First, thak's all for the answers.
I didn't understood what's happend, but I've solved the problem using a
Header("location: ".$PHP_SELF."?celebID=".$ID) after the update command,
passing the same ID I used to update on it. I don't know if it is the best
way, but now I have the "refresh" of the page.
Rodrigo
on 11/14/01 4:01 PM, George Whiffen at [EMAIL PROTECTED] wrote:
> Rodrigo,
>
> I don't quite understand your problem, it might help to see some of the
> php or form html. It might also help if I explain how I usually handle
> updates.
>
> I have a single php page with the form on it which is also the target
> of the form.
>
> Typically users get to the form via a link which includes the "id" on the
> url (i.e. passed as a GET variable). The php picks up this id and uses
> it to query the database to get the current data. Then the form is
> presented with the form values set to the current values and the "id"
> included as a hidden field. The user makes a change and presses a submit
> button with the name "update" (<INPUT TYPE="SUBMIT" NAME="update">)
>
> The php checks to see if "update" is set (isset($update)). If it is
> it uses the form values to update the database before it goes to search
> for the current values. Then it just continues as normal retrieving the
> database (new) values and printing out the form. i.e. the logic is :
>
> if (isset($update))
> {
> update database for "id" record
> e.g. update mytable set myfield = '".$myfield."' where id = '".$id."'
> }
>
> select data for "id" record into myrow
> e.g. select * from mytable where id = '".$id."'
>
> print form including data
> e.g.
> print '
> <HTML><HEAD></HEAD><BODY>
> <FORM ACTION="'.$SCRIPT_NAME.'" METHOD="POST">
> MY FIELD : <INPUT TYPE="TEXT" NAME="myfield" VALUE="'.$myrow['myfield'].'">
> <INPUT TYPE="SUBMIT" NAME="update">
> </FORM>
> </BODY></HTML>
>
> For the user this means they always have visual confirmation that their
> changes have gone to the database after pressing SUBMIT. If they're
> happy they have a link to go wherever they want to next. If they're
> not happy they can correct the data and submit again.
>
> Could you be getting problems because there is confusion between your
> hidden "id" and the "id" on the url? The ACTION="'.$SCRIPT_NAME.'"
> should sort that out since it will remove anything passed on the url
> when the form is submitted.
>
> Or perhaps you have set the values in the form to php variables with
> the same name as the form variables e.g.
>
> print '<INPUT TYPE="TEXT" NAME="myfield" VALUE="'.$myfield.'">
>
> If this is the case, then the form will always come back with the last
> entered details and not blank details since $myfield is continually
> being set to the value of the HTML input variable myfield.
>
> Sorry I can't help more without getting a better idea of what you
> are trying to achieve!
>
> George
>
> Rodrigo Peres wrote:
>>
>> Hi list,
>>
>> I have PHP code to insert the result of a form into mysql. When I nedd to
>> made an update, I pass an "id" in the url and use it to make the update
>> query. The problem is after I click in the update button (input submit) my
>> page refresh and came back in the same state, to see the changes I need to
>> type in the url again with the parameter?? why?? There's a way to avoid this
>> and get a new blank form page after the update?
>>
>> ps: I've stored the "id" in a input type hidden, so I could click the button
>> and still have the id
>>
>> Thank's in advance
>>
>> Rodrigo Peres
>> --
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