Question

$mysqli = new mysqli("localhost", "user", "pass", "db");

$sql = "SELECT OrgNo, CompanyName 
        FROM ematch 
        WHERE CompanyName LIKE '%A%'";

$result = mysqli_query($mysqli, $sql);

while ($row = mysql_fetch_array($result)) 
{
     print $row['OrgNo'] .'<br />';
     print $row['CompanyName'] .'<br />';
}

Error produced is: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in

Answer 1

I think you can’t feed a mysqli resource to a mysql function. Try turning mysql_fetch_array(...) into mysqli_fetch_array(...).