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Posted by Michael G on 09/08/05 04:35
"Michael G" <mike-g@montana.com> wrote in message
news:1126133711_18365@spool6-east.superfeed.net...
>
> "muldoonaz" <donot@spam.me.com> wrote in message
> news:Q0JTe.296169$WN5.93062@fe02.news.easynews.com...
>> Michael G wrote:
>>> The following is from
>>> http://php.mirrors.ilisys.com.au/manual/en/security.database.sql-injection.php .
>>>
>>> Would someone explain the following lines, in particular I don't
>> return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']',
$queryString);
>> }
>>
>
> But I still fail to understand how 'paramArr[\'$1\']' is mapped using $1
> as an index. I've tried printing paramArr['$1'] to see if I might gain
> some understanding but to no avail.
>
ok. I have an explanation, thanks to a Perl book I have. Anyway, the $1 is
called a backreference. Backreferences contain the value that is matched by
each atom of the regular expression. In this case there is only one atom -
(.*?), hence only one backreference. So each time there is a match the value
contained in the curly braces would be copied into the backreference. So
$paramArr['$1'] after the first match would give $paramArr['0'] as per the
regex. Not real sure about what the modifiers 'ei' mean at the end of the
pattern.
Mike
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