sql server - How to pivoting -


i don't understanding pivoting , need on simple query using union. query gives me results fellows:

select    'luft' [source],    sum([salesamount1]) [sales old],    sum([salesamount2]) [sales new] [dbo].[tblsales] ([source] in ('luft')) , [planung] = 'kosten'  union  select    'tgl' [source],    sum([salesamount1]) [sales old],    sum([salesamount2]) [sales new] [dbo].[tblsales] ([source] in ('tgl')) , [planung] = 'kosten'  union  select    'bhw' [source],    sum([salesamount1]) [sales old],    sum([salesamount2]) [sales new] [dbo].[tblsales] ([source] in ('bhw')) , [planung] = 'kosten'  union  select     'wm ' [source],     sum([salesamount1]) [sales old],     sum([salesamount2]) [sales new] [dbo].[tblsales] ([source] in ('wm')) , [planung] = 'kosten' 

results:

[source]      [sales old]      [sales new]  luft           10000            20000  tgl            500              800  bhw            1500             500  wm             700              400 

my expectation using pivot:

[]          [luft]   [tgl]  [bhw]  [wm] sales old    10000    500    1500   700 sales new    20000    800    500    400 

your use of union appends 2 select statements together. reason results how are, you're selecting source , old/new sales columns.

for desired output write below:

if have questions method or query feel free ask.

select    'sales old' [category],   sum(case when [source] = 'luft' [salesamount1] else 0 end) [luft],   sum(case when [source] = 'tgl' [salesamount1] else 0 end) [tgl],   sum(case when [source] = 'bhw' [salesamount1] else 0 end) [bhw],   sum(case when [source] = 'wm' [salesamount1] else 0 end) [wm] [dbo].[tblsales] [planung] = 'kosten'  union  select    'sales new' [category],   sum(case when [source] = 'luft' [salesamount2] else 0 end) [luft],   sum(case when [source] = 'tgl' [salesamount2] else 0 end) [tgl],   sum(case when [source] = 'bhw' [salesamount2] else 0 end) [bhw],   sum(case when [source] = 'wm' [salesamount2] else 0 end) [wm] [dbo].[tblsales] [planung] = 'kosten' 

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