How to Select Top 100 rows in Oracle?

First 10 customers inserted into db (table customers):

select * from customers where customer_id <=
(select  min(customer_id)+10 from customers)

Last 10 customers inserted into db (table customers):

select * from customers where customer_id >=
(select  max(customer_id)-10 from customers)

Hope this helps....

Assuming that create_time contains the time the order was created, and you want the 100 clients with the latest orders, you can:

• add the create_time in your innermost query
• order the results of your outer query by the create_time desc
• add an outermost query that filters the first 100 rows using ROWNUM

Query:

  SELECT * FROM (
     SELECT * FROM (
        SELECT 
          id, 
          client_id, 
          create_time,
          ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn 
        FROM order
      ) 
      WHERE rn=1
      ORDER BY create_time desc
  ) WHERE rownum <= 100

UPDATE for Oracle 12c

With release 12.1, Oracle introduced "real" Top-N queries. Using the new FETCH FIRST... syntax, you can also use:

  SELECT * FROM (
    SELECT 
      id, 
      client_id, 
      create_time,
      ROW_NUMBER() OVER(PARTITION BY client_id ORDER BY create_time DESC) rn 
    FROM order
  ) 
  WHERE rn = 1
  ORDER BY create_time desc
  FETCH FIRST 100 ROWS ONLY)

you should use rownum in oracle to do what you seek

where rownum <= 100

see also those answers to help you

limit in oracle

select top in oracle

select top in oracle 2

As Moneer Kamal said, you can do that simply:

SELECT id, client_id FROM order 
WHERE rownum <= 100
ORDER BY create_time DESC;

Notice that the ordering is done after getting the 100 row. This might be useful for who does not want ordering.

Update:

To use order by with rownum you have to write something like this:

SELECT * from (SELECT id, client_id FROM order ORDER BY create_time DESC) WHERE rownum <= 100;