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I have a requirement in MySQL to take a comma separated string of numbers ("2000, 2001, 2002, 2003, 2004") passed into a stored procedure and compare each one to a another number e.g. 2005. If any of the numbers are greater than or equal to my target 2005, I need to execute code, else leave the loop. I cannot figure out how to iterate over this comma delimited string, turn each into a number using CAST(), and compare it to my target number.
Assuming the argument passed into the procedure is named arg_list ...
If the requirement is to determine if just some (any number other than zero) entries in the list is greater than 2005,
Then we can do something like this in the procedure:
a loop iterator and some work areas:
DECLARE i_ INT DEFAULT 0;
DECLARE ls_number VARCHAR(255) DEFAULT '';
DECLARE ls_greatest VARCHAR(255) DEFAULT '';
initialize, get first number in list, first number in list is greatest so far:
SET i_ := 1;
SET ls_number := TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(CONCAT( arg_list ,','),',',i),',',-1));
SET ls_greatest := ls_number;
WHILE ls_number > '' DO
-- evaluate as numeric and compare
IF ls_number+0 > ls_greatest+0 THEN
-- the one we just got is the biggest one so far, so keep it
SET ls_greatest := ls_number;
END IF;
-- get next number in list
SET i_ := i_ + 1;
SET ls_number := TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(CONCAT( arg_list ,','),',',i),',',-1));
END WHILE;
-- ls_greatest now contains the largest number from the list
IF ls_greatest+0 >= 2005 THEN
-- do some code
END IF;
Note: this assumes that the list in arg_list is well formed, and doesn't contain spurious commas, e.g. if the list was '1,2,3,,4,5' the loop would exit after processing the 3
Defined the procedure as follows.
DELIMITER //
CREATE PROCEDURE getValue(txt varchar(100), abc varchar(100))
BEGIN
SET #i = 1;
SET #txt = REPLACE(txt, '"', '');
SET #last_element = SUBSTRING_INDEX(#txt, ",", -1);
WHILE #txt != '' DO
SET #element = SUBSTRING_INDEX(#txt, ",", #i);
IF CAST(abc AS UNSIGNED) >= CAST(#element AS UNSIGNED) THEN
# execute your code
SET #txt = '';
ELSE
IF !LOCATE(',',#txt) && #element = #last_element THEN
# #element = #last_element comparison is not necessary
SET #txt = '';
ELSE
SET #txt = REPLACE(#txt, CONCAT(#element,","), '');
END IF;
END IF;
END WHILE;
END //
DELIMITER ;
Tried the following two procedure calls.
call getValue("200,400,100","100");
call getValue("200,400,600","100");
Im trying to remove newline characters when the next character is also a newline
using the Function I wrote below, and im running into a problem. When I input this string :
"Line1
Line2
Line3
Line4"
It removes the newlines after line 1 and 2 , but the lines between 3 and 4 remain ? Any ideas as to why the function doesn't work then ?
DELIMITER $$
DROP FUNCTION IF EXISTS removeLines;
CREATE FUNCTION test_dev.removeLines(address varchar(255))
RETURNS varchar(255)
DETERMINISTIC
BEGIN
DECLARE i int;
SET i = 0;
WHILE i < length(address)-1 DO
IF(((SUBSTR(address,i,1) in (CHAR(10) ,CHAR(13))) AND (SUBSTR(address,i+1,1) in (CHAR(10) ,CHAR(13))) ) )
THEN set address = INSERT(address,i,1,'');
SET i= i+1;
ELSE set i= i+1;
END IF;
END WHILE;
RETURN address;
END$$
DELIMITER ;
An alternate way to write your function so you don't have to manually keep track of character indexing.
while instr(address, '\r\n\r\n') do
address = replace(address, '\r\n\r\n', '\r\n');
end while;
I didn't have a chance to test this in a database so the syntax may not be 100% correct.
The problem is that you increment i after you replace a newline. Consider the following simple string:
1\n\n\n2
01 2 3 4 -- indexes
When i is 1, you see that there are newlines at indexes i=1 and i+1=2, so you remove the first one with INSERT(). Now the string is:
1\n\n2
01 2 3
and you do SET i = i + 1. Now i is 2, but indexes i=2 and i+1=3 don't have newlines, so you don't remove the next newline.
Change it so that you only increment i when you don't find a pair of newlines.
IF(((SUBSTR(address,i,1) in (CHAR(10) ,CHAR(13))) AND (SUBSTR(address,i+1,1) in (CHAR(10) ,CHAR(13))) ) )
THEN set address = INSERT(address,i,1,'');
ELSE set i= i+1;
END IF;
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;
Based on the documentation:
User variables can be assigned a value from a limited set of data types: integer, decimal, floating-point, binary or nonbinary string, or NULL value.
As you see, user variables aren't supported BIT datatype. Right? Ok, I have this code:
BEGIN
SELECT active INTO #active FROM users WHERE id = new.user_id;
IF (IFNULL((#active & b'100' > 0), 0) < 1) THEN
SIGNAL SQLSTATE '45000' SET MESSAGE_TEXT = "You cannot vote";
END IF;
END
Note: active column is BIT(10).
You know, my code surprisingly works as well .. Why? I mean how MySQL treats with #active variable as a BIT datatype?
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;