oracle group by一次得到一个组

Question

My Data looks like this

 1. Fname    Lastname   DOB       ADDR   SEX   SSN 
 2. Sanal    Pillai    12/11/85   Pune   M     - 
 3. Sunil    Pillai    05/11/80   Pune   M     123456 
 4. Anju     Pillai    12/11/85   Pune   F     345679 
 5. Arachana Pillai    05/11/80   Pune   M  

-

Now is I do a group by on DOB I would get the data arranged accordingly

select fname,lastname,dob,addr,sex,ssn from data group by dob(..rest of the columns..) having count(dob) > 1

But what I would like to know is, is it possible to get a group at a time, ie

 1. Sanal    Pillai    12/11/85   Pune   M     -
 2. Anju     Pillai    12/11/85   Pune   F     345679

and then the rest two rows

Answer 1

Hi i think you may not know the exact functionality of group by clause. Below i have posted few lines for your understanding.

The Oracle GROUP BY clause is used in a SELECT statement to collect data across multiple records and group the results by one or more columns.

Syntax:

SELECT col1, col2, ... coln, 
       aggregate_function (column_name)
FROM tables
WHERE conditions
GROUP BY col1, col2, ... coln;

The column_names what and all you mentioned in select clause you should also mention in after group by keyword. As per above syntax and your Question The required output query is:

select fname,lastname,dob,addr,sex,sum(ssn) as ssn from data group by fname,lastname,dob,addr,sex  having count(dob) > 1

For More http://www.dba-oracle.com/t_oracle_group_by_having.htm

Answer 2

This is not possible. Sql is used to get data from a database which sometimes has a specific order. Further analysis of the received data is done in the tool you use to show the data. This can be a programming language or another tool that can use sql to receive data.

Answer 3

You can use the RANK or DENSE_RANK analytic functions to assign grouping ID's and order by that:

with t1 as (
  select data.*, dense_rank() over (order by dob) grp from data
)
select t1.* from t1 order by grp;