On Jul 7, 2008, at 3:36 PM, Wolf wrote:
---- Jason Pruim <japruim@xxxxxxxxxx> 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@xxxxxxxxxx
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