I am close, but I need some help to complete a regex. Here is the goal:
Should succeed:
10.05
3.00
50
Should fail:
55.99 (>50)
3.001 (can't have the "1" at the end)
0.50 (< 3)
.99 (< 3)
$50 (can't have "$")
5.2 (if decimal, must have 2 digits after)
Here's the regex I have so far, but it doesn't quite do all the above correctly:
^([1-4][0-9]|50|[3-9])+(\.[0-9][0-9])?$
Can anyone share the answer? Thanks!
^(50(\.00)?|([1-4][0-9]|[3-9])(\.[0-9][0-9])?)$
There were two issues. Firstly, you had allowed non-zero values after the decimal point, even if the value before it was 50. So I separated that out on the top level. Secondly, just remove the +. Because due to it, you can have much larger numbers (by chaining 50 and 43 together, for instance).
However, as Bergi mentioned in a comment, it would be better to just check the format, and do the range check separately (without regex). This would be the format check:
^\d+(\.\d\d)?$
I found a online utility that returns a regex for integers when input the lower and upper limits of the range you want. I used it for the part before the . with limits 3-50 and after the . with limits 0-99. Here is the result:
^0*([3-9]|[1-4][0-9]|50)(\.[0-9]{2})?$
A quick glance... just remove the +
^([1-4][0-9]|50|[3-9])(.[0-9][0-9])?$
You should remove the + before the potential cents. Also, you will need to handle 50$ as a special case, because it can only have .00 after it and not any cent amount.
Also, I changed the [0-9] to the shortcut for digits: \d
/^((0?[3-9]|[1-4]\d)(\.\d\d)?|50(\.00)?)$/
Related
source txt file:
34|Gurla Mandhata|7694|25243|2788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1985|6 (4)|China
command input:
:%s/\(\d\+\)\(\d\d\d\)/\1,\2/g
command output:
34|Gurla Mandhata|7,694|25,243|2,788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1,985|6 (4)|China
Desired output:
34|Gurla Mandhata|7,694|25,243|2,788|Nalakankar Himalaya|30°26'19"N
81°17'48"E|Dhaulagiri|1985|6 (4)|China
Basically 1985 is supposed to be 1985 and not 1,985. I tried to put a \? so every time the pattern matches it stops and a °+ after so it has to detect a ° to match the pattern, but no success. It just replaces the ° and everything before that, complete mess.
My knowledge of regular expressions however combined with the substitute is weak and I'm stuck here.
EDIT
the first 3 numbers represent heights of mountains, those 3 need to change with a (,) and the last number ( 1985 ) represents a year, which must not be changed.
Mathematical solutions are not going to work as loophole since there are mountains with a height off less than 1900
You haven't told us what is the difference between 1985 and other numbers, so I assumed that your "small" numbers are less than 2000.
You almost got it:
:%s/(\d*[2-90])(\d\d\d)/\1,\2/g
Alternatively if that isn't what you want, you can use c flag (:h s_flags):
:%s/\(\d\+\)\(\d\d\d\)/\1,\2/gc
this line will leave the last 3 columns untouched, just do substitution on the content before it:
%s/\v(.*)((\|[^|]*){3}$)/\=substitute(submatch(1),'\v(\d+)(\d{3})','\1,\2','g').submatch(2)/g
Note that the above line will change 1000000 into 1000,000 instead of 1,000,000. Vim's printf() doesn't support %'d, it is pity. If you do have number > 1m, we can find other solutions.
update
I solved it myself, by using 3 seperate commands; one for every number string in the file:
%s/^\(\d*|[^|]*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
:%s/^\(\d*|[^|]*|\d\+,*\d*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
:%s/^\(\d*|[^|]*|\d\+,*\d*|\d\+,*\d*|\)\(\d\+\)\(\d\d\d\)|/\1\2,\3|/g
In case you want to use perl:
:%!perl -F'\|' -lane 'for(#F[2..4]) { s/(\d+)(\d{3})/\1,\2/;} print join "|", #F'
I need help forming regex to limit user input to only numerics and only up to 10 occurrences.
I have regex that is working to keep input to numerics only, but I cannot limit it to up to 10.
Here is what I have:
^(0|[1-9][0-9]*)$
I am okay accepting negative numbers, decimals, and 0's. Any advice?
^\s*([0-9)+){0,10}\s*$
This basically says I want to 0 to 10 things, where each thing is all digits. I added the \s* on either side to allow the user to have put spaces before or after their numbers. This would accept things like
10 1231231 1231 1231 23112 123123
If what you really want is just a single number, that is only up to 10 digits, it is even easier:
^\s[0-9]{1,10}\s$
The regex you're looking for is this:
/(?=^[-+]?\d*\.?\d+$)^.{1,10}$/
Keep in mind that this regex will allow maximum length of input to 10 which includes optional + or - sign at start and a decimal point ..
you can try this:
^(?<=\s)(\-?[\d]{1,10}(?=\s))$
This fails as it has 11 digits
Debug.WriteLine(Regex.IsMatch("12345678901", #"^\d{1,10}$").ToString());
Posted the above answer before you clarified you want up to 10 set of numbers delimited by space.
Tested the accepted answer and in .NET Regex it fails for me.
Even fixing the syntax error it still does not parse by space
Give this a try
Debug.WriteLine(Regex.IsMatch(" 12345 678901 12 ", #"^\s*([+-]?\d+)(\s+[+-]?\d+){1,9}\s*$").ToString());
I'm currently using ([1-9]|1[0-2]) to represent inputs from 1 to 12. (Leading zeros not allowed.)
However it seems rather hacky, and on some days it looks outright dirty.
☞ Is there a proper in-built way to do it?
☞ What are some other ways to represent number ranges?
I tend to go with forms like [2-9]|1[0-2]? which avoids backtracking, though it makes little difference here. I've been conditioned by XML Schema to avoid such "ambiguities", even though regex can handle them fine.
Yes, the correct one:
[1-9]|1[0-2]
Otherwise you don't get the 10.
Here is the better answer, with exact match from 1 - 12.
(^0?[1-9]$)|(^1[0-2]$)
Previous answers doesn't really work well with HTML input regex validation, where some values like '1111' or '1212' will still treat it as a valid input.
You can use:
[1-9]|1[012]
How about:
^[1-9]|10|11|12$
Matches 0-9 or 10 or 11 or 12. thats it, nothing else is matched.
You can try this:
^[1-9]$|^[1][0-2]$
Use the following pattern (0?[1-9]|1[0-2]) use this which will return values from 1 to 12 (January to December) even if it initially starts with 0 (01, 02, 03, ..., 09, 10, 11, 12)
The correct patter to validate numbers from 1 to 12 is the following:
(^[1-9][0-2]$)|(^[1-9]$)
The above expression is useful when you have an input with type number and you need to validate month, for example. This is because the input type number ignores the 0 in front of any number, eg: 01 it returns 1.
You can see it in action here: https://regexr.com/5hk0s
if you need to validate string numbers, I mean, when you use an input with type text but you expect numbers, eg: expiration card month, or months the below expression can be useful for you:
((^0[1-9]$)|(^1[0-2]$))
You can see it in action here https://regexr.com/5hkae
I hope this helps a lot because it is very tricky.
Regards.
In python this matches any number between 1 - 12:
12|11|10|9|8|7|6|5|4|3|2|1
The descending order matters. In ascending order 10, 11 and 12 would match 1 instead as regex usually pick the first matching value.
I am currently trying to validate UK telephone numbers:
The format I'm looking for is: 01234 567891 or 01234567891 - So I need the number to have 5 numbers then a space then 6 numbers or simply a 11 numbers.
The number must start with a 0.
I've had a look at a couple of examples:
/^[0-9]{10,11} - to check that the chars are all numbers
/^0[0-9]{9,10}$/ - to check that the first number is a 0
I'm just unsure how to put all these together and check if there is a space or not.
Could someone help me with this regex?
Thanks
Try this regex:
/^0\d{4}\s?\d{6}$/
Many people try to do input validation and formatting in a single step.
It is better to separate these processes.
Match UK telephone number in any format
^(?:(?:\(?(?:0(?:0|11)\)?[\s-]?\(?|\+)44\)?[\s-]?(?:\(?0\)?[\s-]?)?)|(?:\(?0))(?:(?:\d{5}\)?[\s-]?\d{4,5})|(?:\d{4}\)?[\s-]?(?:\d{5}|\d{3}[\s-]?\d{3}))|(?:\d{3}\)?[\s-]?\d{3}[\s-]?\d{3,4})|(?:\d{2}\)?[\s-]?\d{4}[\s-]?\d{4}))(?:[\s-]?(?:x|ext\.?|\#)\d{3,4})?$
The above pattern allows the user to enter the number in any format they are comfortable with. Don't constrain the user into entering specific formats.
Extract NSN, prefix and extension
^(\(?(?:0(?:0|11)\)?[\s-]?\(?|\+)(44)\)?[\s-]?)?\(?0?(?:\)[\s-]?)?([1-9]\d{1,4}\)?[\d[\s-]]+)((?:x|ext\.?|\#)\d{3,4})?$
Next, extract the various elements.
$2 will be '44' if international format was used, otherwise assume national format with leading '0'.
$4 contains the extension number if present.
$3 contains the NSN part.
Validation and formatting
Use further RegEx patterns to check the NSN has the right number of digits for this number range. Finally, store the number in E.164 format or display it in E.123 format.
There's a very detailed list of validation and display formatting RegEx patterns for UK numbers at:
http://www.aa-asterisk.org.uk/index.php/Regular_Expressions_for_Validating_and_Formatting_UK_Telephone_Numbers
It's too long to reproduce here and it would be difficult to maintain multiple copies of this document.
If you are looking for all UK numbers, I'd look for a bit more than just that number, some are in the format 020 7123 4567 etc.
^\s*\(?(020[7,8]{1}\)?[ ]?[1-9]{1}[0-9{2}[ ]?[0-9]{4})|(0[1-8]{1}[0-9]{3}\)?[ ]?[1-9]{1}[0-9]{2}[ ]?[0-9]{3})\s*$
/\d*(*)*+*-*/
Simple Telephone Regex includes + () and - anywhere, as well as digits
I think ^0[\d]{4}\s?[\d]{5,6}} will work for you. I have used [\d] instead of [0-9].
I find that RegExr is a useful online tool to check and try your regular expressions. It also has a nice library of examples to help point you in the right direction
you should just count the number of digits and check that it's 10,
Some UK numbers have only 9 digits, not 10 (not including the leading 0).
These include 40 of the 01 area codes (using "4+5" format), the 016977 area code (using "5+4" format), all 0500 numbers and some 0800 numbers.
There's a list at: http://www.aa-asterisk.org.uk/index.php/01_numbers
This US numbers pattern accepts following phones as well:
800-432-4500, Opt: 9, Ext: 100316
800-432-4500, Opt: 9, Ext: X100316
800-432-4500, Option #3
(?:(?:\+?1\s*(?:[.-]\s*)?)?(?:\(\s*([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9])\s*\)|([2-9]1[02-9]|[2-9][02-8]1|[2-9][02-8][02-9]))\s*(?:[.-]\s*)?)?([2-9]1[02-9]|[2-9][02-9]1|[2-9][02-9]{2})\s*(?:[.-]\s*)?([0-9]{4}),?(?:\s*(?:#|x\.?|opt(\.|:|\.:)?|option)\s*#?(\d+))?,?(?:\s*(?:#|x\.?|ext(\.|:|\.:)?|extension)\s*(\d+))?
(used this answer in other topic as start point)
I know this is a long shot, but I have a huge text file and I need to add a given number to other numbers matching some criteria.
Eg.
identifying text 1.1200
identifying text 1.1400
and I'd like to transform this (by adding say 1.15) to
identifying text 2.2700
identifying text 2.2900
Normally I'd do this in Python, but it's on a Windows machine where I can't install too many things. I've got Vim though :)
Here is a simplification and a fix on hobbs' solution:
:%s/identifying text \zs\d\+\(.\d\+\)\=/\=(1.15+str2float(submatch(0)))/
Thanks to \zs, there is no need to recall the leading text. Thanks to str2float() a single addition is done on the whole number (in other words, 1.15 + 2.87 will give the expected result, 4.02, and not 3.102).
Of course this solution requires a recent version of Vim (7.3?)
You can do a capturing regex and then use a vimscript expression as a replacement, something like
:%s/\(identifying text \)\(\d\+\)\.\(\d\+\)/
\=submatch(1) . (submatch(2) + 1) . "." . (submatch(3) + 1500)
(only without the linebreak).
Your number format seems to be a fixed one, so it's easy to convert to int and come back (remove the dot) add 11500 and put the dot back.
:%s/\.//
:%normal11500^A " type C-V then C-a
:%s/....$/.&/
If you don't want to do that on all the lines but only the one which match 'identifying text' replace all the % by 'g/indentifying text/'
For integers you can just use n^A to add n to a number (and n^X to subtract it). I doubt whether that works for fractional numbers though.
Well this might not be a solution for vim but I think awk can help:
cat testscript | LC_ALL=C awk '{printf "%s %s %s %s %.3f\n", $1,$2,$3,$4,$5+1.567 }'
and the test
this is a number 1.56
this is a number 2.56
this is a number 3.56
I needed the LC_ALL=C for the correct conversion of the floating point separator, and maybe there is a more elegant solution for printing the beginning/ rest of the string. And the result looks like:
this is a number 3.127
this is a number 4.127
this is a number 5.127
Using macro
qa .......................... start record macro 'a'
/iden<Enter> ................ search 'ident*' press Enter
2w .......................... jump 2 words until number one (before dot)
Ctrl-a ...................... increases the number
2w .......................... jump to number after dot
1500 Ctrl-a ................. perform increases 1500 times
q ........................... stop record to macro 'a'
if you have 300 lines with this pattern just now making
300#a