I have this regex (\w+) replace with \u$0
This makes first letter caps for example: james1 to James1.
But I need a regex to make the first letter caps of each word when it starts with a number for example
12james
1azz4ds
1995brandon
666metal
to
12James
1Azz4ds
1995Brandon
666Metal
How do I solve this problem?
Here, we can also collect the digits, then letters maybe both upper or lowercase, and replace it:
[0-9]+([A-Za-z])
We will be adding a start char to capture only those letters that we wish to replace:
^[0-9]+([A-Za-z])
or:
^([0-9]+)([A-Za-z])
and for this expression our replacement would look like to something similar to:
$1\u$2
RegEx
If this expression wasn't desired, it can be modified or changed in regex101.com.
RegEx Circuit
jex.im visualizes regular expressions:
You could match a word boundary \b, match 1+ digits \d+ and then forget what is matched using \K. Then match a single lowercase a-z:
\b\d+\K[a-z]
Replace with:
\u$0
See a Regex demo
If there can be not a non whitespace before the digits, instead of using \b you might also use:
(?<!\S)\d+\K[a-z]
See another Regex demo
Related
For example we have a string:
asd/asd/asd/asd/1#s_
I need to match this part: /asd/1#s_ or asd/1#s_
How is it possible to do with plain regex?
I've tried negative lookahead like this
But it didn't work
\/(?:.(?!\/))?(asd)(\/(([\W\d\w]){1,})|)$
it matches this '/asd/asd/asd/asd/asd/asd/1#s_'
from this 'prefix/asd/asd/asd/asd/asd/asd/1#s_'
and I need to match '/asd/1#s_' without all preceding /asd/'s
Match should work with plain regex
Without any helper functions of any programming language
https://regexr.com/
I use this site to check if regex matches or not
here's the possible strings:
prefix/asd/asd/asd/1#s
prefix/asd/asd/asd/1s#
prefix/asd/asd/asd/s1#
prefix/asd/asd/asd/s#1
prefix/asd/asd/asd/#1s
prefix/asd/asd/asd/#s1
and asd part could be replaced with any word like
prefix/a1sd/a1sd/a1sd/1#s
prefix/a1sd/a1sd/a1sd/1s#
...
So I need to match last repeating part with everything to the right
And everything to the right could be character, not character, digit, in any order
A more complicated string example:
prefix/a1sd/a1sd/a1sd/1s#/ds/dsse/a1sd/22$$#!/123/321/asd
this should match that part:
/a1sd/22$$#!/123/321/asd
If you want the match only, you can use \K to reset the match buffer right before the parts that you want to match:
^.*\K/a\d?sd/\S+
The pattern will match
^ Start of string
.* Match any char except a newline until end of the line
\K Forget what is matched until now
/a\d?sd/ match a, optional digits and sd between forward slashes
\S+ Match 1+ non whitespace chars
See a regex demo
I know some basics of the RegEx but not a pro in it. And I am learning it. Currently, I am using the following very very simple regex to match any digit in the given sentence.
/d
Now, I want that, all the digits except some patterns like e074663 OR e123444 OR e7736 should be excluded from the match. So for the following input,
Edit 398e997979 the Expression 9798729889 & T900980980098ext to see e081815 matches. Roll over matches or e081815 the expression e081815 for details.e081815 PCRE & JavaScript flavors of RegEx are e081815 supported. Validate your expression with Tests mode e081815.
Only bold digits should be matched and not any e081815. I tried the following without the success.
(^[e\d])(\d)
Also, going forward, some more patterns needs to be added for exclusion. For e.g. cg636553 OR cg(any digits). Any help in this regards will be much appreciated. Thanks!
Try this:
(?<!\be)(?<!\d)\d+
Test it live on regex101.com.
Explanation:
(?<!\be) # make sure we're not right after a word boundary and "e"
(?<!\d) # make sure we're not right after a digit
\d+ # match one or more digits
If you want to match individual digits, you can achieve that using the \G anchor that matches at the position after a successful match:
(?:(?<!\be)(?<=\D)|\G)\d
Test it here
Another option is to use a capturing group with lookarounds
(?:\b(?!e|cg)|(?<=\d)\D)[A-Za-z]?(\d+)
(?: Non capture group
\b(?!e|cg) Word boundary, assert what is directly to the right is not e or cg
| Or
(?<=\d)\D Match any char except a digit, asserting what is directly on the left is a digit
) Close group
[A-Za-z]? Match an optional char a-zA-Z
(\d+) Capture 1 or more digits in group 1
Regex demo
I want to find all expressions that don't end with ":"
I tried to do it like that:
[a-z]{2,}(?!:)
On this text:
foobar foobaz:
foobaz
foobaz:
The problem is, that it just takes away the last character befor the ":" and not the whole match.
Here is the example: https://regex101.com/r/jtLRvz/1
How can I get the negative lookahead work for the whole regular expression?
When [a-z]{2,}(?!:) matches baz:, [a-z]{2,} grabs 2 or more lowercase ASCII letters at once (baz) and the negative lookahead (?!:) checks the char immediately to the right. It is :, so the engine asks itself if there is a way to match the string in a different way. Since {2,} can match two chars, not currently matched three, it backtracks, and finds a valid match.
Add a-z to the lookahead pattern to make sure the char right after 2 or more lowercase ASCII letters is not a letter and not a colon:
[a-z]{2,}(?![a-z:])
^^^
See the regex demo
If your regex engine supports possessive modifiers, or atomic groups, you may use them to prevent backtracking into the [a-z]{2,} subpattern:
[a-z]{2,}+(?!:)
(?>[a-z]{2,})(?!:)
See another regex demo.
I have this regexp:
^[a-z0-9]+([.\-][a-z0-9]+)*$
I need exclude from match only one word "www".
I tried the negative lookahead but without a success.
Use a negative lookahead like this:
^(?!www$)[a-z0-9]+([.-][a-z0-9]+)*$
^^^^^^^^
This will not match a string equal to www.
See the regex demo
If you want to fail a match with strings that contain -www- or .www., use
^(?!.*\bwww\b)[a-z0-9]+([.-][a-z0-9]+)*$
See another regex demo. This pattern contains a (?!.*\bwww\b) lookahead that fails the whole match if there is a www somewhere inside the string and it has no digits or letters round it due to \b word boundaries.
I am using the following regex for detecting negative numbers:
([-]([0-9]*\.[0-9]+|[0-9]+))
But I want to skip the matches which are followed by $.
If i use the folowing regex:
([-]([0-9]*\.[0-9]+|[0-9]+)[^\$])
It will match correctly the positions but will include the following character.
For example in expression:
-0.6+3 - 3.0$
it will match:
-0.6+
I want to match only
-0.6
([-]([0-9]*\.[0-9]+|[0-9]+)(?!\$)
You need a negative lookahead here which will not consume and only make an assertion.
Remove the $ from the group:
([-]([0-9]*\.[0-9]+|[0-9]+))[^\$]
You could use this simplified regex:
(-[0-9]+(?:\.[0-9]+)?)(?!\$)
You can use the regex from Regular-Expressions.info with just minus at the beginning, and a \b added at the end in order to stop before any non-word character:
[-][0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?\b
This regex also captures the numbers with exponent part.
See demo