Do you understand the options that the other people have explained?

-- Rodney

"Ashley M. Kirchner" wrote:

> "Rodney J. Woodruff" wrote:
>
> > http://www.php.net/manual/en/function.msql-fetch-array.php
>
>     Okay, call me dense.  I can't figure this out.  This is what I'm trying to
> do:
>
>     $sql = "select p_id, project from proj where uid=$uid";
>     $result = mysql_db_query($database,$sql);
>
>     (the resulting table in mysql is as follows:
>       +------+-----------+
>       | p_id | project   |
>       +------+-----------+
>       |    0 | Undefined |
>       |    1 | Work      |
>       |    2 | Personal  |
>       +------+-----------+
>       3 rows in set (0.00 sec)
>
>     ...yes, that 'Undefined' IS a valid project, and the p_id's don't
> necessarily start at 0 either.)
>
>     I need that result into the following:
>     $items = array(0 => "Undefined", 1 => "Work", 2 => "Personal");
>
>     Reason is, I pass that $items variable to the following function:
>
>   function MakeSelect($items, $selected) {
>     $str = "";
>     while(list($value, $name) = each($items)) {
>       $str .= "<option value=\"$value\"" . ($value != $selected ? \
>               ">" : " selected>") . "$name\n";
>     }
>     return $str;
>   }
>
>     ...which then creates (assuming the person had 'Work' previously
> selected):
>
>     <select name=whatever_i_specify>
>     <option value="0">Undefined
>     <option value="1" selected>Work
>     <option value="2">Personal
>     </select>
>
>     How do I create that $items array?
>
>     AMK4
>
> --
> W |
>   |  I haven't lost my mind; it's backed up on tape somewhere.
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>   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>   Ashley M. Kirchner <mailto:[EMAIL PROTECTED]>   .   303.442.6410 x130
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