sql query condition check

If I understand correctly, you want to show an error if order_type is not equal to 1. If so, you can do something like this:

$query = "SELECT order_no FROM sales_orders where order_status='Pending' and order_type = 1";

if ($result = $mysqli->query($query)) {
    while ($row = $result->fetch_object()) {
        if ($row->order_type != 1) {
            printf('There is no rail order for order number %d', $row->order_no);
        }
    }

    $result->close();
}

try this:

SELECT IF(order_type=1,order_no,'Error') 
FROM sales_orders 
WHERE order_status='Pending'

The SQL should be then something like this (you can't use order_type=1 in the WHERE as you want to have the "errornous" results back also, thus you need to use an IF there):

SELECT IF(order_type=1,'OK','Error') AS okErrorFlag ,order_no
FROM sales_orders 
WHERE order_status='Pending'

In sql (I take now an associative array as a result where you have only the result of the appropriate row there):

if ($row['okErrorFlag']=='Error')
{
     echo 'ERROR: There is no rail order';
}

try using case statement

select id,
case order_type
    when '1' then order_type
    when '0' then 'error'
end 
from
`sales_orders`
where order_status='pending'

Finally i got it what i need...

This is the query:

• $sql_da_num= "SELECT order_no FROM sales_orders WHERE order_status='Pending' and order_type = '1');

              $result_da_num= db_query($sql_da_num);
              $num_rows = db_num_rows($result_da_num);
              if ($num_rows==0)
              {
              display_error("There is no Rail Order for this selected customer");
              }
              }