I got it, I think:
and this at the beginning of file "B"
$selected == "$selected";
and placed this statement in file "A"
if (isset($selected[$cat]))
{
include ("cat_select.php");
}
I don't know why it works but it does, is there anything wrong with this method?
Gerry wrote:
>
> I hope you understand:
>
> It is too long to include here so I'll give you an idea.
>
> file.B
> -----------------------
> <?php
> $all_cats = "1, 2, 3";
> function html_options($output,
> $values = NULL,
> $selected = NOT_NULL,
> $first_option_output = false)
> }
> return($html_output);
> }
>
> echo"<select name=\"cat\" size=\"1\">";
> //-cats
> $cats = explode(',',$all_cats);
> //--values
> $cats_vals = explode(',',$all_cats);
> print html_options($cats, $cats_vals, $selected, "--Select one--");
> echo"</SELECT>";
> ?>
> ------------------------
>
> file.A code:
> <?php
> $connect to db;
> $get result;
> $cat = $row["cat"];
> $selected = "$cat";
> -----------------------
> include ("file.B");
> -----------------------
> done
> ?>
> From here on I get the menu with --select one-- option first then 1, 2,
> 3, but it ignores the fact the I told it $selected = "$cat". I figure it
> has to do with including the function file instead of having the
> function inside of "A". How can I call file.B from different files and
> have it accept the variable $selected from such files?
>
> Thaks in advance:
>
> Gerry Figueroa
>
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