I think your .? was meant to be .*? instead.
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Jeff[japhy]Pinyan:
Perl,
regex,
and perl
hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??; | [reply] |
$var =~ /n\\(.?)\\/ <BR>
matches for "n\" optionally followed by any character followed by a backslash
$var =~ /n\\([^\\])/ <BR>
matches for n\ not followed by backslash (only one character)
but does also not work for : "/n\123\uu
IMHO following should work better:
$var =~ /n\\([^\\]+)\\/
MP
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I won't claim to be an expert on regexps, but here's my 2 cents.
First, lets go through the ones you posted.
$var =~ /n\\(.?)\\/
will match "{anything}n\{one or zero characters}\{anything}",
and $1 will contain the "{one or zero characters}".
The dot matches one character, the ? modifies that to match
zero or one times. Because it's enclosed in parens, it'll store
it.
The second regexp you showed will match "{anything}n\{anything except \}",
and $1 will contain the "{anything except \}".
Now, to get the match you're looking for, something like:$var =~ /n\\(^\\]+\\/;
which will match "{some text}n\{some more text}\{and even more}"
and store "{some more text}" in $1.
What this regexp does is looks for 'n\' followed by one or more
characters that aren't a '\', followed by a '\'.
Hope this helps. =] | [reply]
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the first regex is going to look for a single character between '\' and '\', which is not quite what you want... this is because of the '?' quantifier after the '.'. '?' means one or more.
the second one should only produce a single character - the first char of 'some text'... the '[^\\]' has an implicit quantifier of match one char only or no match at all
you will need something like this to capture the whole string (Note the '+'):
$var =~ /n\\([^\\]+)\\/;
| [reply]
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? -- optional, zero or one
+ -- required and repeatable, one or more
* -- optional and repeatable, zero or more
All of these can be modified by placing a '?' after them, .+? in this usage however the '?' is affecting the greediness of the operator. Without the '?' it will attempt to match as many as possible (greedy), with the '?' it will attempt to match as few as required (non-greedy).
There is a great book, Mastering Regular Expressions on this topic that is a very worthwhile read. | [reply]
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D'Oh! Must be all these late nights... :-)
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Ah, thanks. The book i had suggested that .? would work in the way .+? does. Plus in the second regexpresion I ment to type:
/\\([^\\]+)/
The thing I was puzzeled about is perhaps better explained with the fact that this doesn't seem to work:
/\\([^\\]+)\\/
Maybe it was just that it was 4 in the morning at the time.
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$var =~ /n\\(.?)\\/
This will match a 'n' followed by a '\' followed by 0 or 1 instance of any char followed by a '\'.
Thus in the case you provide it will not match anything.
There are two possibilities on what your want in your $1:
either what is between two '\' or what is between the first and last '\'
Which is either the regex $var =~ /n\\(.*?)\\/ or $var =~ /n\\(.*)\\/
The difference may be subtle, but look at the strings:
$var='{some text} n\{some text}\{some more text}\';
and
$var='{some text} n\{some text}\{some more text}'; #Your old version
$1 in the first $var will become '{some text}' with the first regex and '{some text}\{some more text}' with the second.
$1 in the second $var will in either case be '{some text}'
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