I think you're confused. There are two parts to this transaction, right.
First, you display the form to the user, using HTML.
Then, the user submits a form, the data is sent back to the server,
usually (99% of the time) via a POST request.
At that point, you'll probably want to do something with the user's
submission (like bump the user to a new category or something).
PHP automatically takes the user's POST response and creates an array
called $_POST.
The keys of the array correspond to the *name* of the form field from
your form.
So, if the name of your field is 'picplace', you'd look in
$_POST['picplace'] for the data.
If you're seeing a value when you var_dump($_POST['picplace']); in the
target PHP page, then your data is making it to the target script just fine.
You've got a problem in the target script.
Look at the part of the code where you're constructing your query -- in
particular, the place where you think you're using the value submitted
for "picplace", and make sure you're accessing it using
$_POST['picplace']. If you're not, that's probably your problem. :)
-Jeromie
>Hi Jeromie,
>
>Thanks for you response.
>
>Can you explain where in the code I should add $_POST [picplace]; ,
>please? Should it be in the select name or option value?
>
>I tried you suggestion and managed to extract the correct value from
>the select box to another PHP page. So the form is OK. It's just the
>transition into the database where the problem is and I guess it's
>linked to me missing out the $POST variable in the <FORM>
>
>Nearly there - thanks all for you responses.
>Chris
>
>
>
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