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Re: PHP upgrade broke my code!

Posted by ZeldorBlat on 01/31/08 14:05

On Jan 30, 7:21 pm, steveg1...@yahoo.com wrote:
> On Jan 30, 1:29 pm, "Paul Lautman" <paul.laut...@btinternet.com>
> wrote:
>
>
>
> > steveg1...@yahoo.com wrote:
> > > Hi, I hope one of you gurus out there can help me. I have a web
> > > application that had been running fine before I upgraded my server,
> > > now it fails on the following snippet of code:
>
> > > function NextOrder($OrderID) {
> > > ConnectDatabase();
> > > $MyQuery = "SELECT OrderID FROM `Orders` WHERE (OrderID > $OrderID)";
> > > $request = mysql_query($MyQuery);
> > > list($Answer) = mysql_fetch_row($request);
> > > echo "DEBUG next is ""$Answer""";
> > > return $Answer;
> > > }
>
> > > When this code is interpreted it outputs this:
>
> > > DEBUG next is ""
>
> > > Not surprisingly, I have a companion routine that looks essentially
> > > the same except for the < operator replacing > and sort order
> > > descending - it fails in the same way. if however, for the purpose of
> > > testing I replace the < with a = then the code works as expected;
>
> > > Did I find a bug in PHP/MySQL somewhere or am I doing something wrong
> > > that I was getting away with before and now getting my hand slapped by
> > > smarter software? ;-)
>
> > > The environment that this worked under is
>
> > > PHP 5.2.0
> > > SuSE 10.2 x586 32 bit
> > > MySQL 5.0.26-12
> > > PHP5-MySQL 5.2.0-10
>
> > > Environment it failed under
>
> > > PHP 5.2.4
> > > SuSE 10.3 x86 64 bit
> > > MySQL 5.0.45-22 x86 64 bit
> > > PHP5-MySQL 5.2.4-10 x86 64 bit
>
> > > Thanks in advance for any light you can shed on this!
>
> > Where/how does the value $OrderID get set?
>
> It's passed to the routine as a parameter. if I echo $MyQuery to the
> output and paste the resultant query to MySQL then it works as
> expected.

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