> Do you need to use ".*?"? If there will be only white characters, use
> \s* instead.
>
unfortunately there's some content either side of $file before
this content is different for each item, so i can't define it in the
pattern.
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> And what about [^>]* -if there are no html tags
>
not thought about that, but XML tags, not HTML tags... does that make any
difference?
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> Not sure if I understand you correct but
> the way you wrote your replace pattern the result should be
> .
> If you want just the file to be replaced you have to use something like
> preg_replace("|.*?($file).*?|si",'\\1',$contents)
>
no, i want just the whole replaced for that particular file
$contents = preg_replace( "|.*?".$file.".*?|si", "",
$contents );
okay, so i figured out that it's matching the first occurence of
which will always be the first record and then going on to match the $file
and deleting everything between. obviously not what i want.
without giving a unique attri
skate wrote:
$contents = preg_replace( "|.*?$file.*?|si", "", $contents );
it's being run on an XML file, where each entry is .. with
a $file pointer in there.
it works okay, except for the fact that it deletes that record, and every
record before it. i can't figure out why it's being greed
> preg_replace("|.*?$file.*?|si","",$contents)
>
>
> fhh
> 1060205191
> hhh
> xml/news/1060205191.xml
>
>
> fgjghjh
> 1060205186
> fgjh
> xml/news/1060205186.xml
>
>
> fgjhh
> 1060205182
> fghh
> xml/news/1060205182.xml
>
>
Not sure if I understand you correct
- Original Message -
From: "skate" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 06, 2003 3:16 PM
Subject: [PHP] greedy preg
>
> $contents = preg_replace( "|.*?$file.*?|si", "",
$contents );
>
> it's be
From: "skate" <[EMAIL PROTECTED]>
> $contents = preg_replace( "|.*?".$file.".*?|si", "",
> $contents );
>
> okay, so i figured out that it's matching the first occurence of
> which will always be the first record and then going on to match the $file
> and deleting everything between. obviously not
Do you need to use ".*?"? If there will be only white characters, use
\s* instead.
skate wrote:
$contents = preg_replace( "|.*?".$file.".*?|si", "",
$contents );
okay, so i figured out that it's matching the first occurence of
which will always be the first record and then going on to match the
>
> You cannot store file pointers. If you output the variable it's going to
> look something like "Resource ID #1" which is meaningless except to the
> instance of the script that created it. Or did you mean something
different
> by "$file pointer"?
sorry, wrong words...
when i meant $file is
And what about [^>]* -if there are no html tags
skate wrote:
Do you need to use ".*?"? If there will be only white characters, use
\s* instead.
unfortunately there's some content either side of $file before
this content is different for each item, so i can't define it in the
pattern.
--
P
$contents = preg_replace( "|.*?$file.*?|si", "", $contents );
it's being run on an XML file, where each entry is .. with
a $file pointer in there.
it works okay, except for the fact that it deletes that record, and every
record before it. i can't figure out why it's being greedy? i know i
skate wrote:
What are the possible values of $file? Are you looking to replace just a
specific occurance of a $file between tags? Maybe this will help:
$contents = preg_replace("|[^<]*$file[^<]*|si","",$contents);
or just use the 'U' modifier for "ungreedy"...
i'm looking to replace the entir
> What are the possible values of $file? Are you looking to replace just a
> specific occurance of a $file between tags? Maybe this will help:
>
> $contents = preg_replace("|[^<]*$file[^<]*|si","",$contents);
>
> or just use the 'U' modifier for "ungreedy"...
>
i'm looking to replace the entire
>
> Yeah, it should be. What kind on content can be around $file and also
> within the tag?
>
here's a snip of the xml file...
it's finding the $file correctly, but deletes all records from the beginning
of the file onwards. it leaves the xml file with valid xml ... at the
start... so for the
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