I've reviewed this question and I'm wondering my output seems to be a little skewed.
From my understanding the REGEXP_REPLACE method, takes a string that you want to replace content with, followed by a pattern to match, then anything that does not match that pattern is replaced with the substitution param.
I've written the following function to extract distance from a text field, in which a spatial query will be performed on the result.
CREATE OR REPLACE FUNCTION extract_distance
(
p_search_string VARCHAR2
)
RETURN VARCHAR2
IS
l_distance VARCHAR2(25);
BEGIN
SELECT REGEXP_REPLACE(UPPER(p_search_string), '(([0-9]{0,4}) ?MILES)', '')
INTO l_distance FROM SYS.DUAL;
RETURN l_distance;
END extract_distance;
When I run this in a block to test:
DECLARE
l_output VARCHAR2(25);
BEGIN
l_output := extract_distance('Stores selling COD4 in 400 Miles');
DBMS_OUTPUT.PUT_LINE(l_output);
END;
I'd expect the output 400 miles but in-fact I get Stores selling COD4 in. Where have I gone wrong?
"REGEXP_REPLACE extends the functionality of the REPLACE function by letting you search a string for a regular expression pattern. By default, the function returns source_char with every occurrence of the regular expression pattern replaced with replace_string." from Oracle docu
You could use, e.g.,
SELECT REGEXP_REPLACE('Stores selling COD4 in 400 Miles', '^.*?(\d+ ?MILES).*$', '\1', 1, 0, 'i') FROM DUAL;
or alternatively
SELECT REGEXP_SUBSTR('Stores selling COD4 in 400 Miles', '(\d+ ?MILES)', 1, 1, 'i') FROM DUAL;
You'll want to use, regexp_substr which returns a substring that matches the regular expression.
REGEX_SUBSTR
Related
I have varchar field in the database that contains text. I need to replace every occurrence of a any 2 letter + 8 digits string to a link, such as VA12345678 will return /cs/page.asp?id=VA12345678
I have a regex that replaces the string but how can I replace it with a string where part of it is the string itself?
SELECT REGEXP_REPLACE ('test PI20099742', '[A-Z]{2}[0-9]{8}$', '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 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 $1.
I need to replace multiple words such as (dog|cat|bird) with nothing in a string where there may be multiple consecutive occurrences of a word. The actual code is to remove salutations and suffixes from a name. Unfortunately the garbage data I get sometimes contains "SNERD JR JR."
I was able to create a regular expression pattern that accomplishes my goal but only for the first occurrence. I implemented a stupid hack to get rid of the second occurrence, but I believe there has to be a better way. I just can't figure it out.
Here is my "hacked" code;
FUNCTION REMOVE_SALUTATIONS(IN_STRING VARCHAR2) RETURN VARCHAR2 DETERMINISTIC
AS
REGEX_SALUTATIONS VARCHAR2(4000) := '(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)';
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 )
...
please, I have in Oracle table this texts (as 2 records)
"Sample text with replace parameter %1%"
"You reached 90% of your limit"
I need replace %1% with specific text from input parameter in Oracle Function. In fact, I can have more than just one replace parameters. I have also record with "Replace this %12% with real value"
This functionality I have programmed:
IF poc > 0 THEN
FOR i in 1 .. poc LOOP
p := get_param(mString => mbody);
mbody := replace(mbody,
'%' || p || '%', parameters(to_number(p, '99')));
END LOOP;
END IF;
But in this case I have problem with text number 2. This functionality trying replace "90%" also and I then I get this error:
ORA-06502: PL/SQL: numeric or value error: NULL index table key value
It's a possible to avoid try replace "90%"? Many thanks for advice.
Best regards
PS: Oracle version: 10g (OCI Version: 10.2)
Regular expressions can work here. Try the following and build them into your script.
SELECT REGEXP_REPLACE( 'Sample text with replace parameter %1%',
'\%[0-9]+\%',
'db_size' )
FROM DUAL
and
SELECT REGEXP_REPLACE( 'Sample text with replace parameter 1%',
'\%[0-9]+\%',
'db_size' )
FROM DUAL
The pattern is pretty simple; look for patterns where a '%' is followed by 1 or more numbers followed by a '%'.
The only issue here will be if you have more than one replacement to make in each string and each replacement is different. In that case you will need to loop round the string each time replacing the next parameter. To do this add the position and occurrence parameters to REGEXP_REPLACE after the replacement string, e.g.
REGEXP_REPLACE( 'Sample text with replace parameter %88888888888%','\%[0-9]+\%','db_size',0,1 )
You are getting the error because at parameters(to_number(p, '99')). Can you please check the value of p?
Also, if the p=90 then then REPLACE will not try to replace "90%". It will replace "%90%". How have you been sure that it's trying to replace "90%"?
I am trying to extract a substring from a text column using a regular expression, but in some cases, there are multiple instances of that substring in the string.
In those cases, I am finding that the query does not return the first occurrence of the substring. Does anyone know what I am doing wrong?
For example:
If I have this data:
create table data1
(full_text text, name text);
insert into data1 (full_text)
values ('I 56, donkey, moon, I 92')
I am using
UPDATE data1
SET name = substring(full_text from '%#"I ([0-9]{1,3})#"%' for '#')
and I want to get 'I 56' not 'I 92'
You can use regexp_matches() instead:
update data1
set full_text = (regexp_matches(full_text, 'I [0-9]{1,3}'))[1];
As no additional flag is passed, regexp_matches() only returns the first match - but it returns an array so you need to pick the first (and only) element from the result (that's the [1] part)
It is probably a good idea to limit the update to only rows that would match the regex in the first place:
update data1
set full_text = (regexp_matches(full_text, 'I [0-9]{1,3}'))[1]
where full_text ~ 'I [0-9]{1,3}'
Try the following expression. It will return the first occurrence:
SUBSTRING(full_text, 'I [0-9]{1,3}')
You can use regexp_match() In PostgreSQL 10+
select regexp_match('I 56, donkey, moon, I 92', 'I [0-9]{1,3}');
Quote from documentation:
In most cases regexp_matches() should be used with the g flag, since
if you only want the first match, it's easier and more efficient to
use regexp_match(). However, regexp_match() only exists in PostgreSQL
version 10 and up. When working in older versions, a common trick is
to place a regexp_matches() call in a sub-select...