In article <[EMAIL PROTECTED]>,
 [EMAIL PROTECTED] ("Shahmat Dahlan") wrote:

> $result=mysql_query($sqlstmt);
> I know the $sqlstmt query does work. But if I do a mysql_fetch_array and
> assign it to a variable called $myrow,
> while ($myrow=mysql_fetch_array($result)) {
>    ...
> }
> how do I display the content array $myrow? When I do a,
> printf("%s&nbsp;", $myrow["equip.equip_type"]);
> nothing is visible.

$result=mysql_query($sqlstmt) or die("Oops! MySQL error: " . 
mysql_error()); //"know" the query is valid

if(mysql_num_rows($result)) //"know" the query gave you something to fetch
   while ($myrow=mysql_fetch_array($result))

Where "..." is:

   printf("%s&nbsp;", $myrow["equip_type"]); //no table name


   printf("%s&nbsp;", $equip_type); //no table name and easier to read ;-)


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