Re: Question before I end up writing alot of extra code...

On Jul 7, 2008, at 3:36 PM, Wolf wrote:

---- Jason Pruim <japruim@raoset.com> wrote:

Hi everyone! So it's been a nice long weekend, I come in to work and try and mess with a project that I'm working on to get some new features added. All was going well until I realized that now my application is breaking... Here's the details... PHP 5.2 MySQL 5.2 I store the info in the database which is submitted from a HTML form.. Some of it text boxes, some check boxes, some radio buttons... I $_POST the info from the form into the processing script. The problem I'm running into though, is when a value has not changed it doesn't get $_POSTed back and my update script erases the info in the database... I'm trying to avoid using $_GET since it can be quite a few variables. Is there anyway I can do it without comparing the original field to what I am displaying?

Gone for a weekend and we have to retrain, at least I'm not the only one... ;) POSTed variables are ALWAYS posted back, changed or not. More then likely you are forgetting a piece of code, but since you didn't post the offending code, I can't point out where you forgot the $ or to restate a variable. :-P

Here is a VERY simplified test :) MAIN PAGE: <?PHP if($row['Tab'] == "done"){ $Tchecked1 = "CHECKED"; $Tchecked2 = NULL; }else{ $Tchecked1 = NULL; $Tchecked2 = "CHECKED"; } echo" <fieldset>Tab<BR> <input type="radio" name="rdoTab" value="done" $Tchecked1>Done <BR> <input type="radio" name="rdoTab" value="on" $Tchecked2>Not Done<BR> </fieldset>"; ?> PROCESSING: <?PHP $tab = $_POST['rdoTab']; $record = $_POST['txtRecord']; $updateQuery = "UPDATE current SET Tab='$tab' WHERE Record='$record'"; mysqli_real_query($link, $updateQuery); ?> -- Jason Pruim Raoset Inc. Technology Manager MQC Specialist 11287 James St Holland, MI 49424 www.raoset.com japruim@raoset.com