Hello Community,
Im trying to reject any newline in a string using regexp.
The string must not be empty.
Yes, this can be done without regexp.
So my Regexp is like:
"Match any character from the begin to the end of a string witch is not
newline, and ensure there is at least one character, and to this match
over multiple lines (so if there are multiple lines reject it)"
Results in /^[^\n]+$/m
My ruby 1.8.6 does:
/^[^\n]+$/m =~ "a\n" -> 0
without multiline flag the same:
/^[^\n]+$/ =~ "a\n" -> 0
But:
/^[^\n]+$/m =~ "\n" -> nil
If I replace "reject newline" with "reject character x" it worked as
expected (x is 'a' in this example):
/^[^a]+$/m =~ "ba" -> nil
Is the multiline-regexp implementation broken or my multiline behaviour
expectation?
Mfg
Markus Schirp
Markus Schirp wrote:
So my Regexp is like:
"Match any character from the begin to the end of a string witch is not
newline, and ensure there is at least one character, and to this match
over multiple lines (so if there are multiple lines reject it)"
Results in /^[^\n]+$/m
My ruby 1.8.6 does:
/^[^\n]+$/m =~ "a\n" -> 0
without multiline flag the same:
/^[^\n]+$/ =~ "a\n" -> 0
The multiline flag only changes the meaning of "." to include the \n character. It doesn't change the meaning of ^ and $ which mean respectively "beginning of line" and "end of line". True true "beginning of string" metacharacter is not ^ but \A, and for "end of string" it's not $ but \z (or \Z if you want to allow a trailing \n on your string)
>> "a\nb\n"[ /^.*$/ ]
=> "a"
>> "a\nb\n"[ /\A.*\z/ ]
=> nil
>> "a\nb\n"[ /\A.*\z/m ]
=> "a\nb\n"
Yeah, it's tricky.
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Daniel