With Inputs As
(
Select 0 As [timestamp], 'john' As Name, 5 As value
Union All Select 1, NULL, 3
Union All Select 8, NULL, 12
Union All Select 12, 'john', 3
Union All Select 33, NULL, 4
Union All Select 54, 'pete', 1
Union All Select 180, NULL, 4
Union All Select 400, 'john', 3
Union All Select 401, NULL, 4
Union All Select 592, 'anna', 2
)
, NamedInputs As
(
Select I.timestamp
, Coalesce (I.Name
, (
Select I3.Name
From Inputs As I3
Where I3.timestamp = (
Select Max(I2.timestamp)
From Inputs As I2
Where I2.timestamp < I.timestamp
And I2.Name Is not Null
)
)) As name
, I.value
From Inputs As I
)
Select NI.name, Sum(NI.Value) As Total
From NamedInputs As NI
Group By NI.name
Кстати, что было бы на несколько порядков быстрее, чем любой запрос, чтобы сначала исправить данные. Т.е. обновите столбец имени, чтобы он имел правильное значение, сделайте его необнуляемым, а затем запустите простую Группировку, чтобы получить ваши итоги.
Дополнительное решение
Select Coalesce(I.Name, I2.Name), Sum(I.value) As Total
From Inputs As I
Left Join (
Select I1.timestamp, MAX(I2.Timestamp) As LastNameTimestamp
From Inputs As I1
Left Join Inputs As I2
On I2.timestamp < I1.timestamp
And I2.Name Is Not Null
Group By I1.timestamp
) As Z
On Z.timestamp = I.timestamp
Left Join Inputs As I2
On I2.timestamp = Z.LastNameTimestamp
Group By Coalesce(I.Name, I2.Name)