I am attempting to configure a regular expression that matches anything numerical value but 1211 so it will still match variations such as 1212 121 1122, 3411, etc
I am unable to test the below at http://regexpal.com/ as it does not seem to support ?
(1(?!2)|1(?!2)|(?<!1)2|(?<!2)1|[^1211])+|[0-9]{1,4})
Am I doing it right and also where can I test it?
EDIT
Please note that I need to implement in a rewrite module/filter.
You can simplify that regex a lot:
^(?!1211)[1-9]\d{0,3}$
As for regexpal, it's not working because your regex is invalid. You can tell right away because it as one more closing parenthesis than opening.
I suggest you take a simple route:
if (data != "1211"){
// your other regex here
}
you can test it using javascript, if you are familiar with it.
Related
Im trying to build a regular expression to capture URLs which contain a certain parameter 7136D38A-AA70-434E-A705-0F5C6D072A3B
Ive set up a simple regex to capture a URL with anything before and anything after this parameter (just just all URLs which contain this parameter). Ive tested this on an online checker: http://scriptular.com/ and seems to work fine. However google analytics is saying this is invalid when i try to use it. Any idea what is causing this?
Url will be in the format
/home/index?x=23908123890123&y=kjdfhjhsfd&z=7136D38A-AA70-434E-A705-0F5C6D072A3B&p=kljdaslkjasd
so i just want to capture URLs that contain that specific "z" parameter.
regex
^.+(?=7136D38A-AA70-434E-A705-0F5C6D072A3B).+$
You just need
^.+=7136D38A-AA70-434E-A705-0F5C6D072A3B.+$
Or (a bit safer):
^.+=7136D38A-AA70-434E-A705-0F5C6D072A3B($|&.+$)
And I think you can even use
=7136D38A-AA70-434E-A705-0F5C6D072A3B($|&)
See demo
Your regex is invalid because GA regex flavor does not support look-arounds (and you have a (?=...) positive look-ahead in yours).
Here is a good GA regex cheatsheet.
To match /home/index?x=23908123890123&y=kjdfhjhsfd&z=7136D38A-AA70-434E-A705-0F5C6D072A3B&p=kljdaslkjasd you can use:
\S*7136D38A-AA70-434E-A705-0F5C6D072A3B\S*
In Eclipse I want to find a string, and using the normal search results in hundreds of irrelevant results. So I'm trying to use regular expressions, but they don't give me the proper results up til now.
This is what I need: find "abc.com", but not "xyz.abc.com" or "#abc.com". To make it clear, it should return www.abc.com.
I've tried the following regex but I'm not sure if this is how it should be:
[^#xyz\.]abc.com
Using a negative lookbehind should suit your needs:
(?<!xyz[.]|#)abc[.]com
Every "abc.com" that is not preceded by "xyz." nor by "#".
Im trying to write a regex expression to match anything (.*)/feed/ with the exception of (.*)/author/feed/
Currently, I have (.*)/feed/(.*) which works well to identify any string /feed/ to redirect. However, I dont want to exlude those that have /author/(.*)/feed/
For example - match http://www.site.com/ANYTHING/feed/ but exclude site.com/author/ANYTHING/feed/
I should clarify that I'm not terribly familiar with regex expressions but this is actually for use within the Redirection plugin for wordpress which states "Full regular expression support."
Any help would be greatly appreciated. Thank you in advance
Depending on the language, you may be able to use a negative look-behind assertion:
(.*)(?<!/author)/feed
The assertion, (?<!/author), ensures that /author does not match behind the text /feed, but does not count it as being matched.
I have a string that looks like:
this is a string [[and]] it is [[awesome|amazing]]
I have the following regular expression so far:
(?<mygroup>(?<=\[\[).+?(?=\]\]))
I am basically trying to capture everything inside the brackets. However, I need to add another condition that says: If the matched result contains a pipe delimiter then only return the word to the right of the pipe delimiter. If there is no pipe then just return everything inside the brackets.
The parsing result I am looking for given the example above should look like:
and
amazing
Any input is appreciated.
(?<mygroup>(?<=\[\[)([^|\]]*|)?([^|]+?)(?=\]\]))
You could use this regex:
(?<=\[\[[^\]]*?)(?!\w+\|)\w+(?=\]\])
it matches both and and amazing words in your test example. You could check it out, I created a test app on Ideone.
From the regex info page:
The tremendous power and expressivity
of modern regular expressions can
seduce the gullible — or the foolhardy
— into trying to use regexes on every
string‐related task they come across.
My advice: Just grab what is between the brackets and parse it after.
Regular expressions are not the answer to everything. May those who follow after you be spared from deciphering the regex you come up with.
I'm having a bit of trouble with my regex and was wondering if anyone could please shed some light on what to do.
Basically, I have this Regex:
\[(link='\d+') (type='\w+')](.*|)\[/link]
For example, when I pass it the string:
[link='8' type='gig']Blur[/link] are playing [link='19' type='venue']Hyde Park[/link]"
It only returns a single match from the opening [link] tag to the last [/link] tag.
I'm just wondering if anyone could please help me with what to put in my (.*|) section to only select one [link][/link] section at a time.
Thanks!
You need to make the wildcard selection ungreedy with the "?" operator. I make it:
/\[(link='\d+')\s+(type='\w+')\](.*?)\[\/link\]/
of course this all falls down for any kind of nesting, in which case the language is no longer regular and regexs aren't suitable - find a parser
Regular Expressions Info a is a fantastic site. This page gives an example of dealing with html tags. There's also an Eclipse plugin that lets you develop expressions and see the matching in realtime.
You need to make the .* in the middle of your regex non-greedy. Look up the syntax and/or flag for non-greedy mode in your flavor of regular expressions.