explanation please

[Michael G]

The following is from
http://php.mirrors.ilisys.com.au/man...-injection.php .

Would someone explain the following lines, in particular I don't understand
'$paramArr[\'$1\']' nor do I understand how the syntax {1} works or how it
is related to arrays?

Thanks, mIke.

<some code snipped>
....
return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']', $queryString);
}

$sqlQuery = 'SELECT col1, col2 FROM tab1 WHERE col1 = {1} AND col3 = {2}
LIMIT {3}';
$stm = mysql_query(prepareSQL($sqlQuery, array('username', 24.3, 20);
?>

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[muldoonaz]

Michael G wrote:

The following is from
http://php.mirrors.ilisys.com.au/man...-injection.php .

Would someone explain the following lines, in particular I don't understand
'$paramArr[\'$1\']' nor do I understand how the syntax {1} works or how it
is related to arrays?

Thanks, mIke.

<some code snipped>
...
return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']', $queryString);
}

$sqlQuery = 'SELECT col1, col2 FROM tab1 WHERE col1 = {1} AND col3 = {2}
LIMIT {3}';
$stm = mysql_query(prepareSQL($sqlQuery, array('username', 24.3, 20);
?>

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----= East and West-Coast Server Farms - Total Privacy via Encryption =----

look at the snippet of code and you'll find your answer. the $paramArr
variable is passed with the calling of the function.

you'd type the following into your script: prepareSQL("something",
"here"); and "here" would become $paramArr.

-- code --

<?php
function prepareSQL($queryString, $paramArr) {
foreach (array_keys($paramArr) as $paramName) {
if (is_int($paramArr[$paramName])) {
$paramArr[$paramName] = (int)$paramArr[$paramName];
}
elseif (is_numeric($paramArr[$paramName])) {
$paramArr[$paramName] = (float)$paramArr[$paramName];
}
elseif (($paramArr[$paramName] != 'NULL') and
($paramArr[$paramName] != 'NOT NULL')) {
$paramArr[$paramName] =
mysql_real_escape_string(stripslashes($paramArr[$paramName]));
$paramArr[$paramName] = '\''.$paramArr[$paramName].'\'';
}
}

return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']', $queryString);
}

[ZeldorBlat]

Ahh, *those* curly brackets are a little different. prepareSQL() is
not a built in PHP function, so I can't say for sure what it does.

Based on your example, I would guess that it does something like
substitute the values passed as the second parameter (the array) into
the string passed as the first parameter ($sqlQuery) where the
parameters in the original string are refered to by {i} where i is
their index in the array.

I'd also venture a guess that your "return preg_replace..." code comes
from prepareSQL(). Without seeing the rest of the function, it looks
like that call preg_replace() is what does the actual replacement of
the variables.

So, in short, those {1}, {2}, etc. are *not* the same as the curly
braces described at http://www.php.net/strings

[muldoonaz]

ZeldorBlat wrote:

Ahh, *those* curly brackets are a little different. prepareSQL() is
not a built in PHP function, so I can't say for sure what it does.

Based on your example, I would guess that it does something like
substitute the values passed as the second parameter (the array) into
the string passed as the first parameter ($sqlQuery) where the
parameters in the original string are refered to by {i} where i is
their index in the array.

I'd also venture a guess that your "return preg_replace..." code comes
from prepareSQL(). Without seeing the rest of the function, it looks
like that call preg_replace() is what does the actual replacement of
the variables.

So, in short, those {1}, {2}, etc. are *not* the same as the curly
braces described at http://www.php.net/strings

i don't know the answer to your question, but here's the function he was
speaking about:

<?php
function prepareSQL($queryString, $paramArr) {
foreach (array_keys($paramArr) as $paramName) {
if (is_int($paramArr[$paramName])) {
$paramArr[$paramName] = (int)$paramArr[$paramName];
}
elseif (is_numeric($paramArr[$paramName])) {
$paramArr[$paramName] = (float)$paramArr[$paramName];
}
elseif (($paramArr[$paramName] != 'NULL') and
($paramArr[$paramName] != 'NOT NULL')) {
$paramArr[$paramName] =
mysql_real_escape_string(stripslashes($paramArr[$paramName]));
$paramArr[$paramName] = '\''.$paramArr[$paramName].'\'';
}
}

return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']', $queryString);
}

[muldoonaz]

ZeldorBlat wrote:

Ahh, *those* curly brackets are a little different. prepareSQL() is
not a built in PHP function, so I can't say for sure what it does.

Based on your example, I would guess that it does something like
substitute the values passed as the second parameter (the array) into
the string passed as the first parameter ($sqlQuery) where the
parameters in the original string are refered to by {i} where i is
their index in the array.

I'd also venture a guess that your "return preg_replace..." code comes
from prepareSQL(). Without seeing the rest of the function, it looks
like that call preg_replace() is what does the actual replacement of
the variables.

So, in short, those {1}, {2}, etc. are *not* the same as the curly
braces described at http://www.php.net/strings

here's the function (taken off the website he quoted):

<?php
function prepareSQL($queryString, $paramArr) {
foreach (array_keys($paramArr) as $paramName) {
if (is_int($paramArr[$paramName])) {
$paramArr[$paramName] = (int)$paramArr[$paramName];
}
elseif (is_numeric($paramArr[$paramName])) {
$paramArr[$paramName] = (float)$paramArr[$paramName];
}
elseif (($paramArr[$paramName] != 'NULL') and
($paramArr[$paramName] != 'NOT NULL')) {
$paramArr[$paramName] =
mysql_real_escape_string(stripslashes($paramArr[$paramName]));
$paramArr[$paramName] = '\''.$paramArr[$paramName].'\'';
}
}

return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']', $queryString);
}

[Michael G]

"muldoonaz" <do***@spam.me.com> wrote in message
news:Q0********************@fe02.news.easynews.com ...

Michael G wrote:

The following is from
http://php.mirrors.ilisys.com.au/man...-injection.php .

Would someone explain the following lines, in particular I don't
understand
'$paramArr[\'$1\']' nor do I understand how the syntax {1} works or how
it is related to arrays?

Thanks, mIke.

<some code snipped>
...
return preg_replace('/\{(.*?)\}/ei','$paramArr[\'$1\']',
$queryString);
}

$sqlQuery = 'SELECT col1, col2 FROM tab1 WHERE col1 = {1} AND col3 = {2}
LIMIT {3}';
$stm = mysql_query(prepareSQL($sqlQuery, array('username', 24.3, 20);
?> ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet
News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+
Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption
=----
look at the snippet of code and you'll find your answer. the $paramArr
variable is passed with the calling of the function.

you'd type the following into your script: prepareSQL("something",
"here"); and "here" would become $paramArr.

Yeah, I understand that. In the OP, $paramArr is an array. I also now
understand that the author of this function uses regular expressions to do
the replacement.

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.

Mike

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[Michael G]

"Michael G" <mi****@montana.com> wrote in message
news:11**************@spool6-east.superfeed.net...

"muldoonaz" <do***@spam.me.com> wrote in message
news:Q0********************@fe02.news.easynews.com ...

Michael G wrote:

The following is from
http://php.mirrors.ilisys.com.au/man...-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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[Justin Koivisto]

Michael G wrote:

"Michael G" <mi****@montana.com> wrote in message
news:11**************@spool6-east.superfeed.net...

"muldoonaz" <do***@spam.me.com> wrote in message
news:Q0********************@fe02.news.easynews.c om...

Michael G wrote:

The following is from
http://php.mirrors.ilisys.com.au/man...-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.

IIRC, the 'e' modifier is for expand or evaluate, and I know that the
'i' modifier is case-insensitive matching.

--
Justin Koivisto, ZCE - ju****@koivi.com
http://koivi.com