Mombu 2012-04-27 16:23:56
Updated by: email@example.com
Reported By: tomas_matousek at hotmail dot com
Bug Type: Strings related
Operating System: *
PHP Version: 5.1.0
Really, where? Do you have a patch to fix this?
No, I needn’t. str_word_count(“bar-var”) returns 1, so ‘-‘ is
considered as a part of the word if it is followed by ‘word’
See the source code. The bug is clear there.
————————————————————————[2005-11-28 09:47:30] firstname.lastname@example.org
Yes, you need to add the – in the list.
————————————————————————[2005-11-28 09:29:10] tomas_matousek at hotmail dot com
Read again the full doc! Don’t stop at the middle.
A list of additional characters which will be considered as ‘word’ ”
————————————————————————[2005-11-28 01:04:24] email@example.com
RTFM: “For the purpose of this function, ‘word’ is defined as a locale
dependent string containing alphabetic characters, which also may
contain, but not start with “‘” and “-” characters.”
————————————————————————[2005-11-27 20:00:54] tomas_matousek at hotmail dot com
By passing “0” as the third parameter, one declares ‘0’ character legal
word character which should be equivalent to any other letter, e.g. ‘x’.
“bar-xbar” is considered to be a word so “bar-0bar” should be word as
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