REGEXP_SUBSTR for portion of string - oracle

I would like to get:
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
from the following expression
LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Does someone know how I can get this using regexp_substr ?
EDIT
Basically I have a field which has 7 sets each separated by _ . The string I gave is just one example. I wanted to retrieve everything after the second _ . There is no fixed character length so I can not use a substr function. Hence I was using regexp_substr. I was able to get away by using a simplified version
Select FILE_NAME, ( (REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,3)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,4)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,5)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,6)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+',1,7)) ) as RegExp
from tbl
Here is some more data from the FILE_NAME field
LAST_FIRST_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
SMITH_JOHN_82961_0130BPQX9QZN9G4P5RDTPA9HR4R_MR_1_1of1
LASTNAME_FIRSTNAME_99999_01V0MU4XUQK0Y24Y9RYTFA7W1CM_MR_3_1of1

To get everything after the second underscore, you do not need regular expressions, but can use something like the following:
select substr(FILE_NAME, instr(FILE_NAME, '_', 1, 2) +1 ) from tbl
The instr returns the position of the second occurrence of '_', starting by the first character; the substr simply gets everything starting from the position given by instr + 1

From your Requirement, you can just go ahead and use the simple SUBSTRfunction. Its faster, and it addresses the simple need to remove the String LASTNAME_FIRSTNAME.
select substr('LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1', 20) data_string
from dual;
Output:
data_string
-----------------
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Unless you have another underlying logic you need to address?
Kindly clarify so i can edit the answer accordingly.

Related

REGEXP_LIKE Oracle equivalent to count characters in Snowflake

I am trying to come up with an equivalent of the below Oracle statement in Snowflake. This would check if the different parts of the string separated by '.' matches the number of characters in the REGEXP_LIKE expression. I have come up with a rudimentary version to perform the check in Snowflake but I am sure there's a better and cleaner way to do it. I am looking to come up with a one-liner regular expression check in Snowflake similar to Oracle. Appreciate your help!
-- Oracle
SELECT -- would return True
CASE
WHEN REGEXP_LIKE('AB.XYX.12.34.5670.89', '^\w{2}\.\w{3}\.\w{2}') THEN 'True'
ELSE NULL
END AS abc
FROM DUAL
-- Snowflake
SELECT -- would return True
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 1), '[A-Z0-9]{2}') AND
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 2), '[A-Z0-9]{3}') AND
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 3), '[A-Z0-9]{2}') AS abc
You need to add a .* at the end as the REGEXP_LIKE adds explicit ^ && $ to string:
The function implicitly anchors a pattern at both ends (i.e. '' automatically becomes '^$', and 'ABC' automatically becomes '^ABC$'). To match any string starting with ABC, the pattern would be 'ABC.*'.
select
column1 as str,
REGEXP_LIKE(str, '\\w{2}\\.\\w{3}\\.\\w{2}.*') as oracle_way
FROM VALUES
('AB.XYX.12.34.5670.89')
;
gives:
STR
ORACLE_WAY
AB.XYX.12.34.5670.89
TRUE
Or in the context of your question:
SELECT IFF(REGEXP_LIKE('AB.XYX.12.34.5670.89', '\\w{2}\\.\\w{3}\\.\\w{2}.*'), 'True', null) AS abc;
Your use of \w seems to suggest you don't need delimited strings to be strictly [A-Z0-9] since word characters allow underscore and period. If all bets were off and the only requirement was to have . at 3rd, 7th and 10th position, you could have used like this way.
select 'AB.XGH.12.34.5670.89' like '__.___.__.%' ;

Need to extract text with REGEXP_SUBSTR - cannot find the right combination

Here is an example of my text- I am trying to get TEXTPART3 AS THE ANSWER:
TEXTPART1 : TEXTPART2: TEXTPART3 - TEXTPART4
I used TRIM(LEADING ':' FROM REGEXP_SUBSTR('textstatementhere', ':.+?-')) - but it is not accounting for the two ":" and the "-" in the text statement I get ' TEXTPART2: TEXTPART3 -'
Can anyone help?
Thanks in advance!
The problem has a more efficient solution using only standard string functions:
with
sample_input (str) as (
select 'TEXTPART1 : TEXTPART2: TEXTPART3 - TEXTPART4' from dual
)
select substr(str, pos, instr(str, '-', pos) - pos - 1) as text_part_3
from (select str, instr(str, ':', 1, 2) + 2 as pos from sample_input)
;
TEXT_PART_3
-----------
TEXTPART3
Leverage the REGEXP_SUBSTR Function to Do All of Your Work
Use a subexpression, (\w), and reference it:
WITH exmple AS (
SELECT
'TEXTPART1 : TEXTPART2: TEXTPART3 - TEXTPART4' txt
FROM
dual
)
SELECT
txt,
regexp_substr(txt, ': (\w*) -', 1, 1, NULL,
1)
FROM
exmple;
I see that you used ., in lieu of \w. Because you chose the meta-character,. (which represents all characters except new line (though that can be included if "n" is set as a pattern matching modifier)), the second colon is thrown in to the matching set.
What does TEXTPART3 include?
Perhaps the meta-character, \w, (which stands for alphanumeric or underscore (_) character), is not what you need.
You could replace it with a non-matching character list to avoid the problems with .:
[^:].
This approach would look like this:
REGEXP_SUBSTR(txt, ': ([^:]*) -',1,1,NULL,1)
Lastly, with the quantiifier associated with this subexpression, I used * which means zero or more matches. I assumed that there would be instances where there could be zero matches for this TEXTPART3. If this is not the case, we can use +.

Oracle regex_replace not working as expected

I have following SQL query (Oracle 18c):
SELECT
--FIRST
translate(
' sOmE tEsT
eNdOfLiNe',
chr(10)||chr(11)||chr(13), 'replText'
) "Result1",
--SECOND
regexp_replace(
' sOmE tEsT
eNdOfLiNe',
'[\x0A|\x0B|`\x0D]', 'replText'
) "Result2",
--THIRD
regexp_replace(
' sOmE tEsT
eNdOfLiNe',
'[\r\n\t]', 'replText', 1, 0
) "Result3"
FROM dual
What I would like to do is replace all tabs, return carriages and new line indicators with new string but it seems like regexp replace is not working (returns initial text). I am really sorry about formatting but I need to handle text in exact format as above with \r \n \t mixed chars.
Here is fiddle: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=63834f9bcab93136635366f18c375b13
I am learning Oracle right now and don't understand why second and third solution returns initial text. The first solution seems to work but I would like to achieve the same effect in SECOND and THIRD solution. What I missed?
I'm pretty sure Oracle does not allow escape sequences in a character class. I believe this is what you have to do. In response to your comment on another answer here and as you are learning, regex is most definitely not regex. Especially Oracle's implementation.
EDIT to explain the regex: The regex pattern is building a string of a regex character class containing 3 characters, hence the concatenation. You can't just have escape characters in the regex as then regex would take those characters as part of the character class pattern itself.
SELECT REGEXP_REPLACE(
' sOmE tEsT
eNdOfLiNe', '['||CHR(9)||CHR(10)||CHR(13)||']', 'X') Result3
FROM dual;
RESULT3
------------------------------
sOmE tEsTXXXXXXXX eNdOfLiNe
1 row selected.
You can try the below using similar format as translate
select regexp_replace(
' sOmE tEsT
eNdOfLiNe',
chr(10)||'|'||chr(11)||'|'||chr(13), 'replText') "Result3"
FROM dual

How to apply regular expression on the below given string

i have a string 'MCDONALD_YYYYMMDD.TXT' i need to use regular expressions and append the '**' after the letter 'D' in the string given . (i.e In the string at postion 9 i need to append '*' based on a column value 'star_len'
if the star_len = 2 the o/p = ''MCDONALD??_YYYYMMDD.TXT'
if the star_len = 1 the o/p = ''MCDONALD?_YYYYMMDD.TXT'
with
inputs ( filename, position, symbol, len ) as (
select 'MCDONALD_20170812.TXT', 9, '*', 2 from dual
)
-- End of simulated inputs (for testing purposes only, not part of the solution).
-- SQL query begins BELOW THIS LINE.
select substr(filename, 1, position - 1) || rpad(symbol, len, symbol)
|| substr(filename, position) as new_str
from inputs
;
NEW_STR
-----------------------
MCDONALD**_20170812.TXT
select regexp_replace('MCDONALD_YYYYMMDD.TXT','MCDONALD','MCDONALD' ||
decode(star_len,1,'*',2,'**'))
from dual
This is how you could do it. I don't think you need it as a regular expression though if it is always going to be "MCDONALD".
EDIT: If you need to be providing the position in the string as well, I think a regular old substring should work.
select substr('MCDONALD_YYYYMMDD.TXT',1,position-1) ||
decode(star_len,1,'*',2,'**') || substr('MCDONALD_YYYYMMDD.TXT',position)
from dual
Where position and star_len are both columns in some table you provide(instead of dual).
EDIT2: Just to be more clear, here is another example using a with clause so that it runs without adding a table in.
with testing as
(select 'MCDONALD_YYYYMMDD.TXT' filename,
9 positionnum,
2 star_len
from dual)
select substr(filename,1,positionnum-1) ||
decode(star_len,1,'*',2,'**') ||
substr(filename,positionnum)
from testing
For the fun of it, here's a regex_replace solution. I went with a star since that what your variable was called even though your example used a question mark. The regex captures the filename string in 2 parts, the first being from the start up to 1 character before the position value, the second the rest of the string. The replace puts the captured parts back together with the stars in between.
with tbl(filename, position, star_len ) as (
select 'MCDONALD_20170812.TXT', 9, 2 from dual
)
select regexp_replace(filename,
'^(.{'||(position-1)||'})(.*)$', '\1'||rpad('*', star_len, '*')||'\2') as fixed
from tbl;

trim value till specified string in oracle pl/sql

i want to trim value of the given string till specified string in oracle pl/sql.
some thing like below.
OyeBuddy$$flex-Flex_Image_Rotator-1443680885520.
In the above string i want to trim till $$ so that i will get "flex-Flex_Image_Rotator-1443680885520".
You can use different ways; here are two methods, with and without regexp:
with test(string) as ( select 'OyeBuddy$$flex-Flex_Image_Rotator-1443680885520.' from dual)
select regexp_replace(string, '(.*)(\$\$)(.*)', '\3')
from test
union all
select substr(string, instr(string, '$$') + length('$$'))
from test
You want to do a SUBSTR where the starting position is going to be the position of '$$' + 2 . +2 is because the string '$$' is of length 2, and we don't want to include that string in the result.
Something like -
SELECT SUBSTR (
'ABCDEF$$some_big_text',
INSTR ('ABCDEF$$some_big_text', '$$') + 2)
FROM DUAL;

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