sql server - Return difference of comparing percentage value of a column -

i have colum 10,000 finincial records. values are: id pk, owingvalue example data

id   owingvalue 1    123.2 2    123.4 3    123.5 4    123.6 5    123.7 6    140.2 7    140.3 

for giving column more 0.7% difference previous record, return records high difference.

in case result column 7 because (column 6 owingvalue - column 5 owingvalue) multiply 100 give 1.650 percent higher our threshold of 0.7% need sql while loop or curso me solve problem.

thanks in advance

try this:

declare @t table ( id int, v money )  insert  @t values  ( 1, 123.2 ),         ( 2, 123.4 ),         ( 3, 123.5 ),         ( 4, 123.6 ),         ( 5, 123.7 ),         ( 6, 140.2 ),         ( 7, 140.3 )  select  t1.id    @t t1         join @t t2 on t1.id = t2.id + 1         join @t t3 on t2.id = t3.id + 1   ( t2.v - t3.v ) / t3.v * 100 > 0.7 

output:

id 7 

if ids have gaps then:

declare @t table ( id int, v money )  insert  @t values  ( 1, 123.2 ),         ( 2, 123.4 ),         ( 3, 123.5 ),         ( 5, 123.6 ),         ( 9, 123.7 ),         ( 16, 140.2 ),         ( 27, 140.3 );      cte           ( select   row_number() on ( order id ) id ,                         v                    @t              )     select  t1.id        cte t1             join cte t2 on t1.id = t2.id + 1             join cte t3 on t2.id = t3.id + 1       ( t2.v - t3.v ) / t3.v * 100 > 0.7