Code: Select all
$query = "SELECT candidateID, category, degree, years_exp, work_time, available FROM js_db1";
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result))
{
{$row['candidateID']} <br>" .
{$row['category']} <br>" .
{$row['degree']} <br><br>";
{$row['years_exp']}
{$row['work_time']}
{$row['available']}
}
$query = "SELECT image_name FROM js_db2 WHERE candidateID = '?'";
$result = mysql_query($query);Thanks in advance
Batoe
You can get "there" from "here", If you know where "there" is!