If result set is empty print "no records"

[DavidPr]

I've tried to incorporate an IF Statement into the below so that if the query returned nothing - a message saying "There are no jobs currently listed" would be displayed, but for some reason it hasn't worked. Right now it just returns a blank table.

[PHP]<?php
$query="SELECT * FROM $nameid WHERE display_ad='yes ' AND submitted > SUBDATE(NOW(), INTERVAL 30 DAY) ORDER BY submitted DESC LIMIT $offset, $rowsPerPage";
$result=mysql_q uery($query) or die('Error, query failed');
$num=mysql_numr ows($result);

$i=0;
while ($i < $num)
{
$id=mysql_resul t($result,$i,"i d");
$company=mysql_ result($result, $i,"company");
$position=mysql _result($result ,$i,"position") ;
$city=mysql_res ult($result,$i, "city");
$date=mysql_res ult($result,$i, "submitted" );
$fdate = explode('-', $date);
?>

<table>
<tr>
<td width="35%"><? echo ("<A HREF='see_job.p hp?nameid=$name id&file=$id'>$c ompany</A>"); ?></td>

<td width="37%"><? echo "$position" ; ?></td>

<td width="16%"><? echo "$city"; ?></td>

<td width="12%"><? echo "$fdate[1]/$fdate[2]/$fdate[0]"; ?></td>
</tr>
</table>

<?php
++$i;
} [/PHP]

[ak1dnar]

First what is this "mysql_numr ows" ? (4th line in the above code)

Then read here please:

<?php
$num=mysql_num_rows($result); 
if($num > 0){
    while($row=mysql_fetch_assoc($result)){
    //Print The $row array elements here
    echo $row['MySQL_COLUMN_NAME1'];
    echo $row['MySQL_COLUMN_NAME2'];
    echo $row['MySQL_COLUMN_NAME3'];
    }
}else{
echo 'No-Records';
}
?>

[DavidPr]

First what is this "mysql_numr ows" ?

It's just something I've always used, but I like yours better. Works fine now, thanks!

David

[ak1dnar]

And David, If you ever need anything post back to the forum. Thanks.