I am using BigQuery on Google Cloud Platform to extract data from GDELT. This uses an SQL syntax and regular expressions.
I have a column of data (called V2Tone), in which each cell looks like this:
1.55763239875389,2.80373831775701,1.24610591900312,4.04984423676012,26.4797507788162,2.49221183800623,299
To select only the first number (i.e., the number before the first comma) using regular expressions, we use this:
regexp_replace(V2Tone, r',.*', '')
How can we select only the second number (i.e., the number between the first and second commas)?
How about the third number (i.e., the number between the second and third commas)?
I understand that re2 syntax (https://github.com/google/re2/wiki/Syntax) is used here, but my understanding of how to put that all together is limited.
If anything is unclear, please let me know. Thank you for your help as I learn to use regular expressions.
Below example is for BigQuery Standard SQL using super simple SPLIT approach
#standardSQL
SELECT
SPLIT(V2Tone)[SAFE_OFFSET(0)] first_number,
SPLIT(V2Tone)[SAFE_OFFSET(1)] second_number,
SPLIT(V2Tone)[SAFE_OFFSET(2)] third_number
FROM `project.dataset.table`
If for some reason you need/want to use regexp here - use below
#standardSQL
SELECT
REGEXP_EXTRACT(V2Tone, r'^(.*?),') first_number,
REGEXP_EXTRACT(V2Tone, r'^(?:(?:.*?),)(.*?),') second_number,
REGEXP_EXTRACT(V2Tone, r'^(?:(?:.*?),){2}(.*?),') third_number,
REGEXP_EXTRACT(V2Tone, r'^(?:(?:.*?),){4}(.*?),') fifth_number
FROM `project.dataset.table`
Note use of REGEXP_EXTRACT instead of REGEXP_REPLACE
You can play, test above options with dummy string from your question as below
#standardSQL
WITH `project.dataset.table` AS (
SELECT '1.55763239875389,2.80373831775701,1.24610591900312,4.04984423676012,26.4797507788162,2.49221183800623,299' V2Tone
)
SELECT
SPLIT(V2Tone)[SAFE_OFFSET(0)] first_number,
SPLIT(V2Tone)[SAFE_OFFSET(1)] second_number,
SPLIT(V2Tone)[SAFE_OFFSET(2)] third_number,
REGEXP_EXTRACT(V2Tone, r'^(.*?),') first_number_re,
REGEXP_EXTRACT(V2Tone, r'^(?:(?:.*?),)(.*?),') second_number_re,
REGEXP_EXTRACT(V2Tone, r'^(?:(?:.*?),){2}(.*?),') third_number_re,
REGEXP_EXTRACT(V2Tone, r'^(?:(?:.*?),){4}(.*?),') fifth_number_re
FROM `project.dataset.table`
with output :
first_number second_number third_number first_number_re second_number_re third_number_re fifth_number_re
1.55763239875389 2.80373831775701 1.24610591900312 1.55763239875389 2.80373831775701 1.24610591900312 26.4797507788162
I don't know of a single regex replace which could be used to isolate a single number in your CSV string, because we need to remove things on both sides of the match, in general. But, we can chain together two calls to regex_replace. For example, if you wanted to target the third number in the CSV string, we could try this:
regexp_replace(regexp_replace(V2Tone, r'^(?:(?:\d+(?:\.\d+)?),){2}', ''),
r',.*', ''))
The pattern I am using to strip of the first n numbers is this:
^(?:(?:\d+(?:\.\d+)?),){n}
This just removes a number, followed by a comma, n times, from the beginning of the string.
Demo
Here is a solution with a single regex replace:
^([^,]+(?:,|$)){2}([^,]+(?:,|$))*|^.*$
Demo
\n is added to the negated character class in the demo to avoid matching accross lines in m|multiline mode.
Usage:
regexp_replace(V2Tone, r'^([^,]+(?:,|$)){2}([^,]+(?:,|$))*|^.*$', '$1')
Explanation:
([^,]+(?:,|$){n} captures everything to the next comma or the end of the string n times
([^,]+(?:,|$))* captures the rest 0 or more times
^.*$ capture everything if we cannot match n times
And then, finally, we can reinsert the nth match using $1.
Related
I have varchar field in the database that contains text. I need to replace every occurrence of a any 2 letter + 8 digits string to a link, such as VA12345678 will return /cs/page.asp?id=VA12345678
I have a regex that replaces the string but how can I replace it with a string where part of it is the string itself?
SELECT REGEXP_REPLACE ('test PI20099742', '[A-Z]{2}[0-9]{8}$', 'link to replace with')
FROM dual;
I can have more than one of these strings in one varchar field and ideally I would like to have them replaced in one statement instead of a loop.
As mathguy had said, you can use backreferences for your use case. Try a query like this one.
SELECT REGEXP_REPLACE ('test PI20099742', '([A-Z]{2}[0-9]{8})', '/cs/page.asp?id=\1')
FROM DUAL;
For such cases, you may want to keep the "text to add" somewhere at the top of the query, so that if you ever need to change it, you don't have to hunt for it.
You can do that with a with clause, as shown below. I also put some input data for testing in the with clause, but you should remove that and reference your actual table in your query.
I used the [:alpha:] character class, to match all letters - upper or lower case, accented or not, etc. [A-Z] will work until it doesn't.
with
text_to_add (link) as (
select '/cs/page.asp?id=' from dual
)
, sample_strings (str) as (
select 'test VA12398403 and PI83048203 to PT3904' from dual
)
select regexp_replace(str, '([[:alpha:]]{2}\d{8})', link || '\1')
as str_with_links
from sample_strings cross join text_to_add
;
STR_WITH_LINKS
------------------------------------------------------------------------
test /cs/page.asp?id=VA12398403 and /cs/page.asp?id=PI83048203 to PT3904
I've some URL's in my cas_fnd_dwd_det table,
casi_imp_urls cas_code
----------------------------------- -----------
www.casiac.net/fnds/CASI/qnxp.pdf
www.casiac.net/fnds/casi/as.pdf
www.casiac.net/fnds/casi/vindq.pdf
www.casiac.net/fnds/CASI/mnip.pdf
how do i copy the letters between last '/' and '.pdf' to another column
expected outcome
casi_imp_urls cas_code
----------------------------------- -----------
www.casiac.net/fnds/CASI/qnxp.pdf qnxp
www.casiac.net/fnds/casi/as.pdf as
www.casiac.net/fnds/casi/vindq.pdf vindq
www.casiac.net/fnds/CASI/mnip.pdf mnip
the below URL's are static
www.casiac.net/fnds/CASI/
www.casiac.net/fnds/casi/
Advise, how do i select the codes between last '/' and '.pdf' ?
I would recommend to take a look at REGEXP_SUBSTR. It allows to apply a regular expression. Db2 has string processing functions, but the regex function may be the easiest solution. See SO question on regex and URI parts for different ways of writing the expression. The following would return the last slash, filename and the extension:
SELECT REGEXP_SUBSTR('http://fobar.com/one/two/abc.pdf','\/(\w)*.pdf' ,1,1)
FROM sysibm.sysdummy1
/abc.pdf
The following uses REPLACE and the pattern is from this SO question with the pdf file extension added. It splits the string in three groups: everything up to the last slash, then the file name, then the ".pdf". The '$1' returns the group 1 (groups start with 0). Group 2 would be the ".pdf".
SELECT REGEXP_REPLACE('http://fobar.com/one/two/abc.pdf','(?:.+\/)(.+)(.pdf)','$1' ,1,1)
FROM sysibm.sysdummy1
abc
You could apply LENGTH and SUBSTR to extract the relevant part or try to build that into the regex.
For older Db2 versions than 11.1. Not sure if it works for 9.5, but definitely should work since 9.7.
Try this as is.
with cas_fnd_dwd_det (casi_imp_urls) as (values
'www.casiac.net/fnds/CASI/qnxp.pdf'
, 'www.casiac.net/fnds/casi/as.pdf'
, 'www.casiac.net/fnds/casi/vindq.pdf'
, 'www.casiac.net/fnds/CASI/mnip.PDF'
)
select
casi_imp_urls
, xmlcast(xmlquery('fn:replace($s, ".*/(.*)\.pdf", "$1", "i")' passing casi_imp_urls as "s") as varchar(50)) cas_code
from cas_fnd_dwd_det
I have to replace a string pattern in SQL with empty string, could anyone please suggest me?
Input String 'AC001,AD001,AE001,SA001,AE002,SD001'
Output String 'AE001,AE002
There are the 4 digit codes with first 2 characters "alphabets" and last two are digits. This is always a 4 digit code. And I have to replace all codes except the codes starting with "AE".
I can have 0 or more instances of "AE" codes in the string. The final output should be a formatted string "separated by commas" for multiple "AE" codes as mentioned above.
Here is one option calling regex_replace multiple times, eliminating the "not required" strings little by little in each iteration to arrive at the required output.
SELECT regexp_replace(
regexp_replace(
regexp_replace(
'AC001,AD001,AE001,SA001,AE002,SD001', '(?<!AE)\d{3},{0,1}', 'X','g'
),'..X','','g'
),',$','','g'
)
See Demo here
I would convert the list to an array, unnest that to rows then filter out those that should be kept and aggregate it back to a string:
select string_agg(t, ',')
from unnest(string_to_array('AC001,AD001,AE001,SA001,AE002,SD001',',') as x(t)
where x.t like 'AE%'; --<< only keep those
This is independent of the number of elements in the string and can easily be extended to support more complex conditions.
This is a good example why storing comma separated values in a single column is not such a good idea to begin with.
I need to replace multiple words such as (dog|cat|bird) with nothing in a string where there may be multiple consecutive occurrences of a word. The actual code is to remove salutations and suffixes from a name. Unfortunately the garbage data I get sometimes contains "SNERD JR JR."
I was able to create a regular expression pattern that accomplishes my goal but only for the first occurrence. I implemented a stupid hack to get rid of the second occurrence, but I believe there has to be a better way. I just can't figure it out.
Here is my "hacked" code;
FUNCTION REMOVE_SALUTATIONS(IN_STRING VARCHAR2) RETURN VARCHAR2 DETERMINISTIC
AS
REGEX_SALUTATIONS VARCHAR2(4000) := '(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)';
BEGIN
RETURN TRIM(REGEXP_REPLACE(REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' '),REGEX_SALUTATIONS,''));
END REMOVE_SALUTATIONS;
I was actually proud that I was able to get this far, as regular expression are not very regular to me. All help is appreciated.
EDIT:
The default for regexp_replace based on my understanding is to do a global replace. But on the outside chance my DB is configured different I did try;
select REGEXP_REPLACE('SNERD JR JR','(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)',' ',1,0) from dual;
and the results are;
SNERD JR
Use occurrence parameter of REGEXP_REPLACE function. The docs says:
occurrence is a nonnegative integer indicating the occurrence of the replace operation:
If you specify 0, then Oracle replaces all occurrences of the match.
If you specify a positive integer n, then Oracle replaces the nth occurrenc
https://docs.oracle.com/cd/B28359_01/server.111/b28286/functions137.htm#SQLRF06302
It should look like:
...
REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' ', 1,0 )
...
I have a column in Oracle which can contain up to 5 separate values, each separated by a '|'. Any of the values can be present or missing. Here are come examples of how the data might look:
100-1
10-3|25-1|120/240
15-1|15-3|15-2|120/208
15-1|15-3|15-2|120/208|STA-2
112-123|120/208|STA-3
The values are arbitrary except for the order. The numerical values separated by dashes always come first. There can be 1 to 3 of these values present. The numerical values separated by a slash (if it is present) is next. The string, 'STA', and a numerical value separated by a dash is always last, if it is present.
What I would like to do is reformat this column to only ever include the first three possible values, those being the three numerical values separated by dashes. Afterwards, I want to replace 2nd numeric in each value (the numeric after the dash) using the following pattern:
1 = A
2 = B
3 = C
I would also like to remove the dash afterwards, but not the '|' that separates the values unless there is a trailing '|'.
To give you an idea, here's how the values at the beginning of the post would look after the reformatting:
100A
10C|25A
15A|15C|15B
15A|15C|15B
112ABC
I'm thinking this can be done with regex expressions but it's got me a little confused. Does anyone have a solution?
If I have to solve this problem I will solve it in following ways.
SELECT
REGEXP_REPLACE(column,'\|\d+\/\d+(\|STA-\d+)?',''),
REGEXP_REPLACE(column,'(\d+)-(1)([^\d])','\1A\3'),
REGEXP_REPLACE(column,'(\d+)-(2)([^\d])','\1B\3'),
REGEXP_REPLACE(column,'(\d+)-(3)([^\d])','\1C\3'),
REGEXP_REPLACE(column,'(\d+)-(123)([^\d])','\1ABC')
FROM table;
Explanation: Let us break down each REGEXP_REPLACE statement one by one.
REGEXP_REPLACE(column,'\|\d+\/\d+(\|STA-\d+)?','')
This will replace the end part like 120/208|STA-2 with empty string so that further processing is easy.
Finding match was easy but replacing A for 1, B for 2 and C for 3 was not possible ( as per my knowledge ) So I did those matching and replacements separately.
In each regex from second statement (\d+)-(yourNumber)([^\d]) first group is number before - then yourNumber is either 1,2,3 or 123 followed by |.
So the replacement will be according to yourNumber.
All demos here from version 1 to 5.
Note:- I have just done replacement for combination of yourNUmber for those present in question. You can do likewise for other combinations too.
you can do this in one line, but you can write simple function to do that
SELECT str, REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?','') cut
, REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4') rep3toC
, REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4'), '(\-)([1,C]*)(2)([1,C]*)', '\1\2B\4') rep2toB
, REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4'), '(\-)([1,C]*)(2)([1,C]*)', '\1\2B\4'), '(\-)([B,C]*)(1)([B,C]*)', '\1\2A\4') rep1toA
, REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(REGEXP_REPLACE(str,'(\|\d+\/\d+)?(\|STA-\d+)?',''), '(\-)([1,2]*)(3)([1,2]*)', '\1\2C\4'), '(\-)([1,C]*)(2)([1,C]*)', '\1\2B\4'), '(\-)([B,C]*)(1)([B,C]*)', '\1\2A\4'), '-', '') "rep-"
FROM (
SELECT '100-1' str FROM dual UNION
SELECT '10-3|25-1|120/240' str FROM dual UNION
SELECT '15-1|15-3|15-2|120/208' str FROM dual UNION
SELECT '15-1|15-3|15-2|120/208|STA-2' str FROM dual UNION
SELECT '112-123|120/208|STA-3' FROM dual
) tab