![]() ![]() Set = regex_replace( ‘’, ”, Surrounds all leters with a ‘.’ # Replaces all non-alphanumerics for empty: # String containing letters, numbers and non-alphanumeric: I did some modifications so that I can use sort of a “last found” $_ flag and “replace only in the beginning” ^ operator: Return replace(original,temp2,replacement) If (left(temp,length(temp)-i) REGEXP pattern)=0 then Set temp=right(original,length(original)-i+1) Įnd if #koniec ifa dla tego czy spelnia pattern If (right(original,length(original)-i) REGEXP pattern)=0 then It leaves you only the part of the string that math pattern.ĬREATE FUNCTION `regex_replace`(`original` vARCHAR(1000), `pattern` vARCHAR(1000), `replacement` vARCHAR(1000)) It simply cuts from the right of the original till it match the pattern I also think it only works for one occurrence of pattern in string but that’s fine by me. Maybe it isn’t fast but it works like a hell for me. I wrote my own function because all previous didn’t work for me. I hope that these examples have helped clarify the issue. …which is correct, and the same as PHP gives, but is not what we are after. SELECT regex_replace(‘’, ‘bar’, ‘foo1 foo2 foo3’) It works, as a character set specifying a single character, but does not do what we want: The following, which is how you use it in your examples, is a character set ( ). SELECT regex_replace(‘fo*’, ‘bar’, ‘foo foo foo’) Ĭlearly illustrates how the SQL function matches only a single character, but the regular expression should match the whole string. The SQL function is acting here like the regex for: ![]() SELECT regex_replace(‘+’, ‘bar’, ‘foo1 foo2 foo3’) Here are some more complex regexes, that illustrate the issue hopefully a little more clearly: SELECT regex_replace(‘/foo/’, ‘bar’, ‘foo1 foo2 foo3’) ![]() Just to show that the ‘/’ required by php is not relevant here: In PHP, which has regex replacements built in, it comes out correctly:Įcho preg_replace(‘/foo/’, ‘bar’, ‘foo1 foo2 foo3’) SELECT regex_replace(‘foo’, ‘bar’, ‘foo1 foo2 foo3’) The simplest regex, which should work, will not: In these examples, the goal is to replace the “foo”s in “foo1 foo2 foo3” with “bar”s, to get “bar1 bar2 bar3”. Here are some examples to explain the problem. Those two lines extract a single character, test that single character, and replace only that one character.Īs I understand it, (ans as my testing below shows) it is impossible, using this, to replace a string of more than one character. ![]() It replaces all instances of single letters that match a regular expression, which is a very different thing. _,') ĭespite your protestations, this function in its current form does not, replace a regular expression. Mysql> select regex_replace('','','2my test3_text-to. If you are using MySQL version 5.0.1 or higher, make sure you set the NO_BACKSLASH_ESCAPES mode ON, before you use the above function to replace any characters which are escaped with back slash “\”, ie: \A,\B,etc… See how to set the NO_BACKSLASH_ESCAPES mode here IF NOT ch REGEXP pattern THEN SET temp = CONCAT(temp,ch) ĮLSE SET temp = CONCAT(temp,replacement) Loop_label: LOOP IF i>CHAR_LENGTH(original) THEN LEAVE loop_label I gave up searching finally and wrote my own “regex_replace” MySQL function.īelow is the source code of my function DELIMITER $$ĬREATE FUNCTION `regex_replace`(pattern VARCHAR(1000),replacement VARCHAR(1000),original VARCHAR(1000))ĭETERMINISTIC BEGIN DECLARE temp VARCHAR(1000) There was one called UDB but that’s also you need to install a module and stuff like that. Recently I came across a requirement where I had to cleanse the data that I’m inserting in to MySQL database using regular expressions, so I started searching for a function to do this on MySQL, but unfortunately I couldn’t find any. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |