Add more columns for each value of another column? - sql

Question

I have this kind of table :
Name Date Value
-----------------------
Test 1/1/2001 10
Test 2/1/2001 17
Test 3/1/2001 52
Foo 5/4/2011 15
Foo 6/4/2011 321
My 15/5/2005 36
My 25/7/2005 75
And I would like to show the results like this :
Name Date Value Name Date Value Name Date Value
---------------------------------------------------------------------
Test 1/1/2001 10 Foo 5/4/2011 15 My 15/5/2005 36
Test 2/1/2001 17 Foo 6/4/2011 321 My 25/7/2005 75
Test 3/1/2001 52
I need to show as many columns as what is present in my Name column
How could I do this in Sql ?

Answer 1

In order to get the result that you want, you are going to have to unpivot the columns in your table and apply the pivot function.
The unpivot can be done using either the UNPIVOT function or you can use CROSS APPLY with VALUES.
UNPIVOT:
select rn,
col +'_'+cast(dr as varchar(10)) col,
new_values
from
(
select name,
convert(varchar(10), date, 101) date,
cast(value as varchar(10)) value,
dense_rank() over(order by name) dr,
row_number() over(partition by name order by date) rn
from yourtable
) d
unpivot
(
new_values
for col in (name, date, value)
) un;
CROSS APPLY:
select rn,
col +'_'+cast(dr as varchar(10)) col,
c.value
from
(
select name,
convert(varchar(10), date, 101) date,
cast(value as varchar(10)) value,
dense_rank() over(order by name) dr,
row_number() over(partition by name order by date) rn
from yourtable
) d
cross apply
(
values
('Name', name), ('Date', date), ('Value', Value)
) c (col, value);
See SQL Fiddle with Demo of both versions. This gives the result:
| RN | COL | NEW_VALUES |
-----------------------------
| 1 | name_1 | Foo |
| 1 | date_1 | 04/05/2011 |
| 1 | value_1 | 15 |
| 2 | name_1 | Foo |
| 2 | date_1 | 04/06/2011 |
| 2 | value_1 | 321 |
| 1 | name_2 | My |
| 1 | date_2 | 05/15/2005 |
| 1 | value_2 | 36 |
These queries take your existing columns values and converts them to rows. Once they are in rows, you create the new column names by using the windowing function dense_rank.
Once the data has been converted to rows, you then use the new column names (created with the dense_rank value) and apply the PIVOT function.
PIVOT with UNPIVOT:
select name_1, date_1, value_1,
name_2, date_2, value_2,
name_3, date_3, value_3
from
(
select rn,
col +'_'+cast(dr as varchar(10)) col,
new_values
from
(
select name,
convert(varchar(10), date, 101) date,
cast(value as varchar(10)) value,
dense_rank() over(order by name) dr,
row_number() over(partition by name order by date) rn
from yourtable
) d
unpivot
(
new_values
for col in (name, date, value)
) un
) src
pivot
(
max(new_values)
for col in (name_1, date_1, value_1,
name_2, date_2, value_2,
name_3, date_3, value_3)
) piv;
See SQL Fiddle with Demo
PIVOT with CROSS APPLY:
select name_1, date_1, value_1,
name_2, date_2, value_2,
name_3, date_3, value_3
from
(
select rn,
col +'_'+cast(dr as varchar(10)) col,
c.value
from
(
select name,
convert(varchar(10), date, 101) date,
cast(value as varchar(10)) value,
dense_rank() over(order by name) dr,
row_number() over(partition by name order by date) rn
from yourtable
) d
cross apply
(
values
('Name', name), ('Date', date), ('Value', Value)
) c (col, value)
) src
pivot
(
max(value)
for col in (name_1, date_1, value_1,
name_2, date_2, value_2,
name_3, date_3, value_3)
) piv;
See SQL Fiddle with Demo.
Dyanmic PIVOT:
The above versions will work great if you have a limited or known number of columns, if not, then you will need to use dynamic SQL:
DECLARE #cols AS NVARCHAR(MAX),
#query AS NVARCHAR(MAX)
select #cols = STUFF((SELECT ',' + QUOTENAME(col +'_'+cast(dr as varchar(10)))
from
(
select dense_rank() over(order by name) dr
from yourtable
) t
cross apply
(
values(1, 'Name'), (2, 'Date'), (3, 'Value')
) c (sort, col)
group by col, dr, sort
order by dr, sort
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set #query = 'SELECT ' + #cols + '
from
(
select rn,
col +''_''+cast(dr as varchar(10)) col,
c.value
from
(
select name,
convert(varchar(10), date, 101) date,
cast(value as varchar(10)) value,
dense_rank() over(order by name) dr,
row_number() over(partition by name order by date) rn
from yourtable
) d
cross apply
(
values
(''Name'', name), (''Date'', date), (''Value'', Value)
) c (col, value)
) x
pivot
(
max(value)
for col in (' + #cols + ')
) p'
execute(#query)
See SQL Fiddle with Demo.
The result for each of the queries is:
| NAME_1 | DATE_1 | VALUE_1 | NAME_2 | DATE_2 | VALUE_2 | NAME_3 | DATE_3 | VALUE_3 |
-------------------------------------------------------------------------------------------------
| Foo | 04/05/2011 | 15 | My | 05/15/2005 | 36 | Test | 01/01/2001 | 10 |
| Foo | 04/06/2011 | 321 | My | 07/25/2005 | 75 | Test | 01/02/2001 | 17 |
| (null) | (null) | (null) | (null) | (null) | (null) | Test | 01/03/2001 | 52 |

Answer 2

By hand or with a program. Most people are accustomed to showing output in a linear fashion (the default). If someone is asking you to do this, you can tell them that's not how the application works. You can export the result set to csv and then import that into something like Excel and reformat it by hand or use a serverside language like ASP.net or PHP to format the results into a table.
When you're parsing the output you could check the last var Name against the current. If they're different then add a column. It would still be tricky to script it because they will more than likely come out of the database in order. So you would have a sequence like test, test, test, foo, foo which would mean that you need to create a multidimensional array to organize the data to get a column count. Then setup the table based on that with a counter that counts row names, then data underneath.
I'm not sure which apps you're familiar with, so in PHP the output would look something like this from the multidimensional array.
row [1]['name']=test
row [1][test][1]['date'] = 1/1/2001
This is more of a visual output though. The databases are designed to hold data and return it in an intuitive fashion.