creating variables which are already stored

[bajer]

i'm POSTING information from previous page, and generating variables:
$a1 = 14;
$a3 = 17;
and so on.....they were generated based on produkt $id number

so for example, I was displaying for array
$id = 1;
and I would generate variable $a1, which would then pass on the information to next page (such as in this case number 14)

But now I run into the problem,
I am trying to display same information, so i'm pulling from aaray
same identical information, but NOW, I cannot create $a1 to display its content, its just shows as "$a1" as text

Would someone have some kind of suggestions?

[vallette]

Do something like this:

$temp='a'.$id;
$$temp=14;
echo 'a'.$id."=".$a1;

If id was equal to 1, the output would be:
'a1=14'

Hope this helps.

[bajer]

You see we don't know what number user inputs, could be 14, could be 2 or 50, as far as we know is that when we generate information from aaray, that for the first item that would be generated $a1, there is a number assigned to it.

When I re-generate information, I want to be able to generate $a1 again, as a VARIABLE, and show then number that was entered on previous page.

For that matter when i'm displaying an item:

 while ($row5 = mysql_fetch_array($result5)) { 
    extract($row5); 

// let's say in this case first id would be 1
echo "$id";

// now i need to create a variable that would get me $a1
??????
// now the value of $a1 should be what user entered on previous page
echo "$a1";
}

perhaps i'm asking for too much? Maybe someone else has another way to SAVE information (units) when i'm trying to pull information of products from 2 different databases, and where one is based on the other. Perhaps I have a bad way of storing values??

Thanks

[zudo]

By a variable that would get you a1, do you mean $_POST ?