I need to check for a link where title="Week 1" or title="Week 13", I have tried somethings what I found on the internet but it won`t work:
http://regexr.com?38bet
So, how to match if something have one number or two numbers?
You can use \d{1,2} This will match between 1 and 2 digits.
I have also found that the following this website is very helpful when I need to write regexes.
Use a zero-or-one quantifier (?), like this:
title="[Ww]eek [1-9][0-9]?"
This will match a string which contains title=" followed by an upper- or lower-case w, followed by eek, a space, and digit from 1 to 9, followed by an optional digit from 0 to 9, followed by a ".
You should also consider using the case-insensitive flag in the language of your choice.
The answer depends on which context you're using the regular expression.
In python, ? will match 0 or 1 characters.
In JavaScript, ? will also match 0 or 1 characters.
In vi, \= will match 0 or 1 characters.
So the answer(s) would be:
Python regex: Week 13?
JavaScript regex: /Week 13?/
vi regex: Week 13\=
Which will match both:
Week 1
Week 13
for java try this one:
String regex = "[\\d{2}]+";
Try this :
title="[Ww]eek (\d+)\.pdf"
RegExr :
http://regexr.com?38bfc
Matches :
1
13
Related
I have the following RegEx that is supposed to do 24 hours time format validation, which I'm trying out in https://rubular.com
/^[0-23]{2}:[0-59]{2}:[0-59]{2}$/
But the following times fails to match even if they look correct
02:06:00
04:05:00
Why this is so?
In character classes, you're supposed to denote the range of characters allowed (in contrast to the numbers you want to match in your example). For minutes and seconds, this is relatively straight-forward - the following expression
[0-5][0-9]
...will match any numerical string from "00" to "59".
But for the hours, you need to two separate expressions:
[01][0-9]|2[0-3]
...one to match "00" to "19" and one to match "20" to "23". Due to the alternative used (| character), these need to be grouped, which adds another bit of syntax (?:...). Finally we're just adding the anchors ^ and $ for beginning and end of string, which you already had where they belong.
^(?:[01][0-9]|2[0-3]):[0-5][0-9]:[0-5][0-9]$
You can check this solution out at regex101, if you like.
Your problem is that you understand characters ranges wrong: 0-23 doesn't mean "match any number from 0 to 23", it means: 0-2- match one digit: 0,1 or 2, then match 3.
Try this pattern: (?:[01][0-9]|2[0-3])(?::[0-5][0-9]){2}
Explanation:
(?:...) - non-capturing group
[01][0-9]|2[0-3] - alternation: match whether 0 or one followed by any digits fro 0 to 9 OR 2 followed by 0, 1, 2 or 3 (number from 00-23)
(?::[0-5][0-9]){2} - match : and [0-5][0-9] (basically number from 00-59) twice
Demo
use this (([0-1]\d|[2][0-3])):(([0-5][0-9])):(([0-5][0-9]))
Online demo
I have files with these filename:
ZATR0008_2018.pdf
ZATR0018_2018.pdf
ZATR0218_2018.pdf
Where the 4 digits after ZATR is the issue number of magazine.
With this regex:
([1-9][0-9]*)(?=_\d)
I can extract 8, 18 or 218 but I would like to keep minimum 2 digits and max 3 digits so the result should be 08, 18 and 218.
How is possible to do that?
You may use
0*(\d{2,3})_\d
and grab Group 1 value. See the regex demo.
Details
0* - zero or more 0 chars
(\d{2,3}) - Group 1: two or three digits
_\d - a _ followed with a digit.
Here is a PCRE variation that grabs the value you need into a whole match:
0*\K\d{2,3}(?=_\d)
See another regex demo
Here, \K makes the regex engine omit the text matched so far (zeros) and then matches 2 to 3 digits that are followed with _ and a digit.
(?:[1-9][0-9]?)?[0-9]{2}(?=_[0-9])
or perhaps:
(?:[1-9][0-9]+|[0-9]{2})(?=_[0-9])
(https://www.freeformatter.com/regex-tester.html, which claims to use the XRegExp library, that you mention in another answer doesn't seem to backtrack into the (?:)? in my first suggestion where necessary, which makes it very different from any regex engine I've encoutered before and makes it prefer to match just the 18 of 218 even though it starts later in the string. But it does work with my second suggestion.
([1-9]\d{2,3})(?=_\d)
{x,y} will match from x to y times the previous pattern, in this case \d
Edit: from your own regex it looked as you wanted the part of the number which starts with a non-zero. However since your examples include leading 0s, maybe you really wanted :
(\d{2,3})(?=_\d)
Which will give you the last 3 digits before underscore unless there are only 2 digits.
I propose you:
^ZATR0*(\d{2,3})_\d+\.pdf$
demo code here. Result:
Match 1 Full match 0-17 ZATR0008_2018.pdf Group 1. 6-8 08
Match 2 Full match 18-35 ZATR0018_2018.pdf Group 1. 24-26 18
Match 3 Full match 36-53 ZATR0218_2018.pdf Group 1. 41-44 218
My regex is weak, in the case of the following string
"OtherId":47
"OtherId":7
"MyId":47 (Match this one)
"MyId":7
I want to pick up the string that has "MyId" and a number that is not 1 - 9
I thought I could just use:
RegEx: How can I match all numbers greater than 49?
Combined using:
Regular Expressions: Is there an AND operator?
But its not happening... you can see my failed attempt here:
https://www.regextester.com/index.php?fam=99753
Which is
\b"MyId":\b(?=.*^[0-10]\d)
What am I doing wrong?
You can use this regex to match any digit >= 10:
^"MyId":[1-9][0-9]+$
RegEx Demo
If leading zeroes are to be allowed as well then use:
^"MyId":0*[1-9][0-9]+$
[1-9] makes sure number starts with 1-9 and [0-9]+ match 1 or more any digits after first digit.
Essentially, you are looking for 2 or more digits:
\"MyId\"\:(\d{2,})
I have escaped the quotes and colon, and {2,} means 2 or more.
If you need exact match to any number greater than 9
^"MyId":[1-9][0-9]+$
I am trying to write a regular expression that ensure if there's a comma then the following text should be 1 or 2 digits numeric.
Here's what I have so far.
(^\d{0,2})+(,\d{0,2})*$
The works in most cases but it is considering the following as valid.
12,22,,,,,,,,,, and 12,22,,,,,,,,,,12,12
What did I do wrong? Thanks!
\d{0,2} means "between 0 and 2 digits". It should be \d{1,2}
You are matching 0 to 2 digits after the comma instead of 1 or 2
the following should do the trick
(^\d{1,2})+(,\d{1,2})*$
Use a negative look-ahead to assert that there aren't 3 digits after a comma, and keep the main regex simply "all commas or digits"
^(?!.*,\d{3})[,\d]+$
I need a regular expression that requires at least ONE digits and SIX maximum.
I've worked out this, but neither of them seems to work.
^[0-9][0-9]\?[0-9]\?[0-9]\?[0-9]\?[0-9]\?$
^[0-999999]$
Any other suggestion?
You can use range quantifier {min,max} to specify minimum of 1 digit and maximum of 6 digits as:
^[0-9]{1,6}$
Explanation:
^ : Start anchor
[0-9] : Character class to match one of the 10 digits
{1,6} : Range quantifier. Minimum 1 repetition and maximum 6.
$ : End anchor
Why did your regex not work ?
You were almost close on the regex:
^[0-9][0-9]\?[0-9]\?[0-9]\?[0-9]\?[0-9]\?$
Since you had escaped the ? by preceding it with the \, the ? was no more acting as a regex meta-character ( for 0 or 1 repetitions) but was being treated literally.
To fix it just remove the \ and you are there.
See it on rubular.
The quantifier based regex is shorter, more readable and can easily be extended to any number of digits.
Your second regex:
^[0-999999]$
is equivalent to:
^[0-9]$
which matches strings with exactly one digit. They are equivalent because a character class [aaaab] is same as [ab].
^\d{1,6}$
....................
You could try
^[0-9]{1,6}$
it should work.
^[0-9]{1,6}$ should do it. I don't know VB.NET good enough to know if it's the same there.
For examples, have a look at the Wikipedia.
\b\d{1,6}\b
Explanation
\b # word boundary - start
\d # any digits between 0 to 9 (inclusive)
{1,6} # length - min 1 digit or max 6 digits
\b # word boundary - end
^[a-zA-Z0-9]{1,6}$
regex 6 digit number and alphabet in angular
/^[0-9][0-9][0-9][0-9]$/
Enter 4 digit number only