Working for the first time with Postgres flavor regular expressions and stumped on it's behavior here.
Input: 'xxxaaabc'
Expected: 'xxxaaazzzbc'
Logic: Match M a's, capture group, and add 'zzz' directly after the sequence
Attempts:
select regexp_replace('aaabc','(?!\Sa+)','zzz','i')
Result: zzzxxxaaabc
select regexp_replace('aaabc','(?!\Sa+)','zzz','i')
Result: xxxazzzaabc
select select regexp_replace('xxxaaabc','(?!=a+)','zzz','i')
Result: zzzxxxaaabc
This one seems to get pretty close regexp_replace('aaabc','(?!\Sa+)','zzz','i'), but the repetition of a's isn't working.
Note the (?!\Sa+) pattern matches an empty location that is not immediately followed with a single non-whitespace char and then one or more a chars.
The (?!=a+) pattern matches a location in a string, that is not immediately followed with = and then one or more as.
You can capture one or more as and then use a backreference to the match and then just append the zzz:
select regexp_replace('xxxaaabc','(a+)','\1zzz','i') AS Result;
See the DB fiddle.
Related
I am using Snowflake SQL. I would like to remove characters from a string after a special character ~. How can I do that?
here is the whole scenario. Let me explain. I do have a string like 'CK#123456~fndkjfgdjkg'. Now, i want only the number after #.And not anything after ~. This is number length varies for that field value. It might be 1 or 5 or 3. And i want to add the condition in where class where this number is equal to check_num from other table after joining. I am trying REGEXP_SUBSTR(A.SRC_TXT, '(?<=CK#)(.+?\b)') = C.CHK_NUM in the where condition. I am getting the error as 'No repititive argument after ?'
You can use a regex for this
-- To remove just the character after a ~
select regexp_replace('fo~o bar','~.', '');
-- returns 'fo bar'
--If you want to keep the ~
select regexp_replace('fo~o bar','~.', '~');
-- returns 'fo~ bar'
--If you want to remove everything after the ~
select regexp_replace('fo~o bar','~.*', '');
--returns 'fo'
If you need to remove other specific character sets after a ~, you can probably do this with a slightly more complicated regex, but I'd need examples of your desired input/output to help with that.
EDIT for updated question
This regex replace should get what you need.
select regexp_replace('CK#123456~fndkjfgdjkg','CK#(\\d*)~.*', '\\1');
-- returns 123456
(\\d*) gets ANY number of digits in a row, and the \\1 causes it to replace the match with what was in the first set of parenthesis, which is your list of digits. the CK# and ~.* are there to make sure the whole string gets matched and replaced.
If the CK# can vary as well, you can use .*? like this.
select regexp_replace('ABCD123HI#123456~fndkjfgdjkg','.*?#(\\d*)~.*', '\\1')
-- returns 123456
I'd probably do something like the following, easy enough but not as cool as RegEx type of functions.
set my_string='fooo~12345';
set search_for_me = '~';
SELECT SUBSTR($my_string, 1, DECODE(position($search_for_me, $my_string), 0, length($my_string), position($search_for_me, $my_string)));
I hope this helps...Rich
It looks like lookahead and lookbehinds are not supported in REGEXP functions, they seem to work in the PATTERN clause of a LIST command. Snowflake documentation makes no mention either way of lookahead or lookbehinds.
In your example:
It seems that the query engine is looking for that repeating argument, where you are attempting a lookbehind
You have not specified what you wanted extracted. You have two capture groups, but in this scenario everything would be returned
Since you are looking to remove everything after ~ you have a delimiter, why not use it in your REGEXP_SUBSTR function?
Try the following:
SELECT $1,REGEXP_SUBSTR($1,'\\w+#(.+?)~',1,1,'is',1)
FROM VALUES
('CK#123456~fndkjfgdjkg')
,('QH#128fklj924~fndkjfgdjkg')
;
This looks for:
One or more word characters
Followed by #
Capturing one or more characters upto and not including ~
Returns the characters within the capture group
You can change the .+? to \\d+? to make sure the pattern is only digits. Backslashes must be escaped with a backslash.
The descriptions for each argument of the function can be found here:
https://docs.snowflake.net/manuals/sql-reference/functions/regexp_substr.html
You could check this!!
select substr('CK#123456~fndkjfgdjkg',4,6) from dual;
OUTPUT
123456
https://docs.snowflake.net/manuals/sql-reference/functions/substr.html
I am doing this in Impala or Hive. Basically let say I have a string like this
f-150:aa|f-150:cc|g-210:dd
Each element is separated by the pipe |. Each has prefix f-150 or whatever. I want to be able to remove the prefix and keep only element that matches specific prefix. For example, if the prefix is f-150, I want the final string after regex_replace is
aa|cc
dd is removed because g-210 is different prefix and not match, therefore the whole element is removed.
Any idea how to do this using string expression in one SQL?
Thanks
UPDATE 1
I tried this in Impala:
select regexp_extract('f-150:aa|f-150:cc|g-210:dd','(?:(?:|(\\|))f-150|keep|those):|(?:^|\\|)\\w-\\d{3}:\\w{2}',0);
But got this output:
f-150:aa
In Hive, I got NULL.
The regexyou in question could look like this:
(?:(?:|(\\|))f-150|keep|those):|(?:^|\\|)\\w-\\d{3}:\\w{2}
I have added some pseudo keywords to retain, but I am sure you get the idea:
Wholy match elements that should be dropped but only match the prefix for those that should be retained.
To keep the separator intact, match | at the beginning of an element in group 1 and put it back in the replacement with $1.
Demo
According to the documentation, your query should be written like a Java regex; likewise, this should perform like this code sample in Java.
You could match the values that you want to remove and then replace with an empty string:
f-150:|\|[^:]+:[^|]+$|[^|]+:[^|]+\|
f-150:|\\|[^:]+:[^|]+$|[^|]+:[^|]+\\|
Explanation
f-150: Match literally
| Or
\|[^:]+:[^|]+$ Match a pipe, not a colon one or more times followed by not a pipe one or more times and assert the end of the line
| Or
[^|]+:[^|]+\| Match not a pipe one or more times, a colon followed by matching not a pipe one or more times and then match a pipe
Test with multiple lines and combinations
You may have to loop through the string until the end to get the all the matching sub string. Look ahead syntax is not supported in most sql so above regexp might not be suitable for SQL syntax. For you purpose you can do something like creating a table to loop through just to mimic Oracle's level syntax and join with your table containing the string.
With loop_tab as (
Select 1 loop union all
Select 2 union all
select 3 union all
select 4 union all
select 5),
string_tab as(Select 'f-150:aa|ade|f-150:ce|akg|f-150:bb|'::varchar(40) as str)
Select regexp_substr(str,'(f\\-150\\:\\w+\\|)',1,loop)
from string_tab
join loop_tab on 1=1
Output:
regexp_substr
f-150:aa|
f-150:ce|
f-150:bb|
How can the matched regular expression be returned from an SQL select? I tried using REGEXP_EXTRACT with no luck (function not available). What I've done that does work is this:
SELECT column ~ '^stuff.*$'
FROM table;
but this gives me a list of true / false. I want to know what is extracted in each case.
If you're trying to capture the regex match that resulted from the expression, then substring would do the trick:
select substring ('I have a dog', 'd[aeiou]g')
Would return any match, in this case "dog."
I think the missing link of what you were trying above was that you need to put the expression you want to capture in parentheses. regexp_matches would work in this case (had you included parentheses around the expression you wanted to capture), but would return an array of text with each match. If it's one match, substring is sort of convenient.
So, circling back to your example, if you're trying to return stuff if and only if it's at the beginning of a column:
select substring (column, '^(stuff)')
or
select (regexp_matches (column, '^(stuff)'))[1]
Use regexp_matches.
SELECT regexp_matches(column,'^stuff.*$')
FROM table
The regexp_matches function returns a text array of all of the captured substrings resulting from matching a POSIX regular expression pattern. It has the syntax regexp_matches(string, pattern [, flags ]). The function can return no rows, one row, or multiple rows (see the g flag below). If the pattern does not match, the function returns no rows. If the pattern contains no parenthesized subexpressions, then each row returned is a single-element text array containing the substring matching the whole pattern. If the pattern contains parenthesized subexpressions, the function returns a text array whose n'th element is the substring matching the n'th parenthesized subexpression of the pattern (not counting "non-capturing" parentheses; see below for details). The flags parameter is an optional text string containing zero or more single-letter flags that change the function's behavior. Flag g causes the function to find each match in the string, not only the first one, and return a row for each such match.
I'm using Amazon Redshift which uses PostgreSQL 8.0.2 (I should have mentioned this in the question). For me what worked was REGEXP_SUBSTR
e.g.
SELECT REGEXP_SUBSTR(column,'^stuff.*$')
FROM table
Im using regexp to find the text after a word appear.
Fiddle demo
The problem is some address use different abreviations for big house: Some have space some have dot
Quinta
QTA
Qta.
I want all the text after any of those appear. Ignoring Case.
I try this one but not sure how include multiple start
SELECT
REGEXP_SUBSTR ("Address", '[^QUINTA]+') "REGEXPR_SUBSTR"
FROM Address;
Solution:
I believe this will match the abbreviations you want:
SELECT
REGEXP_REPLACE("Address", '^.*Q(UIN)?TA\.? *|^.*', '', 1, 1, 'i')
"REGEXPR_SUBSTR"
FROM Address;
Demo in SQL fiddle
Explanation:
It tries to match everything from the begging of the string:
until it finds Q + UIN (optional) + TA + . (optional) + any number of spaces.
if it doesn't find it, then it matches the whole string with ^.*.
Since I'm using REGEXP_REPLACE, it replaces the match with an empty string, thus removing all characters until "QTA", any of its alternations, or the whole string.
Notice the last parameter passed to REGEXP_REPLACE: 'i'. That is a flag that sets a case-insensitive match (flags described here).
The part you were interested in making optional uses a ( pattern ) that is a group with the ? quantifier (which makes it optional). Therefore, Q(UIN)?TA matches either "QUINTA" or "QTA".
Alternatively, in the scope of your question, if you wanted different options, you need to use alternation with a |. For example (pattern1|pattern2|etc) matches any one of the 3 options. Also, the regex (QUINTA|QTA) matches exactly the same as Q(UIN)?TA
What was wrong with your pattern:
The construct you were trying ([^QUINTA]+) uses a character class, and it matches any character except Q, U, I, N, T or A, repeated 1 or more times. But it's applied to characters, not words. For example, [^QUINTA]+ matches the string "BCDEFGHJKLMOPRSVWXYZ" completely, and it fails to match "TIA".
I got this query and want to extract the value between the brackets.
select de_desc, regexp_substr(de_desc, '\[(.+)\]', 1)
from DATABASE
where col_name like '[%]';
It however gives me the value with the brackets such as "[TEST]". I just want "TEST". How do I modify the query to get it?
The third parameter of the REGEXP_SUBSTR function indicates the position in the target string (de_desc in your example) where you want to start searching. Assuming a match is found in the given portion of the string, it doesn't affect what is returned.
In Oracle 11g, there is a sixth parameter to the function, that I think is what you are trying to use, which indicates the capture group that you want returned. An example of proper use would be:
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]', 1,1,NULL,1) from dual;
Where the last parameter 1 indicate the number of the capture group you want returned. Here is a link to the documentation that describes the parameter.
10g does not appear to have this option, but in your case you can achieve the same result with:
select substr( match, 2, length(match)-2 ) from (
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]') match FROM dual
);
since you know that a match will have exactly one excess character at the beginning and end. (Alternatively, you could use RTRIM and LTRIM to remove brackets from both ends of the result.)
You need to do a replace and use a regex pattern that matches the whole string.
select regexp_replace(de_desc, '.*\[(.+)\].*', '\1') from DATABASE;