Normal split works like this:
var a = " a @b c "
console.log(a.split(" "))
["", "a", "b", "c", ""]
But my expected output is: [" a", "@b", "c "] it is possible? And how?
-
1@mplungjan He doesn't want to ignore the leading/trailing whitespace, though, he wants it in his output.CertainPerformance– CertainPerformance2018-08-30 06:54:19 +00:00Commented Aug 30, 2018 at 6:54
-
@ mplungjan It is not duplicate of what you mentioned. Should be reopened.Vikasdeep Singh– Vikasdeep Singh2018-08-30 06:55:04 +00:00Commented Aug 30, 2018 at 6:55
-
Apologies... Here is the link for reference anyway: stackoverflow.com/questions/14912502/…mplungjan– mplungjan2018-08-30 07:40:17 +00:00Commented Aug 30, 2018 at 7:40
Add a comment
|
3 Answers
One option is to use a regular expression and require word boundaries before and after the space:
var a = " a b c "
console.log(a.split(/\b \b/));If non-word characters are allowed as well, you can use match instead - either match spaces at the beginning of the string, followed by non-spaces, or match non-spaces followed by spaces and the end of the string, or match non-spaces without restriction:
const a = " foo @bar c "
console.log(
a.match(/^ *\S+|\S+ *$|\S+/g)
);Lookbehind is another option, but it's not supported enough to be reliable in production code yet.
answered Aug 30, 2018 at 6:51
CertainPerformance
373k55 gold badges354 silver badges359 bronze badges
1 Comment
hiacliok
What about string var a = " foo @bar c ", its not working correctly.
How about
a.split(/(?!^) (?!$)/)
If there may be more than one space and lookbehinds are supported then
a.split(/(?<!^ *) +(?! *$)/)
Comments
You can trim the string before the split, for example:
var a = " a b c ";
a = a.trim();
console.log(a.split(" "));update
i was wrong to read the expected output, the result of my suggested code it's:
["a", "b", "c"] and not [" a", "b", "c "]
