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?
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hiacliok
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1@mplungjan He doesn't want to *ignore* the leading/trailing whitespace, though, he wants it in his output. – CertainPerformance Aug 30 '18 at 06:54
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@ mplungjan It is not duplicate of what you mentioned. Should be reopened. – Vikasdeep Singh Aug 30 '18 at 06:55
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Apologies... Here is the link for reference anyway: https://stackoverflow.com/questions/14912502/how-do-i-split-a-string-by-whitespace-and-ignoring-leading-and-trailing-whitespa – mplungjan Aug 30 '18 at 07:40
3 Answers
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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+ *$|\S+/g)
);Lookbehind is another option, but it's not supported enough to be reliable in production code yet.
CertainPerformance
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How about
a.split(/(?!^) (?!$)/)
If there may be more than one space and lookbehinds are supported then
a.split(/(?<!^ *) +(?! *$)/)
MikeM
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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 "]
mpdev7
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