Re: formating output

[Tatang Widyanto <tatang@xxxxxxxxxxxxxxx>]

print "<table>\n";
print "<tr>\n";

$counter = 0;
while ($result = mysql_fetch_array($pointer)) {
    $counter++;
    if ($counter % 4 === 0) {
        print "<td>".$result[1]."</td>\n";
        print "</tr>\n";
        print "<tr>\n";
    } else {
        print "<td>".$result[1]."</td>\n";
    }
}

// add blank cells if number of cell on last row isn'4 four.
for ($i = 0; $i < $counter % 4; $i++) {
    print "<td>&nbsp;</td>\n";
}

print "</tr>\n";
print "</table>\n";


---------------------------------

Larry L wrote:
> I'm sure this is easy and I'll feel silly when I see how it's done,
> but, my mnd is a blank,
>
> this structure prints out a table with a new row for each item
>
>  <table >
>              <?
>        While ( $results = mysql_fetch_array($pointer)) {
>     print   "<tr><td ><$results[1] </td></tr>" ;
>     }
>     ?>
>         </table>
>
> what I want to do is print out a table 4 columns wide and however many rows
> high needed to run through the data
>
> HOW do I loop through the array and  get <td>data</td>   printed 4 times
> then a new row started and then 4 more <td> cells printed, etc
>
>
> thank you in advance

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