I need to extract first two digits after character "R".
DATA:
165/70 R13 79T
225/75 R16C 118R
205/60 R16 92H
Tried this =REGEXEXTRACT(C2,"^R(.)"), but this is not quite necessary result.
Thank you!
use:
=REGEXEXTRACT(C2; "R(\d{2})")
Related
Library context, using MarcEdit which can also use regex.
I need this:
=773 \\$tEtudes inuit$x0701-1008$1Vol. 44 1-2, $2p. 53-84
to be changed to this:
=773 \\$tEtudes inuit$x0701-1008$1Vol. 44, no. 1-2, $2p. 53-84
Problem is, the 44 in this case and the 1-2 are numbers that will change from one book to the other and I am building commands to automate it.
I tried focusing on changing the space between the 44 and the 1-2 into a ', no. ' with \s but it obiviously changes all spaces characters.
The adding ', no. ' is easy because there is a different box for it but I can't focus on the 2nd space while ignoring the first and last and also keeping every characters before and after.
Thank you for helping, I've been looking/trying all day!
MarcEdit exemple
If the regular expression implementation supports look ahead, you can require that this space is followed by a range and a comma:
Find: \s(?=\d+-\d+,)
Replace: , no.
I am not good in regex and I spent so much time figure out how to search for the below pattern:
,\"16 questions\",
This is what I constructed .\"[0-9,]+ questions\".
I think I am close but not sure how much. Can someone please correct it. The numeric value can have comma in it when the number crosses 1k. e.g 2,500 questions.
,"\d{1,3}(,\d{3,3})*\squestions?",
Explanation:
\d{1,3}= 1~3 decimal digits
(,\d{3,3})* = comma and 3 decimal digits, the whole group repeating 0~N times
\s = whitespace
s? = letter s can be missing
These two parts give you accurate recognition of possible numbers.
▶ Test and visualization.
If the backslashes in your text are true backslashes, then the regex including them would be
,\\"\d{1,3}(,\d{3,3})*\squestions?\\",
This works. You didn't indicate if the numeric value could have more than 1 comma (e.g. 1,000,000)
,\\"((\d{1,3})(,\d{3})*)\squestions?\\",
I quickly found a way to get a working multi-line regular expression for my needs, but having trouble with its conversion into a single line.
So, consider this input with regex /^[2-9]\d{1}(?:\s){0}/gm applied:
4126-54D429-001,
5149-A42102-002,
9251-Z48910-003
...
However, when I turn it to one line, I'm getting only first two digits in ouput:
4126-54D429-001, 5149-A42102-002, 9251-Z48910-003 ...
How can this regexp be written to get this capture:
4126-54D429-001, 5149-A42102-002, 9251-Z48910-003 ... ?
This Should Work.
REGEXP
\b\d{2}(?=\d{2})
INPUT
4126-54D429-001, 5149-A42102-002, 9251-Z48910-003, 7851-Z48910-003
OUTPUT
41
51
92
78
The comma is not essential
If i help u, mark me as correct and vote up
This will capture the first two digits of each in groups:
(\d{2})[^,]*
Am playing around with regexp's but this is my headache. I have a dynamic number which needs a suffix. The suffix is always 0 to 9, 99 or 999.
Example:
I have the number 461200 and now I want to create an regexp that will match 461200 to 461209. What I've learned it should be ^46120[0-9]$? Is this correct or somewhere to the left of hell?
Ok, let us assume it is correct and I now want to match 461200 - 461299? This is where I get lost.
^4612[0-9]{2}?
It cannot be. I am yet to figure this out.
Any help appreciated.
For 1 digit at the end you need:
^4612[0-9]$
2 digits at the end:
^4612[0-9]{2}$
3 digits at the end:
^4612[0-9]{3}$
The number in braces {} means the number of time the preceding character or set has to be repeated.
Ok, let us assume it is correct and I now want to match 461200 -
461299?
You can either repeat the desired character class by saying [0-9][0-9] or use quantifiers [0-9]{2}.
It can be either:
^4612[0-9][0-9]$
or
^4612[0-9]{2}$
Both would work.
maybe try this regex:
^4612\d{2}$
I'm trying to write a regex pattern that will find numbers with two leading 00's in it in a string and replace it with a single 0. The problem is that I want to ignore numbers in parentheses and I can't figure out how to do this.
For example, with the string:
Somewhere 001 (2009)
I want to return:
Somewhere 01 (2009)
I can search by using [00] to find the first 00, and replace with 0 but the problem is that (2009) becomes (209) which I don't want. I thought of just doing a replace on (209) with (2009) but the strings I'm trying to fix could have a valid (209) in it already.
Any help would be appreciated!
Search one non digit (or start of line) followed by two zeros followed by one or more digits.
([^0-9]|^)00[0-9]+
What if the number has three leading zeros? How many zeros do you want it to have after the replacement? If you want to catch all leading zeros and replace them with just one:
([^0-9]|^)00+[0-9]+
Ideally, you'd use negative look behind, but your regex engine may not support it. Here is what I would do in JavaScript:
string.replace(/(^|[^(\d])00+/g,"$10");
That will replace any string of zeros that is not preceded by parenthesis or another digit. Change the character class to [^(\d.] if you're also working with decimal numbers.
?Regex.Replace("Somewhere 001 (2009)", " 00([0-9]+) ", " 0$1 ")
"Somewhere 01 (2009)"