java-script Regex filtering on words

Question

I have the following Regex:

The regex is in a bit of code in our app, I can see it splits words. It obviously removes characters such as $#* and so on. I need it to do the same thing exactly but allow the a hash tag, since the words can now have #hashtags.

"Test #words".toLowerCase().split(/\b/).filter(function(w){return w.match(/^\w+$/) }) // returns ["test", "words"]

The current Regex removes the hash, i want it to remain. So i get:

["test", "#words"]

The regex is in a bit of code in our app, I can see it splits words. It obviously removes characters such as $#* and so on. I need it to do the same thing but allow the hash, since the words can now have #hashtags. — MartinWebb
– MartinWebb, Commented Aug 2, 2016 at 20:09

Wiktor Stribiżew · Accepted Answer · 2016-08-02 20:20:29Z

1

Your "Test #words".toLowerCase().split(/\b/).filter(function(w){return w.match(/^\w+$/) }) does the following:

The whole string is turned to lower case
The string is split at any word boundary (leading and trailing, meaning Test #words is split into [,Test, #,words,])
The parts that match the ^\w+$ regex (1+ word chars from the start till end of string) are kept in the array.

You may use an identical matching approach to also include # with /(?:\B#)?\w+/g:

console.log("Test #words".toLowerCase().match(/(?:\B#)?\w+/g))

The pattern matches:

(?:\B#)? - an optional # preceded with a non-word boundary
\w+ - 1 or more word chars (from [a-zA-Z0-9_] ranges)

If context is not so important, use a simpler /#?\w+/g regex that will match an optional # anywhere in the string, followed with 1+ word chars.

edited Aug 2, 2016 at 20:20

answered Aug 2, 2016 at 20:08

Wiktor Stribiżew

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7 Comments

MartinWebb Over a year ago

I' have added some more detail in the Questions. I want to make sure that the regex does exactly the same thing but allows/passes a hashtag in front of the word. Sorry if i was not completely clear.

Wiktor Stribiżew Over a year ago

That is impossible without code change. The reason is easy: JS regex does not support a lookbehind. Now, the question is: to what extent can the code be changed? I should say that my code above does exactly what you need: it extracts chunks of word chars optionally preceded with #.

MartinWebb Over a year ago

Well I can change the regex. The approach you offer seems to work? I'm not a regex expert so I need to know what else does the regex you suggest other that allowing the # in-front of a word.

Wiktor Stribiżew Over a year ago

I tried to explain what your code does, and I think you now see I am doing the same thing, but now, # is also matched. Note that /#?\w+/g is context-unaware, and - from experience - I know that for the majority of people \B#\w+ worked better. That is why I put the (?:\B#)?\w+ version on top.

MartinWebb Over a year ago

If context is not so important, use a simpler /#?\w+/g regex that will match an optional # anywhere in the string, followed with 1+ word chars. this is the correct solution. Thank you!

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Aleksandr · Accepted Answer · 2016-08-02 20:15:47Z

0

Just add optional # at the beginning of the regexp to support #hashtags.

"Test #words".toLowerCase().match(/#?\w+/g);

answered Aug 2, 2016 at 20:15

Aleksandr

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