Re: why won't my array work?

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On Mar 28, 2008, at 12:40 PM, Eric Butera wrote:
On Fri, Mar 28, 2008 at 12:28 PM, Jason Pruim <japruim@xxxxxxxxxx> wrote:
$chpwold[] = mysqli_query($chpwpostlink, $oldpasswordquery) or
die("Sorry read failed: ". mysqli_error($chpwpostlink));
$chpwresult = $chpwold[0];

Why would you pump that into an array instead of just calling it
result itself?  I'd say you're just making it harder on yourself for
no apparent reason.

The problem seems to be on your other line.

$chpwrow[] = mysqli_fetch_assoc($chpwresult) or die('Sorry it didn\'t
work....' .mysqli_error($chpwpostlink));
echo $chpwrow['loginPassword'];

Just fetch the row into a single variable and not an array.  In your
example  you'd need to access chpwrow[0]['loginPassword'] assuming it
was an empty array up to that point.


Calling things old query and old password isn't really adding any
value to your code.  If you're only going to use it once then throw it
away call it result so it is easier to read and understand.  But then
again feel free to ignore this.

In the scope of my application since I'm checking the currently stored password before updating to a new password $oldpasswordquery makes sense, at least to me :)


Also is there a reason why you aren't
using prepared statements?

a prepared statement seemed like alot of overkill for a simple check to see if the old pass matches what was stored in the database... And I didn't realize that you could use prepared statements for SELECTing rather then UPDATEing... But I'll look into that more, since I know that prepared statements make it much harder to do Sql injection attacks....




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Jason Pruim
Raoset Inc.
Technology Manager
MQC Specialist
3251 132nd ave
Holland, MI, 49424-9337
www.raoset.com
japruim@xxxxxxxxxx




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