Passing ID through form

[Lazandra]

Hiya

I have an order form and once submitted it goes straight to the payment form. I have 2 databases one order and the other payment. So in the order form i need to pass the id to the payment form to tie them together.

Heres my code... can anyone help?

Cheers xx

[PHP]

if ($submit == "Next")

{

$saved = mysql("db691177 98","insert into info values (0, '$name', '$address', '$city', '$county', '$postcode', '$country', '$email_address ', '$tel', '$fax', '$pet_name', '$pet_breed', '$pet_sex', '$pet_steps', '$pet_weight', '$pet_legs', '$pet_frontlegs ', '$pet_vet', '$pet_cause', '$pet_arise', '$pet_comments' , '$pet_aweight', '$pet_bshoulder ', '$pet_ctail', '$pet_dchest', '$pet_echest', '$pet_ffrontche st')");

if ($saved)
$infos = mysql ("db69117798 ", "select * from info");
while ($info = mysql_fetch_arr ay($infos))
{
echo "<script type='text/javascript'>
<!--
window.location = 'payment.php?in foid=$info[id]'
//-->
</script>";
exit;
}
}

else
{
echo mysql_error() ;
}
}
[/PHP]

[Atli]

Hi Lazandra. Welcome to The Scripts!

I see you print some JavaScript code there. Is that not working?
Could we see the code that is accepting the ID?

Also, I have a few thoughts on your code.

First, all the variables in your first query. Where are they coming from? If they are from a form, are you retrieving them from the $_POST super-global or are you perhaps relying on the register_global s constant? (which is very very bad btw!)

Second, your second query, you use * (wildcard) to fetch every single column of the table, when all you need is the ID column. This will increase the database load, which again may cause problems. I would recommend against using the * (wildcard) wherever possible, even if that means typing the names of 45 out of 50 columns. This will also protect your code from failing due to any future table alterations (especially with UPDATE and INSERT statements).

[Lazandra]

Hi Atli!

Thanks for the reply :-)

I'm using a form and its going through POST. Your right about the query using * i will change it forthwith.

In regards to the javascript; that is working and directing to the correct page. But the id that keeps being passed is "1" instead of the actual id.

This is the accepting code.
[PHP]
<input name="infoid" type="hidden" value="<?=$name s[infoid];?>" />
[/PHP]

I hope this makes sense...

xx

[Atli]

In regards to the javascript; that is working and directing to the correct page. But the id that keeps being passed is "1" instead of the actual id.

This is the accepting code.
[PHP]
<input name="infoid" type="hidden" value="<?=$name s[infoid];?>" />
[/PHP]

This is most likely because the second query in your first code will always fetch every single row. Then you loop through them using a while loop, which will echo the JavaScript with the ID of the first row, causing the browser to be redirected and ignore all the other rows.

To fix this you need to put a WHERE clause and specify the row you want to fetch.

Alternately you could use the mysql_insert_id () function to fetch the ID of the row you inserted with your first query and pass that through your JavaScript.

Also, consider using the header() function instead of the JavaScript.
For example:

# Do this after the INSERT query
header("Location: payment.php?id=". mysql_insert_id());
 

[Lazandra]

The header() function function worked perfectly! And so much simpler!

Spank you so much!

xxx

[Atli]

Glad I could help :)
Don't hesitate to post again if you have any more questions or problems (or anything) we can help with!