RE: [PHP-DB] Inserting checkbox data

In addition to what Mihail had to say, I saw a couple of minor things as well. I'm not sure if they're the cause of or solution to the problem you're experiencing. First, I'm pretty sure that a value of value=\"Y\" is not valid HTML. I don't know what it does to your name/value pair when you POST the data from the page, but I'm pretty sure it's invalid HTML. HTML wants value="Y"; it doesn't like the slashes. Second, I noticed you're missing a comma in your SQL statement between $LastName and $Username. That would probably cause your SQL statement to error out. Just my two cents. > -----Original Message----- > From: Graeme McLaren [mailto:mickel@ntlworld.com] > Sent: Friday, October 25, 2002 2:40 PM > To: php-db@lists.php.net > Subject: [PHP-DB] Inserting checkbox data > > > Hi all, I'm having a problem inserting a value from a > checkbox into a MySQL DB. I can print out what the variable > holds, the problem is it just won't go into the DB. Weird > thing is when I login to SSH there isn't a NULL value for it > instead that field is just blank. > > Anyone got any ideas what the problem is? > > > The HTML that is used for the checkbox is: > > Gift Wrapping: <input type=checkbox Name=GiftWrapping[] Value=\"Y\"> > > > So it should theoretically insert "Y" into the DB > > Here is the code that appears to be causing the problem: > > $number=count($GiftWrapping); > for($a=0;$a<=$number;$a++) > { > echo $GiftWrapping[$a]; > $GW = $GiftWrapping[$a]; > echo $GW; > } > > if ($Submit == "Submit") { > $sql = "INSERT INTO SiteMember SET > FirstName = '$FirstName', > LastName = '$LastName' > Username = '$Username', > Password = '$Password', > GiftWrapping = '$GW'"; > } > > > Thank you in advance for any help. > > Graeme :) > > > Public Sub House() > > On Error Resume drink > > If Pint.empty = True Then > Pint.refill > Else > Pint.drink > End if > > stomach.add Pint > > MsgBox " I've had .... " & stomach.count & " Pints" > MsgBox "VERY DRUNK" > > End Sub > >