I'm trying to split a string into an array of words, however I want to keep the spaces after each word. Here's what I'm trying:
var re = /[a-z]+[$\s+]/gi;
var test = "test one two three four ";
var results = test.match(re);
The results I expect are:
[0]: "test "
[1]: "one "
[2]: "two "
[3]: "three "
[4]: "four "
However, it only matches up to one space after each word:
[0]: "test "
[1]: "one "
[2]: "two "
[3]: "three "
[4]: "four "
What am I doing wrong?
Consider:
var results = test.match(/\S+\s*/g);
That would guarantee you don't miss any characters (besides a few spaces at the beginnings, but \S*\s* can take care of that)
Your original regex reads:
[a-z]+ - match any number of letters (at least one)[$\s+] - much a single character - $, + or whitespace. With no quantifier after this group, you only match a single space.Try the following:
test.match(/\w+\s+/g); // \w = words, \s = white spaces
test.match(/\w+\s*/gi)
You are using + inside the char class. Try using * outside the char class instead.
/[a-z]+\s*/gi;
+ inside the char class is treated as a literal + and not as a meta char.
Using * will capture zero or more spaces that might follow any word.
The + is taken literally inside the character class. You have to move it outside: [\s]+ or just \s+ ($ has no meaning inside the class either).
The essential bit of your RegEx that needs changing is the part matching the whitespace or end-of-line.
Try:
var re = /[a-z]+($|\s+)/gi
or, for non-capturing groups (I don't know if you need this with the /g flag):
var re = /[a-z]+(?:$|\s+)/gi
$and+in the second class?