I am looking for a pattern that matches everything until the first occurrence of a specific character, say a ";" - a semicolon.
I wrote this:
/^(.*);/
But it actually matches everything (including the semicolon) until the last occurrence of a semicolon.
You need
/^[^;]*/
The [^;] is a character class, it matches everything but a semicolon.
^ (start of line anchor) is added to the beginning of the regex so only the first match on each line is captured. This may or may not be required, depending on whether possible subsequent matches are desired.
To cite the perlre manpage:
You can specify a character class, by enclosing a list of characters in [] , which will match any character from the list. If the first character after the "[" is "^", the class matches any character not in the list.
This should work in most regex dialects.
Would;
/^(.*?);/
work?
The ? is a lazy operator, so the regex grabs as little as possible before matching the ;.
/^[^;]*/
The [^;] says match anything except a semicolon. The square brackets are a set matching operator, it's essentially, match any character in this set of characters, the ^ at the start makes it an inverse match, so match anything not in this set.
None of the proposed answers did work for me. (e.g. in notepad++)
But
^.*?(?=\;)
did.
Try /[^;]*/
Google regex character classes for details.
sample text:
"this is a test sentence; to prove this regex; that is g;iven below"
If for example we have the sample text above, the regex /(.*?\;)/ will give you everything until the first occurence of semicolon (;), including the semicolon: "this is a test sentence;"
Try /[^;]*/
That's a negating character class.
This was very helpful for me as I was trying to figure out how to match all the characters in an xml tag including attributes. I was running into the "matches everything to the end" problem with:
/<simpleChoice.*>/
but was able to resolve the issue with:
/<simpleChoice[^>]*>/
after reading this post. Thanks all.
this is not a regex solution, but something simple enough for your problem description. Just split your string and get the first item from your array.
$str = "match everything until first ; blah ; blah end ";
$s = explode(";",$str,2);
print $s[0];
output
$ php test.php
match everything until first
This will match up to the first occurrence only in each string and will ignore subsequent occurrences.
/^([^;]*);*/
"/^([^\/]*)\/$/" worked for me, to get only top "folders" from an array like:
a/ <- this
a/b/
c/ <- this
c/d/
/d/e/
f/ <- this
Really kinda sad that no one has given you the correct answer....
In regex, ? makes it non greedy. By default regex will match as much as it can (greedy)
Simply add a ? and it will be non-greedy and match as little as possible!
Good luck, hope that helps.
This works for getting the content from the beginning of a line till the first word,
/^.*?([^\s]+)/gm
I faced a similar problem including all the characters until the first comma after the word entity_id. The solution that worked was this in Bigquery:
SELECT regexp_extract(line_items,r'entity_id*[^,]*')
s="XX.1.1. Accidents"
pattern = re.compile(r'\d|[a-zA-Z]\.\s([a-zA-Z]\S+)')
match=pattern.search(s)
if match:
print(match.group(1))
the output is None. However, I think it should have been "Accidents" Can someone tell me why?
Your | is messing with it - since you're not placing it in a capturing group or anything, it'll match \d or all of [a-zA-Z]\.\s([a-zA-Z]\S+). This is an issue because the regex will act greedily and you'll end up just a single \d.
If you use (?:\d|[a-zA-Z])\.\s([a-zA-Z]\S+), it'll work properly and you'll receive Accidents.
You can put the whole first part in square brackets, then search for the space, and then for the characters and the rest:
pattern = re.compile(r'[a-zA-Z\d.]*\s([a-zA-Z]\S+)')
I have the following phrase:
05/30/2016 07:02 AM (GMT+02:00) added by XXX YYY (PID-000301):\tSome_alphanum_text_Some_alphanum_text_Some_alphanum_text_Some_alphanum_text\t\t*************************************************************************************************\t05/12/2016 02:03 PM (GMT+02:00) added by ZZZ AAA (PID-000301):\tSome_other_alphanum_text_Some_other_alphanum_text_Some_other_alphanum_text_Some_other_alphanum_text\t\t
I would like to write a RegEx which is just going to scoop up for me only 'Some_alphanum_text' and 'Some_other_alphanum_text'.
So far I was trying my luck with something like this:
r'(?:.+\(PID-\d{6}\):)(.+)'
But it is only giving me the 'Some_other_alphanum_text' occurrence.
There can be more than 2 unique strings I will need to scoop out from this mess of a text. Any ideas?
You need to replace .+ with something that only matches what you want to return. Since you only want to match alphanumeric text, use \w instead of .
r'(?:\(PID-\d{6}\):)\s*(\w+)'
You need \s* before the second group because the whitespace before the alphanumeric text won't match \w+.
You also don't need .+ at the beginning. The match will just begin where it finds PID.
DEMO
I believe you need this regex:
\(PID-\d{6}\):\\t(.+?)(?:\\t){2}
regex101
I think you could use this to find all the instances of text occurring between "\t"s
I didn't change the regex area to be a code block so it has not worked.
Now it works! One thing you should consider is that there could be no '\t'. But
every matched text follows a date format such as 05/12/2016 02:03 or ends.
\(PID-\d{6}\)[\n\r\t\s]*:(?:.|[\n\r\t\s])*?(?=[0-9]{2}\/[0-9]{2}\/[0-9]{4}[\n\r\t\s]*[0-9]{2}:[0-9]{2}|$)
I am using Python 2.7 and have a question with regards to regular expressions. My string would be something like this...
"SecurityGroup:Pub HDP SG"
"SecurityGroup:Group-Name"
"SecurityGroup:TestName"
My regular expression looks something like below
[^S^e^c^r^i^t^y^G^r^o^u^p^:].*
The above seems to work but I have the feeling it is not very efficient and also if the string has the word "group" in it, that will fail as well...
What I am looking for is the output should find anything after the colon (:). I also thought I can do something like using group 2 as my match... but the problem with that is, if there are spaces in the name then I won't be able to get the correct name.
(SecurityGroup):(\w{1,})
Why not just do
security_string.split(':')[1]
To grab the second part of the String after the colon?
You could use lookbehind:
pattern = re.compile(r"(?<=SecurityGroup:)(.*)")
matches = re.findall(pattern, your_string)
Breaking it down:
(?<= # positive lookbehind. Matches things preceded by the following group
SecurityGroup: # pattern you want your matches preceded by
) # end positive lookbehind
( # start matching group
.* # any number of characters
) # end matching group
When tested on the string "something something SecurityGroup:stuff and stuff" it returns matches = ['stuff and stuff'].
Edit:
As mentioned in a comment, pattern = re.compile(r"SecurityGroup:(.*)") accomplishes the same thing. In this case you are matching the string "SecurityGroup:" followed by anything, but only returning the stuff that follows. This is probably more clear than my original example using lookbehind.
Maybe this:
([^:"]+[^\s](?="))
Regex live here.
Hi I'm new to regexes.
I have a string that I want to match any number of A-Z a-z 0-9 - and _
I've tried the following in python however it always matches, even the empty space. Can someone tell me why that is?
re.match(r'[A-Za-z0-9_-]+', 'gfds9 41.-=,434')
Your regex matches one or more of those characters. Your text starts with one or more of those characters, hence it matches. If you want it to only match those characters then you have to match them from the beginning to the end of the text.
re.match(r'^[A-Za-z0-9_-]+$', 'gfds9 41.-=,434')
Try the alternative for it maybe it will work for you:
[\w-]+
EDIT:
Although the initial regex you provided also works for me.