Menu
I want to match all characters after 8th character. And not include first 8!
I need exactly a regular expression cause a framework (Ace.js) requires a regexp, not a string. So, this is not an option:
var substring = "123456789".substr(5);
Can I match everything after nth character using regex in JavaScript?
Updates: I can't call replace(), substring() etc because I don't have a string. The string is known at run time and I don't have access to it. As I already said above the framework (Ace.js) asks me for a regex.
671
$ means "Match the end of the string" (the position after the last character in the string). Both are called anchors and ensure that the entire string is matched instead of just a substring.
To match a character having special meaning in regex, you need to use a escape sequence prefix with a backslash ( \ ). E.g., \. matches "." ; regex \+ matches "+" ; and regex \( matches "(" .
The fullmatch() function returns a Match object if the whole string matches the search pattern of a regular expression, or None otherwise. The syntax of the fullmatch() function is as follows: re.fullmatch(pattern, string, flags=0)
A regular expression followed by an asterisk ( * ) matches zero or more occurrences of the regular expression. If there is any choice, the first matching string in a line is used. A regular expression followed by a plus sign ( + ) matches one or more occurrences of the one-character regular expression.
(?<=^.{8}).*
will match everything after the 7th position. Matches 89 in 0123456789
or IJKLM in ABCDEFGHIJKLM
etc
73
console.log("123456789".match(/^.{8}(.*)/)[1])
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
