ID: 35427
User updated by: tomas_matousek at hotmail dot com
Reported By: tomas_matousek at hotmail dot com
-Status: Bogus
+Status: Open
Bug Type: Strings related
Operating System: *
PHP Version: 5.1.0
New Comment:
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'
character.
See the source code. The bug is clear there.
Previous Comments:
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[2005-11-28 09:47:30] [EMAIL PROTECTED]
Yes, you need to add the - in the list.
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[2005-11-28 09:29:10] tomas_matousek at hotmail dot com
Read again the full doc! Don't stop at the middle.
RTFM:
"charlist
A list of additional characters which will be considered as 'word' "
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[2005-11-28 01:04:24] [EMAIL PROTECTED]
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."
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[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
well.
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[2005-11-27 19:28:44] [EMAIL PROTECTED]
"bar-0var" doesn't look like a valid *WORD* to me.
Or is it?
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The remainder of the comments for this report are too long. To view
the rest of the comments, please view the bug report online at
http://bugs.php.net/35427
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Edit this bug report at http://bugs.php.net/?id=35427&edit=1
