sql - Select unique barcode with max timestamp -

sql - Select unique barcode with max timestamp -

i'm trying write sql query using microsoft sql server.

my table looks similar this:

bardcode | products | timestamp 1234 | 12 | 2013-02-19 00:01:00 1234 | 17 | 2013-02-19 00:01:00 432 | 10 | 2013-02-19 00:01:00 432 | 3 | 2013-02-19 00:02:00 643 | 32 | 2013-02-19 00:03:00 643 | 3 | 2013-02-19 00:03:00 123 | 10 | 2013-02-19 00:04:00

i'm trying list of unique barcodes based on recent timestamps.

here not-working query i've been thinking of:

select distinct [barcode], [timestamp] [inventorylocatordb].[dbo].[inventory] max([timestamp])

edit i'd retain additional columns well. do bring together or someting.

for illustration want select the barcode latest timestamp , of other column info come e.g. products column

this should work:

select [barcode], max([timestamp]) [inventorylocatordb].[dbo].[inventory] grouping [barcode]

demo

edit

select [barcode], [products], [timestamp] [inventorylocatordb].[dbo].[inventory] [timestamp] = (select max([timestamp]) [inventorylocatordb].[dbo].[inventory] [barcode] = i.[barcode])

the query retains tuples same barcode / timestamp. depending on granularity of timestamp may not valid.

demo 2

there many ways "filter" above result.

e.g. 1 tuple per barcode, latest timestamp, greatest value of products:

select [barcode], [products], [timestamp] [inventorylocatordb].[dbo].[inventory] [timestamp] = (select max([timestamp]) [inventorylocatordb].[dbo].[inventory] [barcode] = i.[barcode]) , [products] = (select max([products]) [inventorylocatordb].[dbo].[inventory] [barcode] = i.[barcode] , [timestamp] = i.[timestamp])

demo 3

sql sql-server