Develop Reference

Custom regex word boundary (javascript)

I'm trying to create a custom word boundary (like \b) that also takes words starting or ending with the unicode characters "ÆØÅæøå" into consideration.
Now the only thing I can come up with is this ugly thing
((?<![\wÆØÅæøå])(?=[\wÆØÅæøå])|(?![\wÆØÅæøå])(?<=[\wÆØÅæøå]))
Is there a more elegant solution to this? Or is this the only way.

You can use:
(?<!\p{L}\p{M}*|[\p{N}_]) // leading word boundary, similar to \<, [[:<:]] or \m in other flavors
(?![\p{L}\p{N}_]) // trailing word boundary, similar to \>, [[:>:]] or \M
Compile the regex with the u modifier to enable Unicode category classes.
The (?<!\p{L}\p{M}*|[\p{N}_]) is a negative lookbehind that matches a location not immediately preceded with a letter followed with zero or more diacritic marks or a digit or an underscore.
The (?![\p{L}\p{N}_]) is a negative lookahead that matches a location not immediately followed with a letter, digit or an underscore.

Regex allow multiple work in a sentence

I'm trying to parse following sentences with regex (javascript) :
I wish a TV
I want some chocolate
I need fire
Currently I'm trying : I(\b[a-zA-Z]*\b){0,5}(TV|chocolate|fire) but it doesn't work. I also made some test with \w but no luck.
I want to allow any word (max 5 words) between "I" and the last word witch is predefined.

To account for non-word chars in-between words, you may use
/I(?:\W+\w+){0,5}\‌​W+(?:TV|chocolate|fir‌​e)/
See the regex demo
The point is that you added word boundaries, but did not account for spaces, punctuation, etc. (all the other non-word chars) between "words".
Pattern details:
I - matches the left delimiter
(?:\W+\w+){0,5}\‌​W+ - matches 0 to 5 sequences (due to the limiting quantifier {n,m}) of 1+ non-word chars (\W+) and 1+ word chars after them (\w+), and a \W+ at the end matches 1 or more non-word chars that must be present to separate the last matched word chars from the...
(?:TV|chocolate|fir‌​e) - matches the trailing delimiter

You need to add the whitespace after the I. Otherwise it wouldn´t capture the whole sentence.
I(\b[a-zA-Z ]*\b){0,5}(TV|chocolate|fire)
I greate site to test regex expressions is regexr

If you don't care about the spaces, use:
/I(\s[a-zA-Z]*\s?){0,5}(TV|chocolate|fire)/

Try
/I\s+(?:\w+\s+){0,5}(TV|chocolate|fire)/
(Test here)
Based on Stefan Kert version, but rely on right side spaces of each extra word instead of word boundaries.
It also accepts any valid "word" (\w) character words of any length and any valid spacing character (not caring for repetitions).

Regex expression to capitalise first character

For the address field, I need first character of every word to be uppercase. I have been using /\b./g which has eventually resulted in a problem where first character after special characters such as !#*&;' and so on are also capitalised. ie. King'S Street instead of King's Street.
Is there a way to adjust that expression to exclude that behaviour or is changing the entire expression more optimal?

replace \b with (^|[ ])
Your regex will be: /(^|[ ])./g
Explanation:
\b by definition: is used to find a match at the beginning or end of a word.
(^|[ ]) will match with the beginning of the string or any space characters
(^|[ ]). will match every space followed by a character and the first character of the string.
Side note:
Use (^|\s) to match every blank spaces.
Your regex will be: /(^|\s)./g

You could use a lookahead:
\b[a-z](?=\w+)
See a demo on regex101.com.

Regex validation space in the middle of a string

I'm looking for an expression that requires a space in a string, it doesn't have to be dead in the middle just not at the end (or start).
I've had a look on google and stack-overflow, there are quite a few but I haven't found one that does what I need.
Here's what I have at the moment
var re = /^[A-Z]\'?[- a-zA-Z]( [a-zA-Z])*$/igm;

Based on the limited requirements you specified, this will do it. It requires a string to contain ONE space, anywhere but at the start or end.
/^[^ ]+ [^ ]+$/
Explanation: anchoring to the beginning of the string, allow one or more non-space characters, followed by a single space, followed by, again, one or more non-space characters, to the end of the string.
[^ ] is a negated character class. That is, it says "anything but the characters inside [ and ].

Your regex should be: /^[A-Z]\'?[-\sa-zA-Z](\s[a-zA-Z])*$/igm;. According to my idea, regex doesn't recognize a whitespace that why I replace those with \s.

How to extract the last word in a string with a JavaScript regex?

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

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*$

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.

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]

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 '$' character at the end of the line, spaces or a period for instance, you can use \W+ which matches as many non-alphanumeric characters as it can:
\$(\w+)\W+$

Avoid regex - use .split and .pop the result. Use .replace to remove the special characters:
var match = str.split(' ').pop().replace(/[^\w\s]/gi, '');
DEMO