this warning occur and record not display

[9033521295]

<?php
 
//Define your database connections and select your database want to use. In this example I use two connections and two DBs. But you can use more than two.
 
//MySQL Server 1
$dbhost1 = "localhost";
$dbuser1 = "root";
$dbpassword1 = "";
$db1 = "mediteam";
$connection1 = mysql_connect($dbhost1,$dbuser1,$dbpassword1) or die (mysql_error());
mysql_select_db($db1,$connection1);
 
$db2 = "mediteam_medical";
$connection2 = mysql_connect($dbhost1,$dbuser1,$dbpassword1) or die (mysql_error());
mysql_select_db($db2,$connection2);    
 
//The SQL statement
//$sql =" SELECT mediteam.mediteam_area.Area_name AS field1, mediteam_medical.mediteam_medical_medical.fieldname2 AS field2 FROM mediteam.mediteam_area,mediteam_medical.mediteam_medical_medical";
 
$sql =" SELECT mediteam.mediteam_area.Area_name AS, mediteam_medical.mediteam_medical_medical.fieldname2 AS field2  field1 FROM mediteam.mediteam_area,mediteam_medical.mediteam_medical_medical";
 
//Execute query and collect results in $results
$results = mysql_query($sql);
 
//Print result until end of records
while($rows = mysql_fetch_array($results)){
    print $rows["field1"]." | <br>";
}
 
?>

this warning occur and record not display

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\MediTeam\Medical\a.php on line 26

[Dormilich]

your query fails because of an SQL syntax error.

[Dormilich]

with a little formatting, it becomes obvious:

SELECT 
    mediteam.mediteam_area.Area_name AS, 
    mediteam_medical.mediteam_medical_medical.fieldname2 AS field2  field1 
FROM 
    mediteam.mediteam_area,
    mediteam_medical.mediteam_medical_medical