SQL/REGEX puzzle/challenge How to convert ASCII art ranges with multiple characters to relational data? - sql

Question

The motivation here was to easily and accurately generate data samples for the nested ranges challenge.
A table contains a single column of text type.
The text contains one or more lines where each lines contains one or more sections created from letters.
The goal is to write a query that returns a tuple for each section with its start point ,end point and value.
Data sample
create table t (txt varchar (1000));
insert into t (txt) values
(
'
AAAAAAAAAAAAAAAAAAAAAAAAAAAA BBBB CCCCCCCCCCCCCCCCCCCCCCCCC
DDDE FFFFFFFF GGGGGGGGG HHHHHHHH IIIIIII
JJ KKKLLL MM NN OOOOO
P QQ
'
)
;
Requested results
* Only the last 3 columns (section start/end/val) are required, the rest are for debugging purposes.
line_ind section_ind section_length section_start section_end section_val
1 1 28 1 28 A
1 2 4 31 34 B
1 3 25 39 63 C
2 1 3 1 3 D
2 2 1 4 4 E
2 3 8 7 14 F
2 4 9 19 27 G
2 5 8 43 50 H
2 6 7 55 61 I
3 1 2 1 2 J
3 2 3 9 11 K
3 3 3 12 14 L
3 4 2 22 23 M
3 5 2 25 26 N
3 6 5 57 61 O
4 1 1 13 13 P
4 2 2 60 61 Q

Answer 1

Teradata
Currently regexp_split_to_table doesn't seem to support zero-length expression (I've created incident RECGZJKZV). In order to overcome this limitation I'm using regexp_replace to push space between adjacent sequences of letters, e.g. KKKLLL
with l
as
(
select line_ind
,line
from table
(
regexp_split_to_table (-1,t.txt,'\r','')
returns (minus_one int,line_ind int,line varchar(1000))
)
as l
)
select l.line_ind
,r.section_ind
,char_length (r.section) as section_length
,regexp_instr (l.line,'(\S)\1*',1,r.section_ind,0) as section_start
,regexp_instr (l.line,'(\S)\1*',1,r.section_ind,1) - 1 as section_end
,substr (r.section,1,1) as section_val
from table
(
regexp_split_to_table (l.line_ind,regexp_replace (l.line,'(?<=(?P<c>.))(?!(?P=c))',' '),'\s+','')
returns (line_ind int,section_ind int,section varchar(1000))
)
as r
,l
where l.line_ind =
r.line_ind
order by l.line_ind
,r.section_ind
;
Oracle
select regexp_instr (txt,'(\S)\1*',1,level,0) - instr (txt,chr(10),regexp_instr (txt,'(\S)\1*',1,level,0) - length (txt) - 1,1) as section_start
,regexp_instr (txt,'(\S)\1*',1,level,1) - 1 - instr (txt,chr(10),regexp_instr (txt,'(\S)\1*',1,level,0) - length (txt) - 1,1) as section_end
,regexp_substr (txt,'(\S)\1*',1,level,'',1) as section_val
from t
connect by level <= regexp_count (txt,'(\S)\1*')
;

Answer 2

Oracle
This will work even if you have multiple input rows:
WITH lines ( txt, id, line, pos, line_no ) AS(
SELECT txt,
id,
REGEXP_SUBSTR( txt, '.*', 1, 1 ),
REGEXP_INSTR( txt, '.*', 1, 1, 1 ),
1
FROM t
UNION ALL
SELECT txt,
id,
REGEXP_SUBSTR( txt, '.*', pos + 1, 1 ),
REGEXP_INSTR( txt, '.*', pos + 1, 1, 1 ),
line_no + 1
FROM lines
WHERE pos > 0
),
words ( id, line, line_no, section_start, section_end, section_value ) AS (
SELECT id,
line,
line_no,
REGEXP_INSTR( line, '(\S)\1*', 1, 1, 0 ),
REGEXP_INSTR( line, '(\S)\1*', 1, 1, 1 ) - 1,
REGEXP_SUBSTR( line, '(\S)\1*', 1, 1, NULL, 1 )
FROM lines
WHERE pos > 0
AND line IS NOT NULL
UNION ALL
SELECT id,
line,
line_no,
REGEXP_INSTR( line, '(\S)\1*', section_end + 1, 1, 0 ),
REGEXP_INSTR( line, '(\S)\1*', section_end + 1, 1, 1 ) - 1,
REGEXP_SUBSTR( line, '(\S)\1*', section_end + 1, 1, NULL, 1 )
FROM words
WHERE section_end > 0
)
SELECT id,
line_no,
section_start,
section_end,
section_value
FROM words
WHERE section_end > 0
ORDER BY id, line_no, section_start
So, for the input data (with an added id column to be able to easily differentiate the pieces of text):
create table t (id NUMBER(5,0), txt varchar (1000));
insert into t (id, txt) values
(
1,
'
AAAAAAAAAAAAAAAAAAAAAAAAAAAA BBBB CCCCCCCCCCCCCCCCCCCCCCCCC
DDDE FFFFFFFF GGGGGGGGG HHHHHHHH IIIIIII
JJ KKKLLL MM NN OOOOO
P QQ
'
);
insert into t (id, txt) values ( 2, 'RRRSTT UUU V WXYZ' );
This outputs:
ID | LINE_NO | SECTION_START | SECTION_END | SECTION_VALUE
-: | ------: | ------------: | ----------: | :------------
1 | 2 | 1 | 28 | A
1 | 2 | 31 | 34 | B
1 | 2 | 39 | 63 | C
1 | 3 | 1 | 3 | D
1 | 3 | 4 | 4 | E
1 | 3 | 7 | 14 | F
1 | 3 | 19 | 27 | G
1 | 3 | 43 | 50 | H
1 | 3 | 55 | 61 | I
1 | 4 | 1 | 2 | J
1 | 4 | 9 | 11 | K
1 | 4 | 12 | 14 | L
1 | 4 | 22 | 23 | M
1 | 4 | 25 | 26 | N
1 | 4 | 57 | 61 | O
1 | 5 | 13 | 13 | P
1 | 5 | 60 | 61 | Q
2 | 1 | 1 | 3 | R
2 | 1 | 4 | 4 | S
2 | 1 | 5 | 6 | T
2 | 1 | 8 | 10 | U
2 | 1 | 15 | 15 | V
2 | 1 | 17 | 17 | W
2 | 1 | 18 | 18 | X
2 | 1 | 19 | 19 | Y
2 | 1 | 20 | 20 | Z
db<>fiddle here