I am writing a regex to recognize IP address of the form "A.B.C.D", where the value of A, B, C, and D may range from 0 to 255. Leading zeros are allowed. The length of A, B, C, or D can't be greater than 3. I know this regex is easily available on Internet but I am writing it on my own for practice.
First I wrote the regex for A as follows:
a = ^(^0{0,2}\d|^0{0,1}\d\d|[0-1]\d\d|2[0-4]\d|25[0-5])$
It works as expected, then I wrote it for A.B as follows:
ab = ^(^0{0,2}\d|^0{0,1}\d\d|[0-1]\d\d|2[0-4]\d|25[0-5])\.
(^0{0,2}\d|^0{0,1}\d\d|[0-1]\d\d|2[0-4]\d|25[0-5])$
But somehow it isn't working as expected. It's not recognizing strings like "2.3" but recognizing "2.003". This is very weird. I have spent hours figuring it out but have totally given up now. Please help me with this.
As #Jorge pointed out in the comments, the ^ character matches the start of a string/line, which can occur for A, as it is presumably the first group of characters in the line, but cannot occur for B, since it will always be preceded by A. This is why it could match 003 (through the subpattern [0-1]\d\d), but it couldn't match 3 through the subpattern ^0{0,2}\d.
Remove the superfluous ^s, and you should get the desired behavior:
ab = ^(0{0,2}\d|0{0,1}\d\d|[0-1]\d\d|2[0-4]\d|25[0-5])\.
(0{0,2}\d|0{0,1}\d\d|[0-1]\d\d|2[0-4]\d|25[0-5])
Related
I've been fiddling around yesterday but coulnd't get a proper solution. I'm trying to get a regex which maches with floor Inputs (for a building) with all upper case letters.
I want to match either only E, only D, only 1, 2, 3 ect. or only U1, U2, U3 ect., the last one also must be in the correct order the letter comes before the number, not the other way around.
So far I've come up with this regex here: /[UED]|[1-9]/g
But this matches way too many things, for example 2U would also match or ED22 or UD1 and so on. I was trying it out with regexr.com but had no success so far to solve this problem.
Has anyone an idea how I can mach specifically only one of the four above mentionned inputs?
Valid Inputs:
E
U8
D
32
etc..
Invalid Inputs:
2U
ED
EEE
D1
etc.
You may use
^(?:[ED]|U?[1-9]\d*)$
See another regex demo
Details
^ - start of string
(?: - start of a non-capturing group matching either of
[ED] - an E or D
| - or
U?[1-9]\d* - an optional U, a non-zero digit and any 0+ digits
) - end of the group
$ - end of string.
I have a filename like this:
0296005_PH3843C5_SEQ_6210_QTY_BILLING_D_DEV_0000000000000183.PS.
I needed to break down the name into groups which are separated by a underscore. Which I did like this:
(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
So far so go.
Now I need to extract characters from one of the group for example in group 2 I need the first 3 and 8 decimal ( keep mind they could be characters too ).
So I had try something like this :
(.*?)_([38]{2})(.*?) _(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
It didn’t work but if I do this:
(.*?)_([PH]{2})(.*?) _(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
It will pull the PH into a group but not the 38 ? So I’m lost at this point.
Any help would be great
Try the below Regex to match any first 3 char/decimal and one decimal
(.?)_([A-Z0-9]{3}[0-9]{1})(.?)(.*?)(.?)_(.?)(.*?)(.?)_(.?)
Try the below Regex to match any first 3 char/decimal and one decimal/char
(.?)_([A-Z0-9]{3}[A-Z0-9]{1})(.?)(.*?)(.?)_(.?)(.*?)(.?)_(.?)
It will match any 3 letters/digits followed by 1 letter/digit.
If your first two letter is a constant like "PH" then try the below
(.?)_([PH]+[0-9A-Z]{2})(.?)(.*?)(.?)_(.?)(.*?)(.?)_(.?)
I am assuming that you are trying to match group2 starting with numbers. If that is the case then you have change the source string such as
0296005_383843C5_SEQ_6210_QTY_BILLING_D_DEV_0000000000000183.PS.
It works, check it out at https://regex101.com/r/zem3vt/1
Using [^_]* performs much better in your case than .*? since it doesn't backtrack. So changing your original regex from:
(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)_(.*?)(\d{16})(.*)
to:
([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_(.*?)(\d{16})(.*)
reduces the number of steps from 114 to 42 for your given string.
The best method might be to actually split your string on _ and then test the second element to see if it contains 38. Since you haven't specified a language, I can't help to show how in your language, but most languages employ a contains or indexOf method that can be used to determine whether or not a substring exists in a string.
Using regex alone, however, this can be accomplished using the following regular expression.
See regex in use here
Ensuring 38 exists in the second part:
([^_]*)_([^_]*38[^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_(.*?)(\d{16})(.*)
Capturing the 38 in the second part:
([^_]*)_([^_]*)(38)([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_([^_]*)_(.*?)(\d{16})(.*)
I came across the regular expression not containing 101 as follows:
0∗1∗0∗+(1+00+000)∗+(0+1+0+)∗
I was unable to understand how the author come up with this regex. So I just thought of string which did not contain 101:
01000100
I seems that above string will not be matched by above regex. But I was unsure. So tried translating to equivalent pcre regex on regex101.com, but failed there too (as it can be seen my regex does not even matches string containing single 1.
Whats wrong with my translation? Is above regex indeed correct? If not what will be the correct regex?
Here is a bit shorter expression ^0*(1|00+)*0*$
https://www.regex101.com/r/gG3wP5/1
Explanation:
(1|00+)* we can mix zeroes and ones as long as zeroes occur in groups
^0*...0*$ we can have as many zeroes as we want in prefix/suffix
Direct translation of the original regexp would be like
^(0*1*0*|(1|00|000)*|(0+1+0+)*)$
Update
This seems like artificially complicated version of the above regexp:
(1|00|000)* is the same as (1|00+)*
it is almost the solution, but it does not match strings 0, 01.., and ..10
0*1*0* doesn't match strings with 101 inside, but matches 0 and some of 01.., and ..10
we still need to match those of 01.., and ..10 which have 0 & 1 mixed inside, e.g. 01001.. or ..10010
(0+1+0+)* matches some of the remaining cases but there are still some valid strings unmatched
e.g. 10010 is the shortest string that is not matched by all of the cases.
So, this solution is overly complicated and not complete.
read the explanation in the right side tab in regex101 it tells you what your regex does( I think you misunderstood what list operator does) , inside a list operator ( [ ) , the other characters such as ( won't be metacharacters anymore so the expression [(0*1*0*)[1(00)(000)] will be equivalent to [01()*[] which means it matches 0 or 1 or ( or ) or [
The correct translation of the regular expression 0∗1∗0∗+(1+00+000)∗+(0+1+0+)∗
will be as follows:
^((?:0*1*0*)|(?:1|00|000)*|(?:0+1+0+)*)$
regex101
Debuggex Demo
What your regex [(0*1*0*)[1(00)(000)]*(0+1+0+)*] does:
[(0*1*0*)[1(00)(000)]* -> matches any of characters 0,(,),*,[ zero or more times followed by
(0+1+0+)* --> matches the pattern 0+1+0+ 0 or more times followed by
] --> matches the character ]
so you expression is equivalent to
[([)01](0+1+0+)*] which is not a regular expression to match strings that do not contain 101
0* 1* ( (00+000)* 1*)* (ε+0)
i think this expression covers all cases because --
any number apart from 1 can be broken into constituent 2's and 3's i.e. any number n=2*i+3*j. So there can be any number of 0's between 2 consecutive 1's apart from one 0.Hence, 101 cannot be obtained.
ε+0 for expressions ending in one 0.
The RE for language not containing 101 as sub-string can also be written as (0*1*00)*.0*.1*.0*
This may me a smaller one then what you are using. Try to make use of this.
Regular Expression I got (0+10)1. (looks simple :P)
I just considered all cases to make this.
you consider two 1's we have to end up with continuous 1's
case 1: 11111111111111...
case 2: 0000000011111111111111...(once we take two 1's we cant accept 0's so one and only chance is to continue with 1's)
if you consider only one 1 which was followed by 0 So, no issue and after one 1 we can have any number of 0's.
case 3: 00000000 10100100010000100000100000 1111111111
=>(0*+10*)1
final answer (0+10)1.
Thanks for your patience.
So, I've built a regex which follows this:
4!a2!a2!c[3!c]
which is translated to
4 alpha character followed by
2 alpha characters followed by
2 characters followed by
3 optional character
this is a standard format for SWIFT BIC code HSBCGB2LXXX
my regex to pull this out of string is:
(?<=:32[^:]:)(([a-zA-Z]{4}[a-zA-Z]{2})[0-9][a-zA-Z]{1}[X]{3})
Now this is targeting a specific tag (32) and works, however, I'm not sure if it's the cleanest, plus if there are any characters before H then it fails.
the string being matched against is:
:32B:HsBfGB4LXXXHELLO
the following returns HSBCGB4LXXX, but this:
:32B:2HsBfGB4LXXXHELLO
returns nothing.
EDIT
For clarity. I have a string which contains multiple lines all starting with :2xnumber:optional letter (eg, :58A:) i want to specify a line to start matching in and return a BIC from anywhere in the line.
EDIT
Some more example data to help:
:20:ABCDERF Z
:23B:CRED
:32A:140310AUD2120,
:33B:AUD2120,
:50K:/111222333
Mr Bank of Dad
Dads house
England
:52D:/DBEL02010987654321
address 1
address 2
:53B:/HSBCGB2LXXX
:57A://AU124040
AREFERENCE
:59:/44556677
A line which HSBCGB2LXXX contains a BIC
:70:Another line of data
:71A:Even more
Ok, so I need to pass in as a variable the tag 53 or 59 and return the BIC HSBCGB2LXXX only!
Your regex can be simplified, and corrected to allow a character before the H, to:
:32[^:]:.?([a-zA-Z]{6}\d[a-zA-Z]XXX)
The changes made were:
Lost the look behind - just make it part of the match
Inserting .? meaning "optional character"
([a-zA-Z]{4}[a-zA-Z]{2}) ==> [a-zA-Z]{6} (4+2=6)
[0-9] ==> \d (\d means "any digit")
[X]{3} ==> XXX (just easier to read and less characters)
Group 1 of the match contains your target
I'm not quite sure if I understand your question completely, as your regular expression does not completely match what you have described above it. For example, you mentioned 3 optional characters, but in the regexp you use 3 mandatory X-es.
However, the actual regular expression can be further cleaned:
instead of [a-zA-Z]{4}[a-zA-Z]{2}, you can simply use [a-zA-Z]{6}, and the grouping parentheses around this might be unnecessary;
the {1} can be left out without any change in the result;
the X does not need surrounding brackets.
All in all
(?<=:32[^:]:)([a-zA-Z]{6}[0-9][a-zA-Z]X{3})
is shorter and matches in the very same cases.
If you give a better description of the domain, probably further improvements are also possible.
I'm trying to make a expression to verify that the string supplied is a valid format, but it seems that if I don't use regex in a few months, I forget everything I learned and have to relearn it.
My expression is supposed to match a format like this: 010L0404FFCCAANFFCC00M000000XXXXXX
The four delimiters are (L, N, K, M) which arent in the 0-9A-F hexidecimal range to indicate uniqueness must be in that order or not in the list. Each delimiter can only exist once!
It breaks down to this:
Starts off with a 3 digit numbers, which is simply ^([0-9]{3}) and is always required
Second set begins with L, and must be 2 digits + 2 digits + 6 hexdecimal and is optional
Third set begins with N and must be a 6 digit hexdecimal and is optional
The fourth set K is simply any amount of numbers and is optional
The fifth set is M and can be any 6 hexdecimals or XXXXXX to indicate nothing, it must be in multiples of 6 excluding 0, like 336699 (6) or 336699XXXXXXFFCC00 (18) and is optional
The hardest part I cant figure out making it require it in that order, and in multiples, like the L delimiter must come before and K always if it's there (the reason so I don't get variations of the same string which means the same thing with delimiters swapped). I can already parse it, I just want to verify the string is the correct format.
Any help would be appreciated, thanks.
Requiring the order isn't too bad. Just make each set optional. The regex will still match in order, so if the L section, for example, isn't there and the next character is N, it won't let L occur later since it won't match any of the rest of the regex.
I believe a direct translation of your requirements would be:
^([0-9]{3})(L[0-9]{4}[0-9A-F]{6})?(N[0-9A-F]{6})?(K[0-9]+)?(M([0-9A-F]{6}|X{6})+)?$
No real tricks, just making each group optional except for the first three digits, and adding an internal alternative for the two patterns of six digits in the M block.
^([0-9]{3})(L[0-9]{4}[0-9A-F]{6})?(N[0-9A-F]{6})?(K[0-9]+)?(M([0-9A-F]{6})+|MX{6})$