I need is the last match. In the case below the word test without the $ signs or any other special character:
Test String:
$this$ $is$ $a$ $test$
Regex:
\b(\w+)\b
5 Answers
The $ represents the end of the string, so...
\b(\w+)$
However, your test string seems to have dollar sign delimiters, so if those are always there, then you can use that instead of \b.
\$(\w+)\$$
var s = "$this$ $is$ $a$ $test$";
document.body.textContent = /\$(\w+)\$$/.exec(s)[1];If there could be trailing spaces, then add \s* before the end.
\$(\w+)\$\s*$
And finally, if there could be other non-word stuff at the end, then use \W* instead.
\b(\w+)\W*$
4 Comments
Garis M Suero
I think the only valid answer is your last regex. The first three does not meet the confirmation criteria.
six fingered man
@GarisMSuero: I don't know what the actual string may be. If the test string is representative of the actual string format then all but the first should work. If he's using
$ as a weird way to represent the word boundaries, then the first one would work. The question isn't entirely clear, so I gave options for each.Garis M Suero
yeah, agree but he also stated: " without the $ signs or any other special character, how can this be achieved?"
six fingered man
@GarisMSuero: Yes, you're right. The last one would be it then.
In some cases a word may be proceeded by non-word characters, for example, take the following sentence:
Marvelous Marvin Hagler was a very talented boxer!
If we want to match the word boxer all previous answers will not suffice due the fact we have an exclamation mark character proceeding the word. In order for us to ensure a successful capture the following expression will suffice and in addition take into account extraneous whitespace, newlines and any non-word character.
[a-zA-Z]+?(?=\s*?[^\w]*?$)
https://regex101.com/r/D3bRHW/1
We are informing upon the following:
- We are looking for letters only, either uppercase or lowercase.
- We will expand only as necessary.
- We leverage a positive lookahead.
- We exclude any word boundary.
- We expand that exclusion,
- We assert end of line.
The benefit here are that we do not need to assert any flags or word boundaries, it will take into account non-word characters and we do not need to reach for negate.
Comments
var input = "$this$ $is$ $a$ $test$";
If you use var result = input.match("\b(\w+)\b") an array of all the matches will be returned next you can get it by using pop() on the result or by doing: result[result.length]
2 Comments
Maarten Peels
I didn't know that was even possible, will look into it
Your regex will find a word, and since regexes operate left to right it will find the first word.
A \w+ matches as many consecutive alphanumeric character as it can, but it must match at least 1.
A \b matches an alphanumeric character next to a non-alphanumeric character. In your case this matches the '$' characters.
What you need is to anchor your regex to the end of the input which is denoted in a regex by the $ character.
To support an input that may have more than just a '$' character at the end of the line, spaces or a period for instance, you can use \W+ which matches as many non-alphanumeric characters as it can:
\$(\w+)\W+$
Comments
Avoid regex - use .split and .pop the result. Use .replace to remove the special characters:
var match = str.split(' ').pop().replace(/[^\w\s]/gi, '');
1 Comment
ps0604
this solution does not consider special characters in addition to space
.spliton the string instead and grab the last element with.pop. Much simpler.input.match(/\b\w+\b/g).pop();splitto use special characters in addition to space?