I need to find a regular expression to match everu char between * delimiter.
I ca have strings like these:
*1234-567*
**1234-567**
***1234-567***
*1234-567****
**1234-567****
I need to get 1234-567
I did a try with this one. The full match of this regex return also the * chars. I do ot need it.
Can you help me?
Try this pattern:
/\*+([^*]+)\*+/g
Noted that $1 contains what you need.
Online Demo
if you can utilise lookarounds (generally meaning, it isn't javascript), this will match what you want with no match groups needed:
(lookarounds being "zero-width assertions", they do not actually consume any characters)
(?<=\*)[^*\n]+(?=\*)
regex101 demo
Related
I have lines like this:
example.com/p/stuff/...
example.com/page/thing/...
example.com/page/stuff/...
example.com/page/other-stuff/...
etc
where the dots represent continuing URL paths. I want to select URLs that contain /page/ and are NOT followed by thing/. So from the above list we would select:
example.com/page/stuff/...
example.com/page/other-stuff/...
.*?\/page\/[^(thing)].*
this is the regex for matching a string which has /page/ not followed by thing
adding the lazy evalation is suggested because you advance a char at the time, better performance!
You need to use negative lookahead:
example\.com\/page\/(?!thing\/).*
Demo
Use the following regex pattern:
.*?\/page\/(?!thing\/).*
https://regex101.com/r/19wh1w/2
(?!thing\/) - negative lookahead assertion ensures that page/ section is not followed by thing/
Within Notepad++, I want to replace many instances of the type ``string'' by \command{string} where string can be any string of characters. I am fairly close to what I want to achieve with:
Find: (?<=``)(.*?)(?='')
Replace: \\command{\1}
There is still a problem. With the regex code above, instead of \command{string} I get ``\command{string}'' and I am not sure why the `` and '' are not removed?
It is because you are using lookaround assertions. Lookaround (zero-width) assertions only assert that a position can be matched and do not "consume" any characters on the string. You can use the below regular expression.
Find: ``([^']+)''
Replace: \\command{\1}
You need to wrap everything into a capture group and use that. NP++ seems to not support lookahead/behind, but you dont need that for this specific case anyway:
``([^']+)'' -> \\command{\1}
This will make sure it does not match two commands (longest match) in something like:
run ``ls -l'' or ``ls -a''
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'm struggling to come up with the correct regex for the following scenario.
Let's say you have to match a word either starts with http- or nothing
eg : http-test-data, test-data should be a match but xyz-test-data shouldn't be a match
the regex i came up so far is
(?:http-)?(test-data)
but it matches xyz-test-data as well.
You could simply use the following:
(?:http-|^)(test-data)
This tests for either a positive look-behind of http- or for the beginning of the string before test-data.
For example, for the sample data as follows:
http-test-data
xyz-test-data
http-test-data
xyz-test-data
test-data
yes-yes-test-data
-test-data
It yeilds:
http-test-data
http-test-data
test-data
Try this representation
^(http-|)(test-data)
Yes because there is a ? on the (?:http-). Then the regex will also match any string that contains test-data.
Q1: I'm writing a regex in php and not successful. I want to match the following:
so i would
if not then match:
so i
and then:
i would
and
so
i
would
Here is my code:
\b(so i|i would|so i would|(so|i|would))\b
Its only matching the: so, i, would, so i, i would .... but not matching the so i would?
Order your regex correctly.
\b(so i would|so i|i would|(so|i|would))\b
Put the longest string to match to the left.
The | is left-associative and hence, in your version Of the regex, is matching the shorter string.
Just put it at the beginning
\b(so i would|so i|i would|(so|i|would))\b
put longest pattern to left in the group: \b(long|...|short)\b
another solution: \b(so i would|i would|would|so i|so|i)\b
p.s. this is NFA regex engine feature, please refer to "Mastering Regular Expressions"