I am writing a filebeat configuration when I am matching if a line starts with a number like 03:32:33 ( a timestamp). I am currently doing it by-
\d
But its not getting recognised, is there anything else which I should do. I am not particularly good/ have experience with regex. Help will be appreciated.
The real problem is that filebeat does not support \d.
Replace \d by [0-9] and your regular expression will work.
I suggest you to give a look at the filebeat's Supported Patterns.
Also, be sure you've used ^, it stands for the start of the string.
Regex: (^\d)
1st Capturing group (^\d)
^ Match at the start of the string
\d match a digit [0-9]
You can use this regex:
^([0-9]{2}:?){3}
DEMO
Assert position at the beginning of the string «^»
Match the regex below and capture its match into backreference number 1 «([0-9]{2}:?){3}»
Exactly 3 times «{3}»
You repeated the capturing group itself. The group will capture only the last iteration. Put a capturing group around the repeated group to capture all iterations. «{3}»
Or, if you don’t want to capture anything, replace the capturing group with a non-capturing group to make your regex more efficient.
Match a single character in the range between “0” and “9” «[0-9]{2}»
Exactly 2 times «{2}»
Match the character “:” literally «:?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
You can use:
^\d{2}:\d{2}:\d{2}
The character ^ matches the start of a line.
Related
Details={
AwsEc2SecurityGroup={GroupName=m.com-rds, OwnerId=123, VpcId=vpc-123,
IpPermissions=[{FromPort=3306, ToPort=3306, IpProtocol=tcp, IpRanges=[{CidrIp=1.1.1.1/32}, {CidrIp=2.2.2.2/32}, {CidrIp=0.0.0.0/0}, {CidrIp=3.3.3.3/32}],
UserIdGroupPairs=[{UserId=123, GroupId=sg-123abc}]}], IpPermissionsEgress=[{IpProtocol=-1, IpRanges=[{CidrIp=0.0.0.0/0}]}], GroupId=sg-123abc}},
Region=us-east-1, Id=arn:aws:ec2:us-east-1:123:security-group/sg-123abc}]
}
I want to capture exactly arn:aws:ec2:us-east-1:123:security-group/sg-123abc in this example. Generically, I want to capture the value of Id regardless of placement. My current solution is /Details={.*Id=(.*\w)/, but this only works if it's the last object in the data. How can I take into account the following potential scenario:
Id=arn:aws:ec2:us-east-1:123:security-group/sg-123abc, Thing=123abc}]
You have a pattern with 2 times .* which will first match till the end of the line/string (depending on if the dot matches a newline) and it will backtrack to match the last occurrence where this part of the pattern Id=(.*\w) can match.
If you want to use a capture group, you can make the format and the allowed characters a bit more specific:
\bId=(\w+(?:[:\/-]\w+)+)
The pattern in parts
\b A word boundary to prevent a partial word match
Id= Match literally
( Capture group 1
\w+ Match 1+ word chars
(?:[:\/-]\w+)+ Repeat 1+ times either : / - and 1+ word chars
) Close group 1
Regex demo
Or if you know that it starts with Id=arn:
\bId=(arn:[\w:\/-]+)
Regex demo
Note that you don't have to escape the \/ only when the delimiters of the regex are forward slashes, but there is no language tagged.
You can use look-behind to check that there is the Id= prefix, and then match anything that is not a space, comma or closing brace:
(?<=\bId=)[^,}\s]*
In Scala, is it possible to actually insert commas via a regex to separate thousands in numbers where the comma definitely is not there to start with?
For example, I'd like to convert 30000.00 into 30,000.00.
I am not sure this is exactly what you need, but you can use this:
val formatter = java.text.NumberFormat.getNumberInstance
println(formatter.format(30000.00)) // prints 30,000
This is not scala based answer.
You can use regex \d{1,3}(?=(?:\d{3})+\.) to find the matches and substitute each match with the same match plus an extra comma $0,.
See the online demo.
Explanation:
\d{1,3} This matches a decimal character between 1 and 3 times
(?= Positive lookahead starts
(?: This indicates a Non-capturing group
\d{3} matches a digit exactly 3 times
) end of Non-capturing group.
+ matches the previous group one or more times
\. matches the character . literally
) Positive lookahead ends.
I have the following template :
1251 Left Random Text I want to fill
It can go through multiple lines
As you can see
9841 Right Again we see a lot of random text with 3115 numbers
And this also goes
To multiple lines
0121 Right
5151 Right This one is just one line
I was wrong
9731 Left This one is just a line
5123 NA Instruction 5151 was wrong
4113 Right Instr 9841 was correct
We checked
I want to have 3 groups:
1251
Left
Random Text I want to fill
It can go through multiple lines
As you can see
I'm using
(\d+)\s(\w+)\s(.*)
but it stops at the current line only (so I get only Random Text I want to fill in group 3, although I want including As you can see)
If I'm using Single line flag I get only 1 match for each group, group 3 almost being all
Here is live : https://regex101.com/r/W3x0mH/4
You could use a repeating group matching all the lines while asserting that the next line does not start wit 1+ digits followed by Left or Right:
(\d+)\s(\w+)\s(.*(?:\r?\n(?!\d).*)*)
Explanation
(\d+)\s(\w+)\s Match the first 2 groups
(Third capturing group
.* Match 0+ times any char except a newline
(?: Non capturing group
\r?\n(?!\d).* Match newline, assert what is on the right is not a digit
)* Close non capturing group and repeat 0+ times
) Close capturing group
Regex demo
You may use this regex with a lookahead:
^(\d+)\s(\w+)\s(.*?)(?=\n\d|\z)
with DOTALL and MULTILINE modifiers.
Updated Regex Demo
RegEx Details:
^: Line start
(\d+): Match and capture 1+ digits in group #1
\s: match a whitespace
(\w+): Match and capture 1+ word characters in group #2
\s: match a whitespace
(.*?): Match 0 or more of any character (non-greedy) provided next lookahead assertion is satiSfied
(?=\n\d|\z): Lookahead assertion to assert that we have a newline followed by a digit or there is end of input
Faster Regex:
If you are using this regex on a long string then you should also keep overall performance in mind as a regex with DOTALL modifier will tend to get slow on a large size text. For that I suggest using this regex that doesn't need DOTALL modifier:
^(\d+)\s(\w+)\s(.*(?:\n.*)*?)(?=\n\d|\z)
RegEx Demo 2
On regex101 demo this regex takes just 181 steps as compared to first one that takes 1300 steps.
For the third group, repeat any character while using negative lookahead for ^\d, which would indicate the start of a new match:
(\d+)\s(\w+)\s((?:(?!^\d)[\s\S])*)
https://regex101.com/r/W3x0mH/5
You may try with this regex:
^(\d+)\s+(\w+)\s+(.*?)(?=^\d|\z)
^(\d+)\s+ , ^\d+ Line begins with numbers followed by one or more whitespace character \s+
(\w+)\s+ where \w+ one or more characters (left,right,na or something else) followed by one or more whitespace \w+
(.*?) matches everything until it finds a line beginning with number or \z end of string.
I think it fits your requirement....
Regex101
There are a thousand regular expression questions on SO, so I apologize if this is already covered. I did look first.
I have string:
Name Subname 11X22 88X620 AB33(20) YA5619 77,66
I need to capture this string: YA5619
What I am doing is just finding AB33(20) and after this I am capturing until first white space. But AB33(20) can be AB-33(20) or AB33(-20) or AB33(-1).
My preg_match regex is: (?<=\bAB\d{2}\(\d{2}\)\s).+?(?=\s)
Why I am getting error when I change from \d{2} to \d+?
For final result I was thinking this regix will work but no:
(?<=\bAB-?\d+\(-?\d+\)\s).+?(?=\s)
Any ideas what I am doing wrong?
With most regex flavors, lookbehind needs to evaluate to a fixed-length sequence, so you can't use variable quantifiers like * or + or even {1,2}.
Instead of using lookaround, you can simply match your marker pattern and then forget it with \K.
AB-?\d+(?:\(-?\d+\))? \K[^ ]+
demo: https://regex101.com/r/8XXngH/1
It depends on the language. If it is in .NET for example, it matches due to the various length in the lookbehind.
Another solution might be to use a character class and add the character you would allow to match. Then match a whitespace character and capture in a group matching \S+ which matches 1+ times not a whitespace character.
\bAB[()\d-]+\s\K\S+
Explanation
\bAB Match literally prepended with word boundary to prevent AB being part of a larger match.
[()\d-]+ Match 1+ times any of the listed character in the character class
\s Match a whitespace char (or \s+ to match 1 or more)
\K Reset the starting point of the reported match( Forget what was matched)
\S+ Match in a group 1+ times not a whitespace character
Regex demo | Php demo
I noticed some interesting behaviour with some regex work I am doing, and I'd like some insight.
From what I understand, the word character, \w should match the following [a-zA-Z_0-9]
Given this input,
0000000060399301+0000000042456971+0000000
What should this regex
(\d+)\w
Capture?
I would expect it to capture 0000000060399301 but it actually captures 000000006039930
Is there something I am missing? Why is the 1 dropped from the end?
I noticed if I changed the regex to
(\d+\w)
It captures correctly i.e. including the 1
Anyone care to explain? Thanks
You require the regex to match a trailing word character - that would be the 1.
It cannot be another character, because
+ is not a word class character
+ is not a digit
matching is greedy
\d+ - matches one or more digit characters.
\w+ - matches one or more word characters. [A-Za-z\d_]
So with this string 0000000060399301+, \d+ in this (\d+)\w regex matches all the digits (including the 1 before +) at very first, since the following pattern is \w , regex engine tries to find a match, so it backtracks one character to the left and forces \w to match the digit before + . Now the captured group contains 000000006039930 and the last 1 is matched by \w
The 1 is being dropped because \w isn't in the capture group.