Create a procedure named "insert_route"  which has key_id and route_name as the input parameter and status as output parameter. This procedure insert values into the table route. The values are passed as the input parameter to the procedure.Note:If the key_id value does not exists in the table, then set the status parameter as Inserted successfully.If the key_id value already exists in the table then you should handle the exception or warning by displaying the statusas Sorry

Here is a step-by-step guide to creating the procedure "insert_route" in SQL:

1. First, you need to define the procedure. Use the CREATE PROCEDURE statement to do this. The syntax is as follows:

CREATE PROCEDURE insert_route

2. Next, you need to specify the input parameters for the procedure. These are key_id and route_name. You also need to specify the output parameter, status. The syntax is as follows:

(
  IN key_id INT,
  IN route_name VARCHAR(255),
  OUT status VARCHAR(255)
)

3. Now, you need to define the body of the procedure. This is where you specify what the procedure should do. In this case, you want to insert values into the route table. The syntax is as follows:

BEGIN
  DECLARE exist INT;
  SELECT COUNT(*) INTO exist FROM route WHERE id = key_id;
  IF exist = 0 THEN
    INSERT INTO route(id, name) VALUES(key_id, route_name);
    SET status = 'Inserted successfully';
  ELSE
    SET status = 'Sorry';
  END IF;
END;

4. Finally, you need to end the procedure definition with a semicolon. The complete procedure definition is as follows:

CREATE PROCEDURE insert_route
(
  IN key_id INT,
  IN route_name VARCHAR(255),
  OUT status VARCHAR(255)
)
BEGIN
  DECLARE exist INT;
  SELECT COUNT(*) INTO exist FROM route WHERE id = key_id;
  IF exist = 0 THEN
    INSERT INTO route(id, name) VALUES(key_id, route_name);
    SET status = 'Inserted successfully';
  ELSE
    SET status = 'Sorry';
  END IF;
END;

This procedure first checks if the key_id exists in the route table. If it does not exist, it inserts the key_id and route_name into the table and sets the status to 'Inserted successfully'. If the key_id does exist, it sets the status to 'Sorry'.