So create a hidden input field on the page called cur_id and set its value
to the id of the record being displayed. If no record is currently being
displayed, set it to -1.

When you get ready to handle the form data, check the $_POST['cur_id']
variable. If it's set to -1, perform an INSERT. If it's set to anything
else, perform an UPDATE using the value of $_POST['cur_id'] in your WHERE
clause.

I'm sure there are other ways to handle this. This is how I have handled it
in the past though. 

> -----Original Message-----
> From: Jeff [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, August 05, 2003 3:52 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] MySQL Returns Error
> 
> 
> Oops, my mistake, it's not a blank entry  - it re enters the 
> last entry.
> 
> I can hit F5 or click refresh 1000 times and it will enter 
> the last entry
> 1000 times.
> 
> 
> "Jeff" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
> > I tried putting that in a few diffrent places, same result 
> - is there
> > somewhere special I need to put that?
> >
> > "Matt Schroebel" <[EMAIL PROTECTED]> wrote in message
> > news:[EMAIL PROTECTED]
> >
> >
> > > -----Original Message-----
> > > From: Jeff [mailto:[EMAIL PROTECTED]
> > > Sent: Tuesday, August 05, 2003 3:15 PM
> > > To: [EMAIL PROTECTED]
> > > Subject: Re: [PHP-DB] MySQL Returns Error
> > >
> > > Every time I access or refresh the page, it adds one blank
> > > entry to the DB.
> > >
> > > Maybe I have the insert query in the wrong place?
> >
> > if ('POST' == $_SERVER['REQUEST_METHOD']) {
> > //someone submitted the form with method="post"
> > // so validate input, and if okay store it in db
> > }
> >
> >
> 
> 
> 
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