$query = "SELECT 'first_name', 'last_name', 'address1', 'address2', 'city', 'state', 'current_employer', 'entered_by', 'owner', 'email1' FROM candidate INNER JOIN candidate_foreign ON candidate.candidate_id = candidate_foreign.assoc_id WHERE candidate_foreign.field_name = 'IITYear' AND value >= '" .$_POST['from']. "' AND '".$_POST['to']."'"; --------------------------- Now write $query = "SELECT `first_name`, `last_name`, `address1`, address2`, `city`, `state`, `current_employer`, `entered_by`, `owner`, `email1` FROM candidate INNER JOIN candidate_foreign ON candidate.candidate_id = candidate_foreign.assoc_id WHERE candidate_foreign.field_name = 'IITYear' AND value >= " .$_POST['from']. " AND value <= ".$_POST['to']; Or Remove ' from list of the feilds which have you mention after SELECT
<?PHP $loginname = ''; $password = ''; $database = ''; $con = mysql_connect("localhost", "$loginname", "$password") or die("could not connect"); mysql_select_db("$database"); $query = "SELECT * from candidate, candidate_foreign INNER JOIN candidate_foreign ON candidate.candidate_id = candidate_foreign.assoc_id WHERE candidate_foreign.field_name = 'IITYear' AND value >= '" .$_POST['from']. "' AND '".$_POST['to']."'"; $result = mysql_query ("$query"); echo "You are seaching between year '" .$_POST['from']. "' to '" .$_POST['to']. "'"; echo "<TABLE BORDER=1><TH>First Name</TH><TH>Last Name</TH><TH>Year</TH>"; while ($in = mysql_fetch_array($result)) { echo "<TR>"; echo "<TD>" . $in[first_name] . "</TD>"; echo "<TD>" . $in[last_name] ."</TD>"; echo "<TD>" . $in[value] ."</TD>"; echo "</TR>"; } echo "</TABLE>"; mysql_close($con); ?> I want to retrieve year of passing from candidate_foreign table but after run this code i got error as below: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/datalist_year.php on line 21 I tried if (mysql_num_rows($result) before while that time gave and error for mysql_fetch_rows().. How could i get data from both tables? Is there any other code except loop i have to use to retrieve data into php formate. hope you reply soon Regards, Jimi
$query = "SELECT * from candidate INNER JOIN candidate_foreign ON (candidate.candidate_id = candidate_foreign.assoc_id) WHERE candidate_foreign.field_name = 'IITYear' AND value >= " .$_POST['from']. " AND value <= ".$_POST['to'];