Thanks peter for the explanation, but what I need to fix my problem.
Im getting this error code
Notice: iconv(): Detected an illegal character in input string in
/var/www/html/rssfeed/sahafah.php on line 35
When I convert from utf-8 to cp1256
$rss2 = iconv("UTF-8", 'CP1256//TRANSLIT', $rss);
> -----Original Message-----
> From: Peter West [mailto:lists@xxxxxxxxx]
> Sent: Wednesday, February 25, 2015 2:56 PM
> To: Maciek Sokolewicz; PHP General
> Subject: Re: Parsing
>
> > So, what does (.*?) mean? Well, simply said "any character, occuring 0
or
> more times" occuring 0 or 1 times.
>
> I don't think so. ((.*)?) would mean that, but in (.*?), the '?' means
"make
> the preceding pattern non-greedy; that is, make it match the minimum
> number of times. And as the minimum number of matches of (.*) is zero, it
> ends up meaning 'match no character at all. So it will always be true,
> wherever it occurs in a match string.
>
> For instance,
>
> $ php -a
> Interactive shell
>
> php > $test = "aabbcc";
> php > $re = '/.+?(bb?).*/';
> php > preg_match($re, $test, $match);
> php > print_r($match);
> Array
> (
> [0] => aabbcc
> [1] => bb
> )
> Note here that the initial pattern piece '.+?' is limited to the minimum
> match.
> The minimum match is a single character, but that is overruled by the
> attempt to match
> capturing sub-expression '(bb?)' so it in fact matches 'aa'. Note that
in
> this regexp, (bb?) means "...a 'b' char followed by zero or one 'b'
chars."
> Now change that initial sub-expression.
> php > $re = '/.*?(bb?).*/';
> php > preg_match($re, $test, $match);
> php > print_r($match);
> Array
> (
> [0] => aabbcc
> [1] => bb
> )
> The minimum match here is no characters, constrained to the minimum.
> But again,
> the minimum match must be extended to accommodate '(bb?)'.
> Now remove the minimising constraint.
> php > $re = '/.*(bb?).*/';
> php > preg_match($re, $test, $match);
> php > print_r($match);
> Array
> (
> [0] => aabbcc
> [1] => b
> )
> Only one 'b'! Which 'b' is matched? It's the second 'b'. The minimum
match
> for (bb?) is a single 'b' followed by zero 'b's; so the second 'b'
satisfies
> the capturing expression, and the now-greedy initial subexpression can
> gobble up all of the character to that second 'b'.
>
> Don't believe me?
> php > $re = '/(.*)(bb?).*/';
> php > preg_match($re, $test, $match);
> php > print_r($match);
> Array
> (
> [0] => aabbcc
> [1] => aab
> [2] => b
> )
> Let's back up.
> php > $re = '/(.*?)(bb?).*/';
> php > preg_match($re, $test, $match);
> php > print_r($match);
> Array
> (
> [0] => aabbcc
> [1] => aa
> [2] => bb
> )
> As before, with a non-greedy initial sub-expression,
> except that we now capture that initial sub-expression.
> (bb?) means "...a 'b' followed by zero or one 'b's, greedily.
> Can we force that to be non-greedy?
> php > $re = '/(.*?)(bb??)(.*)/';
> php > preg_match($re, $test, $match);
> php > print_r($match);
> Array
> (
> [0] => aabbcc
> [1] => aa
> [2] => b
> [3] => bcc
> )
> Yes we can, by appending a moderating '?' which curbs the
> appetite of the capturing sub-expression: (bb??)
>
> Peter West
> "...and behold, something greater than Jonah is here."
>
> > On 23 Feb 2015, at 10:48 pm, Maciek Sokolewicz
> <maciek.sokolewicz@xxxxxxxxx> wrote:
> >
> > Secondly, the above two regexp rules are slightly bloated. What they
> actually mean is:
> > ( = start new catchable pattern
> > . = any character
> > * = 0 or more of the previous pattern
> > ? = 0 or 1 of the previous pattern
> > ) = end catchable pattern
> > \s = any whitespace character
> >
> > So, what does (.*?) mean? Well, simply said "any character, occuring 0
or
> more times" occuring 0 or 1 times. But since the any character pattern
> already occurs 0 or more times, the pattern as a whole will either be
matched
> (1 time) or not (0 times). Making the ? metacharacter useless. Now if it
were
> (.+?) then it would state "one or more of any character, with the entire
> pattern optional.
> > So in practice, the following patterns are equal in what they
> > represent: (.*?), (.*), (.+?)
> >
>
>
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