Regular expression match any character at the beginning end with specific string

Question

I have the following string:

"bar foo test-line drop-line"

I need to replace the words that starts with anything and ends with '-line'.

basically having:

"bar foo new_word new_word"

I tried:

string.replace(/\.-line$/g,'new_word')

but it doesn't work.

Answer 1

Try this as a regex:

/\S+-line(?![-\w])/

The word anchor is not suitable here since dashes are not considered part of a word, so /\\S+-line\\b/ would mistakenly match text-with-line-not-to-be-replaced . Hence the lookahead construct.

Of course, according to your use case, \\S may seem a little coarse. If your words really only consist of letters then dashes etc, then you can use the normal* (special normal*)* pattern:

/[a-z]+(-[a-z]+)*-line(?![-\w])/i

(normal: [az] , special: - )

(edit: changed the lookahead construct, thanks to @thg435)

Answer 2

In your regular expression, you can use \\b to detect beginning of new words and \\w for "word characters":

"your string".replace(/\b\w+-line\b/g,'new_word')

See http://www.w3schools.com/jsref/jsref_obj_regexp.asp

Answer 3

How about

string.replace(/\b\S+-line\b/g,'new_word')

\\b Word boundary

\\S non whitespace

As fge astutely notes, we need a \\b after line so we don't catch things ending lines liner etc etc etc