How to find out the index of the first number encountered in the reverse direction of a string?
For example: 'CUSTOMC23VBA' and 'CUSTOMC245BA'.
So, function should return as '2' or '3' from reverse or the index value as '9' or '10'.
I could get the value by hard-coding the SUBSTR('CUSTOMC23VBA', -3) but I would want it to be generic as regular expressions.
You can try:
select regexp_instr(reverse('CUSTOM123XYZ'), '[[:digit:]]',1,1) from dual
Output: 4
Zero based index would be:
select regexp_instr(reverse('CUSTOM123XYZ'), '[[:digit:]]',1,1)-1 from dual
Output: 3
If you want the rest of the string from the last number, you can use substr and take advantage of the negative position to count from end of string:
select substr('CUSTOM123XYZ', -1 * (regexp_instr(reverse('CUSTOM123XYZ'), '[[:digit:]]',1,1)-1)) from dual;
Output: XYZ
An example testing multiple input strings:
with d as (
select 'CUSTOM123XYZ' as input_str from dual
union
select 'CUSTOM123XZ' as input_str from dual
union
select 'CUSTOM 1 X 3YZ' as input_str from dual
)
select input_str,
substr(input_str, -1 * (regexp_instr(reverse(input_str), '[[:digit:]]',1,1)-1)) as result
from d
Output:
INPUT_STR RESULT
CUSTOM 1 X 3YZ YZ
CUSTOM123XYZ XYZ
CUSTOM123XZ XZ
Is there only one number in the string? if so, you could go for something like this:
select REGEXP_REPLACE('CUSTOMC23VBA', '[[:alpha:]]','') from dual
But this will fail when there are multiple numbers in the string.
Related
I've seen examples of how to separate comma-separated strings into rows like this:
select distinct id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
from tbl1
connect by regexp_substr(value, '[^,]+', 1, level) is not null
order by id, level;
but, my question is, how do I do this on double quote and comma delimited strings?
Ex: the above works for strings like "1,2,3,4,5,6,7", but what about "1","2","3","4,5","6,7,8","9" so that the rows end up like:
1
2
3
4,5
6,7,8
9
edit: I'm on Oracle 11.2.0.4, 11gR2.
There is a hack. Replace the pattern "," with # and use it in regular expression.It works like a charm.
Input String : "1","2","3","4,5","6,7,8","9","Ant,B","Gradle","E,F","G"(Can be number/Character doesn't matter)
with temp as (
select replace(replace('"1","2","3","4,5","6,7,8","9","Ant,B","Gradle","E,F","G"','","','#'),'"') Colmn from dual
)
SELECT trim(regexp_substr(str, '[^#]+', 1, level)) str
FROM (SELECT Colmn str FROM temp) t
CONNECT BY instr(str, '#', 1, level - 1) > 0
Output :
STR
1
2
3
4,5
6,7,8
9
Ant,B
Gradle
E,F
G
10 rows
Refer DBFiddle link for demo.
https://dbfiddle.uk/?rdbms=oracle_18&fiddle=d09c326f614d10f5d3c407fdfd3a44c5
here is the another solution.it must be working all string with have number.
i used number values index as base.
with temp as (
select '"1","2","3",x"4,5","6,73,8","9"' Colmn from dual
)
SELECT regexp_substr(Colmn, '\d{1}', REGEXP_INSTR(Colmn, '\d{1}', REGEXP_INSTR(Colmn, '\d{1}') ,level),1 ) from temp
CONNECT BY REGEXP_COUNT (Colmn,'\d{1}')+1> level
how to multiple replacement string using regexp_replace in oracle?
formula = 1*3
formula detail (1->value1, 2->value3, 3->value3)
I want the result
formula = value1*value3
I don't understand what this:
formula = 1*3
formula detail (1->value1, 2->value3, 3->value3)
actually represents. What is the first row? Is it a string, stored in some table? Is it a string stored in a variable? Is it the complete string, or is the string just 1*3 (without formula =)?
What is the second row? The same doubts as for the first one.
Anyway: if we pretend that the first row represents a string, while the second represents your wish and not some code, then nested REPLACE (i.e. no regular expressions at all) does the job:
SQL> create or replace function f_rep
2 (par_1 in varchar2, par_2 in varchar2, par_3 in varchar2)
3 return varchar2
4 is
5 l_str varchar2(200) := 'formula = 1*3';
6 begin
7 l_str := replace(replace(replace(l_str, '1', par_1),
8 '2', par_2),
9 '3', par_3);
10 return l_str;
11 end;
12 /
Function created.
SQL> select f_rep('value1', null, 'value3') result from dual;
RESULT
--------------------------------------------------------------------------------
formula = value1*value3
SQL>
I have seen from from oracle documentation I can do a|b which will match all instances of character 'a 'or character 'b'. I wondered if it was possible in regexp to have 'a' and not 'b' e.g. match all instances of 'a' except where a is followed by 'b'.
so with these 3 strings
AAAA
ACAD
AAAB
I want to match 1 and 2 but not 3.
You could try:
with x as (
select 'AAAA' as str from dual
union all
select 'ACAD' as str from dual
union all
select 'AAAB' as str from dual
)
select * from x
where regexp_like(str, '[aA][^bB]')
and NOT regexp_like(str, '[aA][bB]')
Output:
AAAA
ACAD
Note: If you don't care if A is followed by some other non-B char, you can simply do:
with x as (
select 'AAAA' as str from dual
union all
select 'ACAD' as str from dual
union all
select 'AAAB' as str from dual
)
select * from x
where regexp_like(str, '[aA]')
and NOT regexp_like(str, '[aA][bB]')
which would match "AAAA" or "XYZA"
I have a long string.
message := 'I loooove my pet animal';
This string in 23 chars long. If message is greater that 15 chars, I need to find the length of message where I can break the string into 2 strings. For example, in this case,
message1 := 'I loove my'
message2 := 'pet animal'
Essentially it should find the position of a whole word at the previous to 15 chars and the break the original string into 2 at that point.
Please give me ideas how I can do this.
Thank you.
Here is a general solution - with possibly more than one input string, and with inputs of any length. The only assumption is that no single word may be more than 15 characters, and that everything between two spaces is considered a word. If a "word" can be more than 15 characters, the solution can be adapted, but the requirement itself would need to state what the desired result is in such a case.
I make up two input strings in a CTE (at the top) - that is not part of the solution, it is just for testing and illustration. I also wrote this in plain SQL - there is no need for PL/SQL code for this type of problem. Set processing (instead of one row at a time) should result in much better execution.
The approach is to identify the location of all spaces (I append and prepend a space to each string, too, so I won't have to deal with exceptions for the first and last substring); then I decide, in a recursive subquery, where each "maximal" substring should begin and where it should end; and then outputting the substrings is trivial. I used a recursive query, that should work in Oracle 11.1 (or 11.2 with the syntax I used, with column names in CTE declarations - it can be changed easily to work in 11.1). In Oracle 12, it would be easier to rewrite the same idea using MATCH_RECOGINZE.
with
inputs ( id, str ) as (
select 101, 'I loooove my pet animal' from dual union all
select 102, '1992 was a great year for - make something up here as needed' from dual
),
positions ( id, pos ) as (
select id, instr(' ' || str || ' ', ' ', 1, level)
from inputs
connect by level <= length(str) - length(replace(str, ' ')) + 2
and prior id = id
and prior sys_guid() is not null
),
r ( id, str, line_number, pos_from, pos_to ) as (
select id, ' ' || str || ' ', 0, null, 1
from inputs
union all
select r.id, r.str, r.line_number + 1, r.pos_to,
( select max(pos)
from positions
where id = r.id and pos - r.pos_to between 1 and 16
)
from r
where pos_to is not null
)
select id, line_number, substr(str, pos_from + 1, pos_to - pos_from - 1) as line_text
from r
where line_number > 0 and pos_to is not null
order by id, line_number
;
Output:
ID LINE_NUMBER LINE_TEXT
---- ----------- ---------------
101 1 I loooove my
101 2 pet animal
102 1 1992 was a
102 2 great year for
102 3 - make
102 4 something up
102 5 here as needed
7 rows selected.
First you reverse string.
SELECT REVERSE(strField) FROM DUAL;
Then you calculate length i = length(strField).
Then find the first space after the middle
j := INSTR( REVERSE(strField), ' ', i / 2, i)`
Finally split by i - j (maybe +/- 1 need to test it)
DEMO
WITH parameter (id, strField) as (
select 101, 'I loooove my pet animal' from dual union all
select 102, '1992 was a great year for - make something up here as needed' from dual union all
select 103, 'You are Supercalifragilisticexpialidocious' from dual
), prepare (id, rev, len, middle) as (
SELECT id, reverse(strField), length(strField), length(strField) / 2
FROM parameter
)
SELECT p.*, l.*,
SUBSTR(strField, 1, len - INSTR(rev, ' ', middle)) as first,
SUBSTR(strField, len - INSTR(rev, ' ', middle) + 2, len) as second
FROM parameter p
JOIN prepare l
ON p.id = l.id
OUTPUT
How do I use reg expression to found last number in a string and then everything on the right hand side in to column c1 and from the last number everything on the left + 1 character go into column c2 ?
e.g 1
string = 1234john4345 this is a test.
Result
c1 = 1234john4345
c2 = this is a test.
e.g 2
string = 1234john4345a this is a test.
Result
c1 = 1234john4345a
c2 = this is a test.
select test
--Group 1: Match everything up to the last digit, and one other character
--Group 2: Everything after group 1
,regexp_replace(test, '(.*[[:digit:]].)(.*)', '\1') c1
,regexp_replace(test, '(.*[[:digit:]].)(.*)', '\2') c2
from
(
select '1234john4345 this is a test.' test from dual union all
select '1234john4345a this is a test' test from dual
);