Parsing Integers out of a string in SQL - mysql

I have a fairly large database with with a column that has strings that are for the most part really just ints, e.g. "1234" or "345". However some of them have strings prepended to them (of varying length), so e.g. "a123" or "abc123".
Is there a smart way to create a new column with just the integer values? Thus, "abc123" would become "123"? I know I can read all of the rows in PHP and then use a regex to do it pretty easily but I wanted to see if there was a way to let SQL do this for me.

Unfortunately you can't do this in MySQL just by itself. MySQL has regex matching capabilities, but no regex replacement capabilities. Your best option would be to use a regex in PHP to perform your replacement. (Source)
$sql = "SELECT `id`, `myMixedColumn` FROM `myTable` "
. "WHERE `myMixedColumn` NOT RLIKE '^[[:digit:]]+$'";
$r = mysql_query($sql);
$updates = array();
while ($row = mysql_fetch_assoc($r)) {
$updates = sprintf("UPDATE `myTable` SET `myIntField` = %s WHERE `id` = %d",
preg_replace("#\\D#", "", $row['myMixedColumn']),
$row['id']
);
}

If you really want to do it in SQL, just for the sake of not doing in PHP, you could make a small function that does it until MySQL implements a regex replace, but I wouldn't bet anything on the performance.
Something like that would only work if the letters were at the beginning and that there would be no other characters than [a-zA-Z]. And you'd have to check how it runs in different charsets.
CREATE FUNCTION last_letter(s VARCHAR(100)) RETURNS INT
BEGIN
DECLARE last, current INT default 0;
DECLARE letter_a INT;
DECLARE letter_z INT;
DECLARE letter_iter INT;
SELECT ord('a') INTO letter_a;
SELECT ord('z') INTO letter_z;
SET letter_iter = letter_a;
# Will loop for all letters a to z
WHILE letter_iter <= letter_z DO
# Will get the last case-insensitive occurrence of a letter
SELECT LOCATE(CHAR(letter_iter), REVERSE(LOWER(s))) INTO current;
IF current > 0 THEN
SELECT LENGTH(s) - current + 1 INTO current;
END IF;
# Was that the rightmost letter?
IF current > last THEN
SET last = current;
END IF;
SET letter_iter = letter_iter + 1;
END WHILE;
# Return the max we found
RETURN last;
END; //
And then to get the integer values:
UPDATE test_table SET int_result =
CAST(SUBSTR(str_value, last_letter(str_value) + 1) AS SIGNED);

Related

mysql SELECT part of string with Regex (find and extract number) [duplicate]

I have a MySQL database and I have a query as:
SELECT `id`, `originaltext` FROM `source` WHERE `originaltext` regexp '[0-9][0-9]'
This detects all originaltexts which have numbers with 2 digits in it.
I need MySQL to return those numbers as a field, so i can manipulate them further.
Ideally, if I can add additional criteria that is should be > 20 would be great, but i can do that separately as well.
If you want more regular expression power in your database, you can consider using LIB_MYSQLUDF_PREG. This is an open source library of MySQL user functions that imports the PCRE library. LIB_MYSQLUDF_PREG is delivered in source code form only. To use it, you'll need to be able to compile it and install it into your MySQL server. Installing this library does not change MySQL's built-in regex support in any way. It merely makes the following additional functions available:
PREG_CAPTURE extracts a regex match from a string. PREG_POSITION returns the position at which a regular expression matches a string. PREG_REPLACE performs a search-and-replace on a string. PREG_RLIKE tests whether a regex matches a string.
All these functions take a regular expression as their first parameter. This regular expression must be formatted like a Perl regular expression operator. E.g. to test if regex matches the subject case insensitively, you'd use the MySQL code PREG_RLIKE('/regex/i', subject). This is similar to PHP's preg functions, which also require the extra // delimiters for regular expressions inside the PHP string.
If you want something more simpler, you could alter this function to suit better your needs.
CREATE FUNCTION REGEXP_EXTRACT(string TEXT, exp TEXT)
-- Extract the first longest string that matches the regular expression
-- If the string is 'ABCD', check all strings and see what matches: 'ABCD', 'ABC', 'AB', 'A', 'BCD', 'BC', 'B', 'CD', 'C', 'D'
-- It's not smart enough to handle things like (A)|(BCD) correctly in that it will return the whole string, not just the matching token.
RETURNS TEXT
DETERMINISTIC
BEGIN
DECLARE s INT DEFAULT 1;
DECLARE e INT;
DECLARE adjustStart TINYINT DEFAULT 1;
DECLARE adjustEnd TINYINT DEFAULT 1;
-- Because REGEXP matches anywhere in the string, and we only want the part that matches, adjust the expression to add '^' and '$'
-- Of course, if those are already there, don't add them, but change the method of extraction accordingly.
IF LEFT(exp, 1) = '^' THEN
SET adjustStart = 0;
ELSE
SET exp = CONCAT('^', exp);
END IF;
IF RIGHT(exp, 1) = '$' THEN
SET adjustEnd = 0;
ELSE
SET exp = CONCAT(exp, '$');
END IF;
-- Loop through the string, moving the end pointer back towards the start pointer, then advance the start pointer and repeat
-- Bail out of the loops early if the original expression started with '^' or ended with '$', since that means the pointers can't move
WHILE (s <= LENGTH(string)) DO
SET e = LENGTH(string);
WHILE (e >= s) DO
IF SUBSTRING(string, s, e) REGEXP exp THEN
RETURN SUBSTRING(string, s, e);
END IF;
IF adjustEnd THEN
SET e = e - 1;
ELSE
SET e = s - 1; -- ugh, such a hack to end it early
END IF;
END WHILE;
IF adjustStart THEN
SET s = s + 1;
ELSE
SET s = LENGTH(string) + 1; -- ugh, such a hack to end it early
END IF;
END WHILE;
RETURN NULL;
END
There isn't any syntax in MySQL for extracting text using regular expressions. You can use the REGEXP to identify the rows containing two consecutive digits, but to extract them you have to use the ordinary string manipulation functions which is very difficult in this case.
Alternatives:
Select the entire value from the database then use a regular expression on the client.
Use a different database that has better support for the SQL standard (may not be an option, I know). Then you can use this: SUBSTRING(originaltext from '%#[0-9]{2}#%' for '#').
I think the cleaner way is using REGEXP_SUBSTR():
This extracts exactly two any digits:
SELECT REGEXP_SUBSTR(`originalText`,'[0-9]{2}') AS `twoDigits` FROM `source`;
This extracts exactly two digits, but from 20-99 (example: 1112 return null; 1521 returns 52):
SELECT REGEXP_SUBSTR(`originalText`,'[2-9][0-9]') AS `twoDigits` FROM `source`;
I test both in v8.0 and they work. That's all, good luck!
I'm having the same issue, and this is the solution I found (but it won't work in all cases) :
use LOCATE() to find the beginning and the end of the string you wan't to match
use MID() to extract the substring in between...
keep the regexp to match only the rows where you are sure to find a match.
I used my code as a Stored Procedure (Function), shall work to extract any number built from digits in a single block. This is a part of my wider library.
DELIMITER $$
-- 2013.04 michal#glebowski.pl
-- FindNumberInText("ab 234 95 cd", TRUE) => 234
-- FindNumberInText("ab 234 95 cd", FALSE) => 95
DROP FUNCTION IF EXISTS FindNumberInText$$
CREATE FUNCTION FindNumberInText(_input VARCHAR(64), _fromLeft BOOLEAN) RETURNS VARCHAR(32)
BEGIN
DECLARE _r VARCHAR(32) DEFAULT '';
DECLARE _i INTEGER DEFAULT 1;
DECLARE _start INTEGER DEFAULT 0;
DECLARE _IsCharNumeric BOOLEAN;
IF NOT _fromLeft THEN SET _input = REVERSE(_input); END IF;
_loop: REPEAT
SET _IsCharNumeric = LOCATE(MID(_input, _i, 1), "0123456789") > 0;
IF _IsCharNumeric THEN
IF _start = 0 THEN SET _start = _i; END IF;
ELSE
IF _start > 0 THEN LEAVE _loop; END IF;
END IF;
SET _i = _i + 1;
UNTIL _i > length(_input) END REPEAT;
IF _start > 0 THEN
SET _r = MID(_input, _start, _i - _start);
IF NOT _fromLeft THEN SET _r = REVERSE(_r); END IF;
END IF;
RETURN _r;
END$$
If you want to return a part of a string :
SELECT id , substring(columnName,(locate('partOfString',columnName)),10) from tableName;
Locate() will return the starting postion of the matching string which becomes starting position of Function Substring()
I know it's been quite a while since this question was asked but came across it and thought it would be a good challenge for my custom regex replacer - see this blog post.
...And the good news is it can, although it needs to be called quite a few times. See this online rextester demo, which shows the workings that got to the SQL below.
SELECT reg_replace(
reg_replace(
reg_replace(
reg_replace(
reg_replace(
reg_replace(
reg_replace(txt,
'[^0-9]+',
',',
TRUE,
1, -- Min match length
0 -- No max match length
),
'([0-9]{3,}|,[0-9],)',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
'^[0-9],',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
',[0-9]$',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
',{2,}',
',',
TRUE,
1, -- Min match length
0 -- No max match length
),
'^,',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
',$',
'',
TRUE,
1, -- Min match length
0 -- No max match length
) AS `csv`
FROM tbl;

phpMyAdmin - mariaDB roman numerals function

can anybody help me with my sorting function - seriously I don't know how can I make it work as supposed to. :( Database is in MariaDB in Xampp. I use phpMyAdmin to execute the query.
DELIMITER $$
DROP FUNCTION IF EXISTS convRomanNumeral$$
CREATE FUNCTION convRomanNumeral (numeral CHAR(4))
RETURNS INT
BEGIN
DECLARE intnum INT;
CASE numeral
WHEN "I" THEN intnum = 1;
WHEN "II" THEN intnum = 2;
END CASE;
RETURN intnum;
END;
$$
SET #iteration = -1;
UPDATE `st0gk_docman_documents`
SET created_on = DATE('2016-06-14') + INTERVAL(#iteration := #iteration + 1) SECOND
WHERE `docman_category_id` = 141 ORDER BY convRomanNumeral(SUBSTRING(SUBSTRING_INDEX(title,'/',1),' ',-2) ASC, SUBSTRING_INDEX(title,'/',-2)+0 ASC;
So what I want to achieve is to sort documents by title. Example titles are:
Document Nr I/36/2006
Document Nr II/36/2006
Document Nr I/32/2006
Document Nr II/19/2006
After sorting them by first Roman number and then by second Arabic number I want to update the date. Code below for updating by only second Arabic number works properly:
SET #iteration = -1;
UPDATE `st0gk_docman_documents`
SET created_on = DATE('2016-06-14') + INTERVAL(#iteration := #iteration + 1) SECOND
WHERE `docman_category_id` = 141 ORDER BY SUBSTRING_INDEX(title,'/',-2)+0 ASC;
I would like to use CASE to return proper variable for Roman values. I know it's not perfect but I can't even make the CASE and FUNCTION work. What I am doing wrong? All suggestions are welcome.
The best way to do this is to add another column that has a sortable equivalent of that string. And use non-SQL code to do the parsing and building of that column before inserting into the table.
First mistake that I was making it was trying to execute the whole query at once... After taking the first lodge out of the way the debugging seemed way simpler. :D
So I created my case function to convert Roman numerals:
DELIMITER $$
DROP FUNCTION IF EXISTS convRomanNumeralSubFunction$$
CREATE FUNCTION convRomanNumeralSubFunction (numeral CHAR(1))
RETURNS INT
BEGIN
DECLARE intnum INT;
CASE numeral
WHEN "I" THEN SELECT 1 INTO intnum;
WHEN "X" THEN SELECT 10 INTO intnum;
WHEN "C" THEN SELECT 100 INTO intnum;
WHEN "M" THEN SELECT 1000 INTO intnum;
WHEN "V" THEN SELECT 5 INTO intnum;
WHEN "L" THEN SELECT 50 INTO intnum;
WHEN "D" THEN SELECT 500 INTO intnum;
END CASE;
RETURN intnum;
END;
$$
After that I declared the second function needed for conversion. I don't know if You can declare function inside function... and I didn't want to waste more time on this. For sure You can declare Function inside Procedure. Anyhow. WARNING: This function is not proof of BAD numerals like IIX. Numerals like that or will be badly counted. Also AXI will not count.
DELIMITER $$
DROP FUNCTION IF EXISTS convRomanNumeral$$
CREATE FUNCTION convRomanNumeral (numeral CHAR(10))
RETURNS INT
BEGIN
DECLARE currentintnum, previntnum, intnum, counter, numerallength INT;
SET numerallength = LENGTH(numeral);
SET counter = numerallength;
SET intnum = 0;
SET previntnum = 0;
WHILE counter > 0 DO
SET currentintnum = CAST(convRomanNumeralSubFunction(SUBSTRING(numeral,counter, 1)) as integer);
IF currentintnum < previntnum THEN
SET intnum = intnum - currentintnum;
ELSE
SET intnum = intnum + currentintnum;
END IF;
SET previntnum = currentintnum;
SET counter = counter - 1;
END WHILE;
RETURN intnum;
END;
$$
So that's it. Now You can convert all kind of Roman numerals and sort them up.
Use this to test the conversion:
SELECT convRomanNumeral("XIX");
This is example sorting code that I in the end used:
SET #iteration = -1;
UPDATE `st0gk_docman_documents`
SET created_on = DATE('2016-06-07') + INTERVAL(#iteration := #iteration + 1) SECOND
WHERE `docman_category_id` = 67 ORDER BY convRomanNumeralBreak(SUBSTRING_INDEX(SUBSTRING_INDEX(title,'/',1),' ',-1)) ASC, SUBSTRING_INDEX(title,'/',-2)+0 ASC;
Also one more thing - if You'll try to excecute this on mySQL then You have to fix this line:
SET currentintnum = CAST(convRomanNumeralSubFunction(SUBSTRING(numeral,counter, 1)) as integer);
into this:
SET currentintnum = CAST(convRomanNumeralSubFunction(SUBSTRING(numeral,counter, 1)) as SIGNED);
This code could be improved but as the #Rick James stated this should be done differently - not in as db update but in different table structure and sorting mechanism.

Generate a random string in MySQL

I'm trying to get a random string in phpmyadmin using a function.
I have the following code:
CREATE FUNCTION randomPassword()
RETURNS varchar(128)
BEGIN
SET #chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
SET #charLen = length(#chars);
SET #randomPassword = '';
WHILE length(#randomPassword) < 12
SET #randomPassword = concat(#randomPassword, substring(#chars,CEILING(RAND() * #charLen),1));
END WHILE;
RETURN #randomPassword ;
END;
Now I get the error:
1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 5
Does anyone know how I can fix this?
This is faster than concat + substring routine.
select substring(MD5(RAND()),1,20);
As I've tested inserting 1M random data, md5 routine consumes only 1/4 (even less) time of concat + substring routine;
The problem is a md5 string contains only 32 chars so if you need a longer one you'd have to manually generate more md5 strings and substring it yourself.
Try this more simple solution:
SELECT CONV(FLOOR(RAND() * 99999999999999), 10, 36)
SELECT SUBSTRING(REPLACE(REPLACE(REPLACE( TO_BASE64(MD5(RAND())), '=',''),'+',''),'/',''), 2, 40)
This solution to generate a fixed length random string that contains all lower- and upper-case chars and digits.
SELECT SUBSTRING(REPLACE(REPLACE(REPLACE( TO_BASE64(MD5(RAND())), '=',''),'+',''),'/',''), 2, FLOOR(10+RAND()*31))
If you need a random length string (from 10 to 40 symbols in this example)
It's solved by using the Delimiter, i don't know for sure how, but it works
Thanks
DELIMITER $$
CREATE FUNCTION randomPassword()
RETURNS varchar(128)
BEGIN
SET #chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
SET #charLen = length(#chars);
SET #randomPassword = '';
WHILE length(#randomPassword) < 12
DO
SET #randomPassword = concat(#randomPassword, substring(#chars,CEILING(RAND() * #charLen),1));
END WHILE;
RETURN #randomPassword ;
END$$
DELIMITER ;
CREATE FUNCTION randomPassword()
RETURNS varchar(128)
AS
BEGIN
declare #chars nvarchar(25);
declare #charlen int;
declare #randomPassword nvarchar(128);
SET #chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
SET #charLen = len(#chars);
SET #randomPassword = '';
WHILE(LEN(#randomPassword) < 12)
BEGIN
SET #randomPassword = concat(#randomPassword, substring(#chars,CEILING(RAND() * #charLen),1));
END
RETURN #randomPassword
END

MySQL selecting string with multi special characters

I'm having a problem selecting strings from database. The problem is if you have +(123)-4 56-7 in row and if you are searching with a string 1234567 it wouldn't find any results. Any suggestions?
You can use the REPLACE() method to remove special characters in mysql, don't know if it's very efficient though. But it should work.
There is already another thread in SO which covers a very similar question, see here.
If it is always this kind of pattern you're searching, and your table is rather large, I advice against REPLACE() or REGEX() - which ofc will do the job if tweaked properly.
Better add a column with the plain phone numbers, which doesn't contain any formatting character data at all - or even better, a hash of the phone numbers. This way, you could add an index to the new column and search against this. From a database perspective, this is much easier, and much faster.
You can use User Defined Function to get Numeric values from string.
CREATE FUNCTION GetNumeric (val varchar(255)) RETURNS tinyint
RETURN val REGEXP '^(-|\\+){0,1}([0-9]+\\.[0-9]*|[0-9]*\\.[0-9]+|[0-9]+)$';
CREATE FUNCTION GetNumeric (val VARCHAR(255))
RETURNS VARCHAR(255)
BEGIN
DECLARE idx INT DEFAULT 0;
IF ISNULL(val) THEN RETURN NULL; END IF;
IF LENGTH(val) = 0 THEN RETURN ""; END IF;
SET idx = LENGTH(val);
WHILE idx > 0 DO
IF IsNumeric(SUBSTRING(val,idx,1)) = 0 THEN
SET val = REPLACE(val,SUBSTRING(val,idx,1),"");
SET idx = LENGTH(val)+1;
END IF;
SET idx = idx - 1;
END WHILE;
RETURN val;
END;
Then
Select columns from table
where GetNumeric(phonenumber) like %1234567%;
Query using replace function as -
select * from phoneTable where replace(replace(replace(phone, '+', ''), '-', ''), ')', '(') LIKE '%123%'

How to extract two consecutive digits from a text field in MySQL?

I have a MySQL database and I have a query as:
SELECT `id`, `originaltext` FROM `source` WHERE `originaltext` regexp '[0-9][0-9]'
This detects all originaltexts which have numbers with 2 digits in it.
I need MySQL to return those numbers as a field, so i can manipulate them further.
Ideally, if I can add additional criteria that is should be > 20 would be great, but i can do that separately as well.
If you want more regular expression power in your database, you can consider using LIB_MYSQLUDF_PREG. This is an open source library of MySQL user functions that imports the PCRE library. LIB_MYSQLUDF_PREG is delivered in source code form only. To use it, you'll need to be able to compile it and install it into your MySQL server. Installing this library does not change MySQL's built-in regex support in any way. It merely makes the following additional functions available:
PREG_CAPTURE extracts a regex match from a string. PREG_POSITION returns the position at which a regular expression matches a string. PREG_REPLACE performs a search-and-replace on a string. PREG_RLIKE tests whether a regex matches a string.
All these functions take a regular expression as their first parameter. This regular expression must be formatted like a Perl regular expression operator. E.g. to test if regex matches the subject case insensitively, you'd use the MySQL code PREG_RLIKE('/regex/i', subject). This is similar to PHP's preg functions, which also require the extra // delimiters for regular expressions inside the PHP string.
If you want something more simpler, you could alter this function to suit better your needs.
CREATE FUNCTION REGEXP_EXTRACT(string TEXT, exp TEXT)
-- Extract the first longest string that matches the regular expression
-- If the string is 'ABCD', check all strings and see what matches: 'ABCD', 'ABC', 'AB', 'A', 'BCD', 'BC', 'B', 'CD', 'C', 'D'
-- It's not smart enough to handle things like (A)|(BCD) correctly in that it will return the whole string, not just the matching token.
RETURNS TEXT
DETERMINISTIC
BEGIN
DECLARE s INT DEFAULT 1;
DECLARE e INT;
DECLARE adjustStart TINYINT DEFAULT 1;
DECLARE adjustEnd TINYINT DEFAULT 1;
-- Because REGEXP matches anywhere in the string, and we only want the part that matches, adjust the expression to add '^' and '$'
-- Of course, if those are already there, don't add them, but change the method of extraction accordingly.
IF LEFT(exp, 1) = '^' THEN
SET adjustStart = 0;
ELSE
SET exp = CONCAT('^', exp);
END IF;
IF RIGHT(exp, 1) = '$' THEN
SET adjustEnd = 0;
ELSE
SET exp = CONCAT(exp, '$');
END IF;
-- Loop through the string, moving the end pointer back towards the start pointer, then advance the start pointer and repeat
-- Bail out of the loops early if the original expression started with '^' or ended with '$', since that means the pointers can't move
WHILE (s <= LENGTH(string)) DO
SET e = LENGTH(string);
WHILE (e >= s) DO
IF SUBSTRING(string, s, e) REGEXP exp THEN
RETURN SUBSTRING(string, s, e);
END IF;
IF adjustEnd THEN
SET e = e - 1;
ELSE
SET e = s - 1; -- ugh, such a hack to end it early
END IF;
END WHILE;
IF adjustStart THEN
SET s = s + 1;
ELSE
SET s = LENGTH(string) + 1; -- ugh, such a hack to end it early
END IF;
END WHILE;
RETURN NULL;
END
There isn't any syntax in MySQL for extracting text using regular expressions. You can use the REGEXP to identify the rows containing two consecutive digits, but to extract them you have to use the ordinary string manipulation functions which is very difficult in this case.
Alternatives:
Select the entire value from the database then use a regular expression on the client.
Use a different database that has better support for the SQL standard (may not be an option, I know). Then you can use this: SUBSTRING(originaltext from '%#[0-9]{2}#%' for '#').
I think the cleaner way is using REGEXP_SUBSTR():
This extracts exactly two any digits:
SELECT REGEXP_SUBSTR(`originalText`,'[0-9]{2}') AS `twoDigits` FROM `source`;
This extracts exactly two digits, but from 20-99 (example: 1112 return null; 1521 returns 52):
SELECT REGEXP_SUBSTR(`originalText`,'[2-9][0-9]') AS `twoDigits` FROM `source`;
I test both in v8.0 and they work. That's all, good luck!
I'm having the same issue, and this is the solution I found (but it won't work in all cases) :
use LOCATE() to find the beginning and the end of the string you wan't to match
use MID() to extract the substring in between...
keep the regexp to match only the rows where you are sure to find a match.
I used my code as a Stored Procedure (Function), shall work to extract any number built from digits in a single block. This is a part of my wider library.
DELIMITER $$
-- 2013.04 michal#glebowski.pl
-- FindNumberInText("ab 234 95 cd", TRUE) => 234
-- FindNumberInText("ab 234 95 cd", FALSE) => 95
DROP FUNCTION IF EXISTS FindNumberInText$$
CREATE FUNCTION FindNumberInText(_input VARCHAR(64), _fromLeft BOOLEAN) RETURNS VARCHAR(32)
BEGIN
DECLARE _r VARCHAR(32) DEFAULT '';
DECLARE _i INTEGER DEFAULT 1;
DECLARE _start INTEGER DEFAULT 0;
DECLARE _IsCharNumeric BOOLEAN;
IF NOT _fromLeft THEN SET _input = REVERSE(_input); END IF;
_loop: REPEAT
SET _IsCharNumeric = LOCATE(MID(_input, _i, 1), "0123456789") > 0;
IF _IsCharNumeric THEN
IF _start = 0 THEN SET _start = _i; END IF;
ELSE
IF _start > 0 THEN LEAVE _loop; END IF;
END IF;
SET _i = _i + 1;
UNTIL _i > length(_input) END REPEAT;
IF _start > 0 THEN
SET _r = MID(_input, _start, _i - _start);
IF NOT _fromLeft THEN SET _r = REVERSE(_r); END IF;
END IF;
RETURN _r;
END$$
If you want to return a part of a string :
SELECT id , substring(columnName,(locate('partOfString',columnName)),10) from tableName;
Locate() will return the starting postion of the matching string which becomes starting position of Function Substring()
I know it's been quite a while since this question was asked but came across it and thought it would be a good challenge for my custom regex replacer - see this blog post.
...And the good news is it can, although it needs to be called quite a few times. See this online rextester demo, which shows the workings that got to the SQL below.
SELECT reg_replace(
reg_replace(
reg_replace(
reg_replace(
reg_replace(
reg_replace(
reg_replace(txt,
'[^0-9]+',
',',
TRUE,
1, -- Min match length
0 -- No max match length
),
'([0-9]{3,}|,[0-9],)',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
'^[0-9],',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
',[0-9]$',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
',{2,}',
',',
TRUE,
1, -- Min match length
0 -- No max match length
),
'^,',
'',
TRUE,
1, -- Min match length
0 -- No max match length
),
',$',
'',
TRUE,
1, -- Min match length
0 -- No max match length
) AS `csv`
FROM tbl;