (
select
number,
(select count(*) from (select distinct number from numbers) b where b.number <= a.number) as cnt
from (select distinct number from numbers) a
) t1 where number > cnt
union
select max(number) + 1 number from numbers
) t1
http://sqlfiddle.com/#!7/720cc/3
Just another method, using EXCEPT this time:
SELECT a + 1 AS missing FROM T
EXCEPT
SELECT a FROM T
ORDER BY missing
LIMIT 1;

How to split and display distinct letters from a word in SQL?

Yesterday in a job interview session I was asked this question and I had no clue about it. Suppose I have a word "Manhattan " I want to display only the letters 'M','A','N','H','T'
in SQL. How to do it?
Any help is appreciated.
Well, here is my solution (sqlfiddle) - it aims to use a "Relational SQL" operations, which may have been what the interviewer was going for conceptually.
Most of the work done is simply to turn the string into a set of (pos, letter) records as the relevant final applied DQL is a mere SELECT with a grouping and ordering applied.
select letter
from (
-- All of this just to get a set of (pos, letter)
select ns.n as pos, substring(ss.s, ns.n, 1) as letter
from (select 'MANHATTAN' as s) as ss
cross join (
-- Or use another form to create a "numbers table"
select n from (values (1),(2),(3),(4),(5),(6),(7),(8),(9)) as X(n)
) as ns
) as pairs
group by letter -- guarantees distinctness
order by min(pos) -- ensure output is ordered MANHT
The above query works in SQL Server 2008, but the "Numbers Table" may have to be altered for other vendors. Otherwise, there is nothing used that is vendor specific - no CTE, or cross application of a function, or procedural language code ..
That being said, the above is to show a conceptual approach - SQL is designed for use with sets and relations and multiplicity across records; the above example is, in some sense, merely a perversion of such.
Examining the intermediate relation,
select ns.n as pos, substring(ss.s, ns.n, 1) as letter
from (select 'MANHATTAN' as s) as ss
cross join (
select n from (values (1),(2),(3),(4),(5),(6),(7),(8),(9)) as X(n)
) as ns
uses a cross join to generate the Cartesian product of the string (1 row) with the numbers (9 rows); the substring function is then applied with the string and each number to obtain each character in accordance with its position. The resulting set contains the records-
POS LETTER
1 M
2 A
3 N
..
9 N
Then the outer select groups each record according to the letter and the resulting records are ordered by the minimum (first) occurrence position of the letter that establishing the grouping. (Without the order by the letters would have been distinct but the final order would not be guaranteed.)
One way (if using SQL Server) is with a recursive CTE (Commom Table Expression).
DECLARE #source nvarchar(100) = 'MANHATTAN'
;
WITH cte AS (
SELECT SUBSTRING(#source, 1, 1) AS c1, 1 as Pos
WHERE LEN(#source) > 0
UNION ALL
SELECT SUBSTRING(#source, Pos + 1, 1) AS c1, Pos + 1 as Pos
FROM cte
WHERE Pos < LEN(#source)
)
SELECT DISTINCT c1 from cte
SqlFiddle for this is here. I had to inline the #source for SqlFiddle, but the code above works fine in Sql Server.
The first SELECT generates the initial row(in this case 'M', 1). The second SELECT is the recursive part that generates the subsequent rows, with the Pos column getting incremented each time until the termination condition WHERE Pos < LEN(#source) is finally met. The final select removes the duplicates. Internally, SELECT DISTINCT sorts the rows in order to facilitate the removal of duplicates, which is why the final output happens to be in alphabetic order. Since you didn't specify order as a requirement, I left it as-is. But you could modify it to use a GROUP instead, that ordered on MIN(Pos) if you needed the output in the characters' original order.
This same technique can be used for things like generating all the Bigrams for a string, with just a small change to the general structure above.
declare #charr varchar(99)
declare #lp int
set #charr='Manhattan'
set #lp=1
DECLARE #T1 TABLE (
FLD VARCHAR(max)
)
while(#lp<=LEN(#charr))
begin
if(not exists(select * from #T1 where FLD=(select SUBSTRING(#charr,#lp,1))))
begin
insert into #T1
select SUBSTRING(#charr,#lp,1)
end
set #lp=#lp+1
end
select * from #T1
check this it may help u
Here's an Oracle version of #user2864740's answer. The only difference is how you construct the "numbers table" (plus slight differences in aliasing)
select letter
from (
select ns.n as pos, substr(ss.s, ns.n, 1) as letter
from (select 'MANHATTAN' as s from dual) ss
cross join (
SELECT LEVEL as n
FROM DUAL
CONNECT BY LEVEL <= 9
ORDER BY LEVEL) ns
) pairs
group by letter
order by min(pos)