Regex: don't match if string contains whitespace

Question

I can't seem to figure out the regex pattern for matching strings only if it doesn't contain whitespace. For example

"this has whitespace".match(/some_pattern/)

should return nil but

"nowhitespace".match(/some_pattern/)

should return the MatchData with the entire string. Can anyone suggest a solution for the above?

Answer 1

>> "this has whitespace".match(/^\S*$/)
=> nil
>> "nospaces".match(/^\S*$/)
=> #<MatchData "nospaces">

^ = Beginning of string

\S = non-whitespace character, * = 0 or more

$ = end of string

Answer 2

In Ruby I think it would be

/^\S*$/

This means "start, match any number of non-whitespace characters, end"

Answer 3

   "nowhitespace".match(/^[^\s]*$/)

Answer 4

You want:

/^\S*$/

That says "match the beginning of the string, then zero or more non-whitespace characters, then the end of the string." The convention for pre-defined character classes is that a lowercase letter refers to a class, while an uppercase letter refers to its negation. Thus, \s refers to whitespace characters, while \S refers to non-whitespace.

Answer 5

Not sure you can do it in one pattern, but you can do something like:

"string".match(/pattern/) unless "string".match(/\s/)

Answer 6

str.match(/^\S*some_pattern\S*$/)

Answer 7

You could always search for spaces, an then negate the result:

"str".match(/\s/).nil?