I have an Ordernumber that looks like this: 1001-00001.2 but I want it without the extra .2
I tried to use the following in my code:
{$sArticle.ordernumber|regex_replace:"/'.'/\d":" "}
it didn't work because I don't know how I can use the dot.
Dot matches any single character. Try escaping it:
{$sArticle.ordernumber|regex_replace:"/\.\d+$/":""}
There's actually a few changes, here.
Everything was moved to be within the /.../ that mark a regex
The dot was escaped with \.
A quantifier (+) was added to \d so it matches one or more digits.
An anchor ($) is used to make sure it doesn't match anywhere but the end of the string.
The replacement is now an empty string
Related
I am trying to match a word with regex. for example, I want to match only first 2 folders in below string
/folder1/folder2/filder3/folder4/folder5
I wrote a below regex to match first two folders but it matches everything till /folder5 but I wanted to match only till /folder2
/(\w.+){2}
I guess .+ matches everything. Any idea how to handle this?
You can use
^/[^/]+/[^/]+
^(?:/[^/]+){2}
Or, if you need to escape slashes:
^\/[^\/]+\/[^\/]+
^(?:\/[^\/]+){2}
See the regex demo. [^/] is a negated character class that matches any char other than a / char.
I would like to match a String with a look-ahead using the following regex: /A20.(?!4)/. This string should match:
A20.1
A20.2
A20.3
A20.41
A20.42
A20.400
...
The only A20* string that should not match is
A20.4
It works fine, expect A20.41 or A20.42.. How can I terminate the regex?
I have tried /A20.(?!4)$/, but it did not work.
You could use negated character class such as [^4], which would mean "match everything except four". But I think you still want to match only digits, so I'd simply use character class [123567890] for that (note that 4 is excluded).
So pattern would be:
A20\.[123567890]
Also, you use . (dot) to match the dot, but dot is special regex character, so you need to escape it to treat it literally: \.
you have to look ahead more. If there is another digit behind the 4 then match it, so:
A20.((?!4)|(?=\d\d))
I have a comma separated string which I want to validate using a regex. What I have written is gives me a match if there a part wrong later in the string. I want to discard it completely if any part is wrong.
My regex : ^(?:[\w\.]+,{1}(?:STR|INT|REAL){1},{1}(\s*|$))+
Positive Case : Component,STR,YoungGenUse,STR,YoungGenMax,STR,OldGenUse,INT,OldGenMax,INT,PermGenUse,INT,PermGenMax,INT,MajCollCnt,INT,MinCollDur,REAL,MinCollCnt,INT,
Negative Case :
Component,STR,YoungGenUse,STR,YoungGenMax,TEST,OldGenUse,INT,OldGenMax,INT,PermGenUse,INT,PermGenMax,INT,MajCollCnt,INT,MinCollDur,REAL,MinCollCnt,INT,
For the second case, my regex gives a match for the bold portion eventhough, later there is an incorrect part (TEST). How can I modify my regex to discard the entire string?
The pattern that you tried would not match TEST in YoungGenMax,TEST because the alternatives STR|INT|REAL do not match it.
It would show until the last successful match in the repetition which would be Component,STR,YoungGenUse,STR,
You have to add the anchor at the end, outside of the repetition of the group, to indicate that the whole pattern should be followed by asserting the end of the string.
There are no spaces or dots in your string, so you might leave out \s* and use \w+ without the dot in the character class. Note that \s could also possibly match a newline.
^(?:\w+,(?:STR|INT|REAL),)+$
Regex demo
If you want to keep matching optional whitespace chars and the dot:
^(?:[\w.]+,(?:STR|INT|REAL),\s*)+$
Regex demo
Note that by repeating the group with the comma at the end, the string should always end with a comma. You can omit {1} from the pattern as it is superfluous.
your regex must keep matching until end of the string, so you must use $ to indicate end of the line:
^(?:[\w.]+,{1}(?:STR|INT|REAL){1},{1}(\s*|$))+$
Regex Demo
I'm trying to create a simple Grammar correction tool.
I want to create a regular expression that finds fullstops (" . ") that are not followed by a space so I can replace that with a fullstop and space.
For e.g. This is a sentence.This is another sentence.
Only the first fullstop in the above example should be matched in the expression.
I've tried /\.[^\s]/g but it returns an additional character after the matched fullstop. I would like to match only the fullstop.
How can I do this?
The negated character class [^\s] in the pattern expects a match (any character except a whitespace character), that is why you have the additional character.
If you want to match the dot only, you could use a negative lookahead to assert what is on the right is not a whitspace char or the end of the string:
\.(?!\s|$)
Regex demo
To not match a dot that is not followed by a whitespace char excluding a newline:
\.(?![^\S\r\n])
Regex demo
You can look for all dots using:
(\.)
This will match all dots on below examples:
This is a sentence.This is another sentence.
i am looking. for dots. . ...
You can add a |$ to seek for end of line, and with a little tweak, you get a regex that match all dots not followed by whitespace nor being on the end of a line:
(\.(?!\ |$))
Note that there's a whitespace as literal here. The "must-work-everywhere" example will be like:
(\.(?![[:space:]]|$))
If not, search on the regex reference on the language you use.
My use case is as follows: I would like to find all occurrences of something similar to this /name.action, but where the last part is not .action eg:
name.actoin - should match
name.action - should not match
nameaction - should not match
I have this:
/\w+.\w*
to match two words separated by a dot, but I don't know how to add 'and do not match .action'.
Firstly, you need to escape your . character as that's taken as any character in Regex.
Secondly, you need to add in a Match if suffix is not present group - signified by the (?!) syntax.
You may also want to put a circumflex ^ to signify the start of a new line and change your * (any repetitions) to a + (one or more repititions).
^/\w+\.(?!action)\w+ is the finished Regex.
^\w+\.(?!action)\w*
You need to escape the dot character.
\w+\.(?!action).*
Note the trailing .* Not sure what you want to do after the action text.
See also Regular expression to match string not containing a word?
You'll need to use a zero-width negative lookahead assertion. This will let you look ahead in the string, and match based on the negation of a word.
So the regex you'd need (including the escaped . character) would look something like:
/name\.(?!action)/