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%'
Related
Requirements: Before, I used instr() in Oracle to achieve the requirements, but now I want to use MySQL to achieve the same effect, and try to use the functions in MySQL to achieve it.
INSTR(A.SOME_THING.B,".",1,2)<>0 --ORACLE
As far as I can tell, that's not that difficult for simple cases. But, as number of parameters raises, MySQL "replacement" for the same Oracle functionality gets worse.
As your code:
instr(some_thing, '.', 1, 2)
means
search through some_thing
for a dot
starting from the first position
and find dot's second occurrence
you can't do that in a simple manner using MySQL, as you'll need a user-defined function. Something like this (source is INSTR Function - Oracle to MySQL Migration; I suggest you have a look at the whole document. I'm posting code here because links might get broken):
DELIMITER //
CREATE FUNCTION INSTR4 (p_str VARCHAR(8000), p_substr VARCHAR(255),
p_start INT, p_occurrence INT)
RETURNS INT
DETERMINISTIC
BEGIN
DECLARE v_found INT DEFAULT p_occurrence;
DECLARE v_pos INT DEFAULT p_start;
lbl:
WHILE 1=1
DO
-- Find the next occurrence
SET v_pos = LOCATE(p_substr, p_str, v_pos);
-- Nothing found
IF v_pos IS NULL OR v_pos = 0 THEN
RETURN v_pos;
END IF;
-- The required occurrence found
IF v_found = 1 THEN
LEAVE lbl;
END IF;
-- Prepare to find another one occurrence
SET v_found = v_found - 1;
SET v_pos = v_pos + 1;
END WHILE;
RETURN v_pos;
END;
//
DELIMITER ;
Use it as
SELECT INSTR4('abcbcb', 'b', 3, 2);
and get 6 as a result.
In OracleDB the code
INSTR(column, ".", 1, 2) <> 0 --ORACLE
checks does the column contains at least 2 point chars in the value.
In MySQL this can be replaced with, for example,
LENGTH(column) - LENGTH(REPLACE(column, '.', '')) >= 2
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;
I have a field that contains version information such as:
V12.0
V1.0
BE0.50
VV24
I want to query for version greater then n. The number I want to do the comparison on is preceded by a non-fixed number of characters. Is it possible to do something like:
SELECT version FROM table WHERE int_part(version) > 10
V12.0
VV24
In this case it seems that version number appears in end only, alphabets are in beginning. So, probable solution will be to reverse the string, type cast it to get numbers, then reverse back the number obtained.
Try following solution
SELECT version
FROM table
HAVING CAST(REVERSE(IF(LOCATE(".", version), (CAST(REVERSE(version) AS DECIMAL(4,2))), (CAST(REVERSE(version) AS UNSIGNED INTEGER)))) AS DECIMAL(4,2)) > 12;
Hope it helps...
Seem like you have to create you own function to do this:
CREATE FUNCTION IsNumeric (val varchar(255)) RETURNS tinyint
RETURN val REGEXP '^(-|\\+){0,1}([0-9]+\\.[0-9]*|[0-9]*\\.[0-9]+|[0-9]+)$';
CREATE FUNCTION NumericOnly (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;
select NumericOnly('vv24');
+---------------------+
| NumericOnly('vv24') |
+---------------------+
| 24 |
+---------------------+
Big downside of MySQL not supporting Regular Expressions in REPLACE method. If I have to do any matching of sorts, I usually pull all data to my PHP and process it there. In long run it is better solution. Of course you could always use User Defined Functions if you prefer.
http://dev.mysql.com/doc/refman/5.0/en/adding-udf.html
#shubhansh
Would 24 not change to 0.24 and failed > 12 ?
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;
i have a table with address as column.
values for address is "#12-3/98 avenue street", which has numbers, special characters and alphabets.
i want to write my sql query usng regex to remove special characters from the address value
ex: "12398avenuestreet" will be the value i want after removing the special characters
thank you.
maybe this function help you
CREATE FUNCTION strip_non_alpha(
_dirty_string varchar(40)
)
RETURNS varchar(40)
BEGIN
DECLARE _length int;
DECLARE _position int;
DECLARE _current_char varchar(1);
DECLARE _clean_string varchar(40);
SET _clean_string = '';
SET _length = LENGTH(_dirty_string);
SET _position = 1;
WHILE _position <= _length DO
SET _current_char = SUBSTRING(_dirty_string, _position, 1);
IF _current_char REGEXP '[A-Za-z0-9]' THEN
SET _clean_string = CONCAT(_clean_string, _current_char);
END IF;
SET _position = _position + 1;
END WHILE;
RETURN CONCAT('', _clean_string);
END;
so you need to call this like
update mytable set address = strip_non_alpha(address);
You don't need RegExp for simple character replacement.
MySQL string functions
Unfortunately, MySQL regular expressions are "match only", you cannot do a replace in your query. This leaves you with only something like this (witch is very-very stupid):
SELECT REPLACE(REPLACE(address, '?', ''), '#', '') -- and many many other nested replaces
FROM table
Or put this logic inside your application (the best option here).
MySQL regular expressions is only for pattern matching and not replacing, so your best bet is to create a function or a repetative use of Replace().
As far as I know, it is not possible to replace via MySQL regex, since these functions are only used for matching.
Alternatively, you can use MySQL Replace for this:
SELECT REPLACE(REPLACE(REPLACE(REPLACE(address, '#', ''), '-', ''), '/', ''), ' ', '') FROM table;
Which will remove #, -, / and spaces and result in the string you want.
You may use this MySQL UDF. And then simply,
update my_table set my_column = PREG_REPLACE('/[^A-Za-z0-9]/' , '' , my_column);