I want to understand how this regular expression (aka regex) stored in "regex" variable works?
regex='^([1-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(\.([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3}$'
I am new to bash scripting and having hard time work with regular expression!
Numbers from 1-9, 0-9, 0-4 and 0-5 are repeated at least twice, which is creating confusion!
Thank you!
Look at this part alone:
[1-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5]
It's a series of alternatives (separated by |), here on separate lines:
[1-9] # Matches 1-9
[1-9][0-9] # Matches 10-99
1[0-9][0-9] # Matches 100-199
2[0-4][0-9] # Matches 200-249
25[0-5] # Matches 250-255
In other words, it matches any number from 1 to 255 inclusive. It's a bit roundabout because regex has no concept of numbers, only of character strings.
The regex attempts to match a four of these numbers with periods between, in order to match a whole IPv4 address.
It looks like someone was trying to match an IPv4 address. The group
([1-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])
Matches a number from 1 to 255, it then matches a number from 0 to 255 three more times.
(\.([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3}
They tried to separate the four numbers with dots. The original regex you posted didn't escape the "." so it would match any character between the four groups. Someone has since edited the regex to fix that character.
The regex is wrapped in ^ and $ to make sure the string contains that and only that. ^ matches the beginning of a string. $ matches the end.
Related
i want to match the strings which is listed below other than than that whatever the string is it should not match
rahul2803
albert1212
ra456
r1
only the above mentioned strings should match in the following group of data
rahul
2546rahul
456
rahul2803
albert1212
ra456
r1
rahulrenjan
r4ghyk
i tried with ([a-z]*[0-9]) but it's not working.
In regular expressions * means zero or more so your regex matches zero letters. If you want one or more use + (\d means digit).
^[a-zA-Z]+\d+$
Regular expressions are fun to solve once you get the hang of the syntax.
This one should be pretty straight:
Start with a letter. ^[a-z] (I am not taking the case of capital
letters here, if they are then ^[a-zA-Z] )
Have multiple letters/digits in between .*
End the string with a digit [0-9]$
Combine all 3 and you get:
^[a-z].*[0-9]$
My RegExp is very rusty! I have two questions, related to the following RegExp
Question Part 1
I'm trying to get the following RegExp to work
^.*\d{1}\.{1}\d{1}[A-Z]{5}.*$
What I'm trying to pass is x1.1SMITHx or x1.1.JONESx
Where x can be anything of any length but the SMITH or JONES part of the input string is checked for 5 upper case characters only
So:
some preamble 1.1SMITH some more characters 123
xyz1.1JONES some more characters 123
both pass
But
another bit of string1.1SMITHABC some more characters 123
xyz1.1ME some more characters 123
Should not pass because SMITH now contains 3 additional characters, ABC, and ME is only 2 characters.
I only pass if after 1.1 there are 5 characters only
Question Part 2
How do I match on specific number of digits ?
Not bothered what they are, it's the number of them that I can't get working
if I use ^\d{1}$ I'd have thought it'll only pass if one digit is present
It will pass 5 but it also passes 67
It should fail 67 as it's two digits in length.
The RegExp should pass only if 1 digit is present.
For the first one, check out this regex:
^.*\d\.\d[A-Z]{5}[^A-Z]*$
Before solving the problem, I made it easier to read by removing all of the {1}. This is an unnecessary qualifier since regex will default to looking for one character (/abc/ matches abc not aaabbbccc).
To fix the issue, we just need to replace your final .*. This says match 0+ characters of anything. If we make this "dot-match-all" more specific (i.e. [^A-Z]), you won't match SMITHABC.
I came up with a number of solution but I like these most. If your RegEx engine supports negative look-ahead and negative look-behind, you can use this:
Part 1: (?<![A-Z])[A-Z]{5}(?![A-Z])
Part 2: (?<!\d)\d(?!\d)
Both have a pattern of (?<!expr)expr(?!expr).
(?<!...) is a negative look-behind, meaning the match isn't preceded by the expression in the bracket.
(?!...) is a negative look-ahead, meaning the match isn't followed by the expression in the bracket.
So: for the first pattern, it means "find 5 uppercase characters that are neither preceded nor followed by another uppercase character". In other words, match exactly 5 uppercase characters.
The second pattern works the same way: find a digit that is not preceded or followed by another digit.
You can try it on Regex 101.
I need a regular expression for 4 characters. The first 3 characters must be a number and the last 1 must be a letter or a digit.
I formed this one, but it not working
^([0-9]{3}+(([a-zA-Z]*)|([0-9]*)))?$
Some valid matches: 889A, 777B, 8883
I need a regular expression for first 3 will be a number and the last 1 will be a alphabet or digit
This regex should work:
^[0-9]{3}[a-zA-Z0-9]$
This assumes string is only 4 characters in length. If that is not the case remove end of line anchor $ and use:
^[0-9]{3}[a-zA-Z0-9]
Try this
This will match it anywhere.
\d{3}[a-zA-Z0-9]
This will match only beginning of a string
^\d{3}[a-zA-Z0-9]
You can also try this website: http://gskinner.com/RegExr/
It makes it very easy to create and test your regex.
Just take the stars out...
^([0-9]{3}+(([a-zA-Z])|([0-9])))?$
The stars mean zero or more of something before it. You are already using an or (|) so you want to match exactly one of the class, or one of the other, not zero or more of the class, or zero or more of the other.
Of course, it can be simplified further:
^\d{3}[a-zA-Z\d]$
Which literally means... three digits, followed by a character from either lowercase or uppercase a-z or any digit.
as I go through the regex101 quiz/lessons, I am supposed to match an IP address (without leading zeros).
Now the following
^[^0]+[0-9]+\\.[^0]+[0-9]+\\.[^0]+[0-9]+\\.[^0]+[0-9]+$
matches 23.34.7433.33
but fails to match single digit numbers like 1.2.3.4
Why is this so, when my + is supposed to match "1 to infinite" times...?
You are in fact matching more than 2 digits for each number in the IP address because you have:
[^0]+[0-9]+
[^0]+ matches at least one character, and [0-9]+ matches at least 1 character. Both will match 'at least 2 characters' (characters being in scope of the character classes).
Also 23.34.7433.3 doesn't match your regex for the reason I stated above.
And you might try this regex for the purpose you stated:
^(?:[1-9][0-9]{0,2}\.){3}[1-9][0-9]{0,2}$
[1-9][0-9]{0,2} will match up to 3 digits, with a non leading 0.
EDIT: You mentioned in a comment that 0.0.0.0 (single digit zeroes) are to be accepted as well. The modified regex from above would be:
^(?:(?:[1-9][0-9]{0,2}|0)\.){3}(?:[1-9][0-9]{0,2}|0)$
Assuming you want to check an IPv4, I suggest you this pattern:
^(?<nb>2(?>[0-4][0-9]|5[0-5])|1[0-9]{2}|[1-9]?[0-9])(?>\.\g<nb>){3}$
I have defined a named subpattern nb to make the pattern shorter, but if you prefer, you can rewrite all and replace \g<nb>:
^(?>2(?>[0-4][0-9]|5[0-5])|1[0-9]{2}|[1-9]?[0-9])(?>\.(?>2(?>[0-4][0-9]|5[0-5])|1[0-9]{2}|[1-9]?[0-9])){3}$
Numbers greater than 255 are not allowed.
pattern details:
The goal is to describe what is allowed:
numbers with 3 digits that begins with "2" can be followed by a digit in [0-4] and a digit in [0-9] OR by 5 and a digit in [0-5] because it can exceed 255.
numbers with 3 digits that begins with "1" can be followed by any two digits.
any number with 2 digits that doesn't begin with "0"
any number with 1 digit (zero included)
If I add one by one these rules, I obtain
2(?>[0-4][0-9]|5[0-5])
2(?>[0-4][0-9]|5[0-5]) | 1[0-9]{2}
2(?>[0-4][0-9]|5[0-5]) | 1[0-9]{2} | [1-9][0-9]
2(?>[0-4][0-9]|5[0-5]) | 1[0-9]{2} | [1-9][0-9] | [0-9]
Now I have a definition for allowed numbers. I can reduce a little the size of the pattern replacing [1-9][0-9] | [0-9] by [1-9]?[0-9]
Then you only have to add the dot repeat the subpattern four times: x.x.x.x
But since there is only three dots, I write the first number and I repeat 3 times a group that contains a dot and a number:
2(?>[0-4][0-9]|5[0-5])|1[0-9]{2}|[1-9]?[0-9] # the first number
(?>\.2(?>[0-4][0-9]|5[0-5])|1[0-9]{2}|[1-9]?[0-9]){3} # the group repeated 3 times
To be sure that the string doesn't contain anything else that the IP I described, I add anchors for the start of string ^ and for the end of string $, then the string begins and ends with the IP.
To reduce the size of a pattern you can define a named group which allows to reuse the subpattern it contains,
Then you can rewrite the pattern like this:
^
(?<nb> 2(?>[0-4][0-9]|5[0-5])|1[0-9]{2}|[1-9]?[0-9] ) # named group definition
(?> \. \g<nb> ){3} # \g<nb> is the reference to the subpattern named nb
$
[0-9]+ should be [0-9]*
* matches 0 or more.
+ matches 1 or more.
You already have the case [^0] <--- this actually wrong because it will match letters also.
besides that it will match the first character that's NOT zero then at least one number after that.
It should be written as
[1-9][0-9]*
This essentially checks the first letter and sees if its a number that's between 1-9 then the next numbers(0 nums to infinite nums) after that is a number 0-9.
Then this will come out to.
^[1-9][0-9]*\.[1-9][0-9]*\.[1-9][0-9]*\.[1-9][0-9]*$
Edit live on Debuggex
cleaning it up.
^(?:[1-9][0-9]*\.){3}[1-9][0-9]*$
this should work...
^(?:[1-9][0-9]*\.|[0-9])(?:[1-9][0-9]*\.|[0-9])(?:[1-9][0-9]*\.|[0-9])(?:[1-9][0-9]*|[0-9])$
cleaned up.
^(?:(?:[1-9][0-9]*|0)\.){3}(?:[1-9][0-9]*|0)$
Your regex would match ABCDEFG999.FOOBSR888 etc, because [^0] is any character other than a zero, and bith character classes are required by the +.
I think you want this:
^[1-9]\d*(\\.[1-9]\d*){3}$
having replaced various verbose expressions with their equivalent, this is 4 groups of digits each starting with a non-zero.
Actually the problem is far more complicated, because your approach (once corrected) allows 999.999.999.999, which is not a valid IP.
It might be because you need at least two digits between two dots '.'
try using this pattern: ^[^0]+[0-9]*\.[^0]+[0-9]*\.[^0]+[0-9]*\.[^0]+[0-9]*$
to match ip address you should use this pattern:
\b(?:\d{1,3}.){3}\d{1,3}\b
taken from here:
http://www.regular-expressions.info/examples.html
I'm trying to come up with some regex to match against 1 hyphen per any number of digit groups. No characters ([a-z][A-Z]).
123-356-129811231235123-1235612346123451235
/[^\d-]/g
The one above will match the string below, but it will let the following go through:
1223--1235---123123-------
I was looking at the following post How to match hyphens with Regular Expression? for an answer, but I didn't find anything close.
#Konrad Rudolph gave a good example.
Regular expression to match 7-12 digits; may contain space or hyphen
This tool is useful for me http://www.gskinner.com/RegExr/
Assuming it can't ever start with a hyphen:
^\d(-\d|\d)*$
broken down:
^ # match beginning of line
\d # match single digit
(-\d|\d)+ # match hyphen & digit or just a digit (0 or more times)
$ # match end of line
That makes every hyphen have to have a digit immediately following it. Keep in mind though, that the following are examples of legal patterns:
213-123-12314-234234
1-2-3-4-5-6-7
12234234234
gskinner example
Alternatively:
^(\d+-)+(\d+)$
So it's one or more group(s) of digits followed by hyphen + final group of digits.
Nothing very fancy, but in my tests it matched only when there were hyphen(s) with digits on both sides.