How to group by a column - sql

Question

Hi I know how to use the group by clause for sql. I am not sure how to explain this so Ill draw some charts. Here is my original data:
Name Location
----------------------
user1 1
user1 9
user1 3
user2 1
user2 10
user3 97
Here is the output I need
Name Location
----------------------
user1 1
9
3
user2 1
10
user3 97
Is this even possible?

Answer 1

The normal method for this is to handle it in the presentation layer, not the database layer.
Reasons:
The Name field is a property of that data row
If you leave the Name out, how do you know what Location goes with which name?
You are implicitly relying on the order of the data, which in SQL is a very bad practice (since there is no inherent ordering to the returned data)
Any solution will need to involve a cursor or a loop, which is not what SQL is optimized for - it likes working in SETS not on individual rows

Answer 2

Hope this helps
SELECT A.FINAL_NAME, A.LOCATION
FROM (SELECT DISTINCT DECODE((LAG(YT.NAME, 1) OVER(ORDER BY YT.NAME)),
YT.NAME,
NULL,
YT.NAME) AS FINAL_NAME,
YT.NAME,
YT.LOCATION
FROM YOUR_TABLE_7 YT) A
As Jirka correctly pointed out, I was using the Outer select, distinct and raw Name unnecessarily. My mistake was that as I used DISTINCT , I got the resulted sorted like
1 1
2 user2 1
3 user3 97
4 user1 1
5 3
6 9
7 10
I wanted to avoid output like this.
Hence I added the raw id and outer select
However , removing the DISTINCT solves the problem.
Hence only this much is enough
SELECT DECODE((LAG(YT.NAME, 1) OVER(ORDER BY YT.NAME)),
YT.NAME,
NULL,
YT.NAME) AS FINAL_NAME,
YT.LOCATION
FROM SO_BUFFER_TABLE_7 YT
Thanks Jirka

Answer 3

If you're using straight SQL*Plus to make your report (don't laugh, you can do some pretty cool stuff with it), you can do this with the BREAK command:
SQL> break on name
SQL> WITH q AS (
SELECT 'user1' NAME, 1 LOCATION FROM dual
UNION ALL
SELECT 'user1', 9 FROM dual
UNION ALL
SELECT 'user1', 3 FROM dual
UNION ALL
SELECT 'user2', 1 FROM dual
UNION ALL
SELECT 'user2', 10 FROM dual
UNION ALL
SELECT 'user3', 97 FROM dual
)
SELECT NAME,LOCATION
FROM q
ORDER BY name;
NAME LOCATION
----- ----------
user1 1
9
3
user2 1
10
user3 97
6 rows selected.
SQL>

Answer 4

I cannot but agree with the other commenters that this kind of problem does not look like it should ever be solved using SQL, but let us face it anyway.
SELECT
CASE main.name WHERE preceding_id IS NULL THEN main.name ELSE null END,
main.location
FROM mytable main LEFT JOIN mytable preceding
ON main.name = preceding.name AND MIN(preceding.id) < main.id
GROUP BY main.id, main.name, main.location, preceding.name
ORDER BY main.id
The GROUP BY clause is not responsible for the grouping job, at least not directly. In the first approximation, an outer join to the same table (LEFT JOIN below) can be used to determine on which row a particular value occurs for the first time. This is what we are after. This assumes that there are some unique id values that make it possible to arbitrarily order all the records. (The ORDER BY clause does NOT do this; it orders the output, not the input of the whole computation, but it is still necessary to make sure that the output is presented correctly, because the remaining SQL does not imply any particular order of processing.)
As you can see, there is still a GROUP BY clause in the SQL, but with a perhaps unexpected purpose. Its job is to "undo" a side effect of the LEFT JOIN, which is duplication of all main records that have many "preceding" ( = successfully joined) records.
This is quite normal with GROUP BY. The typical effect of a GROUP BY clause is a reduction of the number of records; and impossibility to query or test columns NOT listed in the GROUP BY clause, except through aggregate functions like COUNT, MIN, MAX, or SUM. This is because these columns really represent "groups of values" due to the GROUP BY, not just specific values.

Answer 5

If you are using SQL*Plus, use the BREAK function. In this case, break on NAME.
If you are using another reporting tool, you may be able to compare the "name" field to the previous record and suppress printing when they are equal.

Answer 6

If you use GROUP BY, output rows are sorted according to the GROUP BY columns as if you had an ORDER BY for the same columns. To avoid the overhead of sorting that GROUP BY produces, add ORDER BY NULL:
SELECT a, COUNT(b) FROM test_table GROUP BY a ORDER BY NULL;
Relying on implicit GROUP BY sorting in MySQL 5.6 is deprecated. To achieve a specific sort order of grouped results, it is preferable to use an explicit ORDER BY clause. GROUP BY sorting is a MySQL extension that may change in a future release; for example, to make it possible for the optimizer to order groupings in whatever manner it deems most efficient and to avoid the sorting overhead.
For full information - http://academy.comingweek.com/sql-groupby-clause/

Answer 7

SQL GROUP BY STATEMENT
SQL GROUP BY clause is used in collaboration with the SELECT statement to arrange identical data into groups.
Syntax:
1. SELECT column_nm, aggregate_function(column_nm) FROM table_nm WHERE column_nm operator value GROUP BY column_nm;
Example :
To understand the GROUP BY clauserefer the sample database.Below table showing fields from “order” table:
1. |EMPORD_ID|employee1ID|customerID|shippers_ID|
Below table showing fields from “shipper” table:
1. | shippers_ID| shippers_Name |
Below table showing fields from “table_emp1” table:
1. | employee1ID| first1_nm | last1_nm |
Example :
To find the number of orders sent by each shipper.
1. SELECT shipper.shippers_Name, COUNT (orders.EMPORD_ID) AS No_of_orders FROM orders LEFT JOIN shipper ON orders.shippers_ID = shipper.shippers_ID GROUP BY shippers_Name;
1. | shippers_Name | No_of_orders |
Example :
To use GROUP BY statement on more than one column.
1. SELECT shipper.shippers_Name, table_emp1.last1_nm, COUNT (orders.EMPORD_ID) AS No_of_orders FROM ((orders INNER JOIN shipper ON orders.shippers_ID=shipper.shippers_ID) INNER JOIN table_emp1 ON orders.employee1ID = table_emp1.employee1ID)
2. GROUP BY shippers_Name,last1_nm;
| shippers_Name | last1_nm |No_of_orders |
for more clarification refer my link
http://academy.comingweek.com/sql-groupby-clause/