We need to analyze how long your orders take to be shipped from the date that the order is placed. To do this, you must create a report that displays the customer number, date ordered, date shipped, and the number of months in whole numbers from the time the order is placed to the time the order is shipped. Which statement produces the required results?Select one:a.SELECT custid, orderdate, shipdate,ROUNDOFF(shipdate - orderdate) "Time Taken"FROM ord;b.SELECT custid, orderdate, shipdate,ROUND(DAYS_BETWEEN (shipdate, orderdate))/ 30) "Time Taken"FROM ord;c.SELECT custid, orderdate, shipdate,ROUND(MONTHS_BETWEEN (shipdate, orderdate))"Time Taken" FROM ORD;d.SELECT custid, orderdate, shipdate,MONTHS_BETWEEN (shipdate, orderdate)"Time Taken"FROM ord;

We need to create a report to display the order id, ship date and order total of your ORDER table. If the order has not been shipped, your report must display 'Not Shipped'. If the total is not available, your report must display 'Not Available'.In the ORDER table, the SHIPDATE column has a datatype of DATE. The TOTAL column has a datatype of INT.Which statement do you use to create this report?Select one:a.SELECT ordid,TO_CHAR(shipdate, 'Not Shipped'),TO_CHAR(total,'Not Available')FROM order;b.SELECT ordid, IFNULL(shipdate, 'Not Shipped') as SHIPDATE,Total FROM order;c.SELECT ordid, NVL(to_char(shipdate), 'Not Shipped') SHIPDATE,NVL(total,'Not Available')TOTAL FROM order;d.SELECT ordid, shipdate "Not Shipped",total "Not Available"FROM order;

Get the time range between which the orders were placed.

Select distinct customers who have placed orders where the Amount is greater than $25 AND OrderDate is between '2024-06-01' and '2024-07-15' OR RestaurantID is 2.

Problem StatementFrom the following table, write a query to count all the orders generated on "2012-08-17." Return the number of orders.Table name: ordersNote: Table names are case sensitive.Input format :The input table is already created, and records are already prepopulated, as mentioned in the problem statement.Output format :The output displays the count of all the orders generated on '2012-08-17' as shown below.Count2

Query the customer_number from the orders table for the customer who has placed the largest number of orders.It is guaranteed that exactly one customer will have placed more orders than any other customer.The orders table is defined as follows:ColumnTypeorder_number (PK)intcustomer_numberintorder_datedaterequired_datedateshipped_datedatestatuschar(15)commentchar(200)Sample Inputorder_numbercustomer_numberorder_daterequired_dateshipped_date112017-04-092017-04-132017-04-12222017-04-152017-04-202017-04-18332017-04-162017-04-252017-04-20442017-04-182017-04-282017-04-25Sample Outputcustomer_number3ExplanationThe customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them only has one order.So the result is customer_number '3'. Follow up: What if more than one customer have the largest number of orders, can you find all the customer_number in this case?Optionsselect customer_numberfrom ( select customer_number, count(*) as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_number select customer_number, count(*) as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_numberfrom ( select customer_number, count(*) as cnt from orders group by customer_number) as elimit 1;select customer_numberfrom ( select customer_number, count(*) as cnt group by customer_number) as eorder by e.cnt desclimit 1;