I have the following string: LLLTTTLTLLLTT. The number of L's and T's are dynamic.
I tried to use regex "L+T+". It's not working as the number of occurrences is not only once. What is the right regex pattern to match?
Or try this regexp:
([LT]+)
The ( ) are not necessary depending what you want to achieve
Unless there is a clear requirement that the string cannot be empty, it is [LT]*.
Try this:
(L+T+)+
Basically, you are matching the patter "L+T+" more than one time.
How about
(L?T?)+
I am assuming that the string can start either with a L or a T
Or as user1290772 has pointed out, the equivalant would be
[LT]+
Try this:
(L+T+)+
Which means a combination of Ls and then Ts one or more times.
Bu this promises that it starts with L.
If you want either, try:
[LT]+
Related
Ok, I know that it is a question often asked, but I did not manage to get what I wanted.
I am looking for a regular expression in order to find a pattern that does not contain a particular substring.
I want to find an url that does not contains the b parameter.
http://www.website.com/a=789&c=146 > MATCH
http://www.website.com/a=789&b=412&c=146 > NOT MATCH
Currently, I have the following Regex:
\bhttp:\/\/www\.website\.com\/((?!b=[0-9]+).)*\b
But I am wrong with the \b, the regex match the beginning of th string and stop when it find b=, instead of not matching.
See: http://regex101.com/r/fN3zU5/3
Can someone help me please?
Just use a lookahead to check anything following the URL must be a space or line end.
\bhttp:\/\/www\.website\.com\/(?:(?!b=[0-9]+).)*?\b(?= |$)
DEMO
use this:
^http:\/\/www\.website\.com\/((?!b=[0-9]+)).*$
\b only matches word endings.
^ matches start and end of string
and you dont even need to do it that complicated, If you dont want the url with the b parameter use this:
^http:\/\/www\.website\.com\/(?!b).*$
demo here : http://regex101.com/r/fN3zU5/5
import re
pattern=re.compile(r"(?!.*?b=.*).*")
print pattern.match(x)
This will look ahead if there is a "b=" present.A negative lookahead means it will not match that string.
You had a look at this possibility:
http://regex101.com/r/fN3zU5/6
^http:\/\/www\.website\.com\/[ac\=\d&]*$
only allow &,=,a,c and digits
complete url in group and there should not be a "b=" parameter
if you have more options and you dont want to list them all:
you dont allow a 'b' to be part of your parameters
^http:\/\/www\.website\.com\/[^b]*$
http://regex101.com/r/fN3zU5/7
^http:\/\/www\.website\.com\/(?!.*?b=.*?).*$ works too here "b=" is permitted at any position of the parameter string so you could even have the "b" string as a value of a parameter.
See
http://regex101.com/r/fN3zU5/8
This is what you want. ^http:\/\/www\.website\.com\/(([^b]=[0-9]+).)*$
Its a simple pattern not flexible but it works :
http:\/\/www\.website\.com\/+a=+\w+&+c=+\w+
I want a regex which return me only characters before first point.
Ex :
T420_02.DOMAIN.LOCAL
I want only T420_02
Please help me.
You can use the following regex: ^(.*?)(?=\.)
The captured group contains what you need (T420_02 in your example).
This simple expression should do what you need, assuming you want to match it at the beginning of the string:
^(.+?)\.
The capture group contains the string before (but not including) the ..
Here's a fiddle: http://www.rexfiddle.net/s8l0bn3
Use regex pattern ^[^.]+(?=[.])
I am doing pattern match for some names below:
ABCD123_HH1
ABCD123_HH1_K
Now, my code to grep above names is below:
($name, $kind) = $dirname =~ /ABCD(\d+)\w*_([\w\d]+)/;
Now, problem I am facing is that I get both the patterns that is ABCD123_HH1, ABCD123_HH1_K in $dirname. However, my variable $kind doesn't take this ABCD123_HH1_K. It does take ABCD123_HH1 pattern.
Appreciate your time. Could you please tell me what can be done to get pattern with _k.
You need to add the _K part to the end of your regex and make it optional with ?:
/ABCD(\d+)_([\w\d]+(_K)?)/
I also erased the \w*, which is useless and keeps you from correctly getting the HH1_K.
You should check for zero or more occurrences of _K.
* in Perl's regexp means zero or more times
+ means atleast one or more times.
Hence in your regexp, append (_K)*.
Finally, your regexp should be this:
/ABCD(\d+)\w*_([\w\d]+(_K)*)/
\w includes letters, numbers as well as underscores.
So you can use something as simple as this:
/ABCD\w+/
Is it possible to have this done with one regex?
I need to match only those strings that have exactly one period/dot but the restriction is that that period/dot must not be at the end of the string.
Example:
abc.d will match
.abcd will match
abcd. will not match
Yes, you can do it in one regex:
^[^.]*\.[^.]+$
I really like #codaddict's answer, but how about something without Regex? ( C# code below )
if(a.Split('.').Length>2 || a.EndsWith("."))
{
Console.WriteLine("invalid");
}
What I like is that it is much more clear that you don't want a string with two . and also a . should not be in the end. And this might actually be faster than using a regex.
I'm working wih a regular expression and have some lines in javascript. My expression should deliver two matches but recognizes only one and I don't know whats the problem.
The Lines in javascript look like this:
if(mode==1) var adresse = "?APPNAME=CampusNet&PRGNAME=ACTION&ARGUMENTS=-A7uh6sBXerQwOCd8VxEMp6x0STE.YaNZDsBnBOto8YWsmwbh7FmWgYGPUHysiL9u0.jUsPVdYQAlvwCsiktBzUaCohVBnkyistIjCR77awL5xoM3WTHYox0AQs65SoHAhMXDJVr7="; else var adresse = "?APPNAME=CampusNet&PRGNAME=ACTION&ARGUMENTS=-AHMqmg-jXIDdylCjFLuixe..udPC2hjn6Kiioq7O41HsnnaP6ylFkQLhaUkaWKINEj4l2JqL2eBSzOpmG.b5Av2AvvUxEinUhMBTt5awdgAL4SkBEgYXGejTGUxcgPE-MfiQjefc=";
My expression looks like this:
(?<Popup>(popUp\(')|(adresse...")).*\?((?<Parameters>APPNAME=CampusNet[^>"']*["']))
I want to have two matches with APPNAME...... as Parameters.
[UPDATE] Like Tim Pietzcker wrote i used the greedy version and should have used the lazy version. while he wrote that i solved it myself by using .? instead of . in the middle so the expression looks like this:
(?<Popup>(popUp\(')|(adresse...")).*?\\?((?<Parameters>APPNAME=CampusNet[^>"']*["']))
That worked. Thanks to Tim Pietzcker
Your regex matches too much - from the very first adresse until the very last " because it uses a greedy quantifier .*.
If you make that quantifier lazy, i. e.
(?<Popup>(popUp\(')|(adresse...")).*?\?((?<Parameters>APPNAME=CampusNet[^>"']*["']))
you get two matches.
Alternatively, if your data allows this, use a different quantifier that only matches non-space characters. This will match faster (but will fail of course if the text you're trying to match could possibly contain spaces):
(?<Popup>(popUp\(')|(adresse..."))\S*\?((?<Parameters>APPNAME=CampusNet[^>"']*["']))
Usually you must apply the regex with the "global" flag to find all matches. I can't really say more until I see the complete code sample you are working with.