sql - Count how many times an integer is appears in a column -

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i'm attempting find way count end of integer , display trailing value. example, have following table:

credit ====== 1051000 10000 2066 16000

i'd able count amount of times '0' appears , end result following

credit              cntoccurancevals ======              ====== 1051000             3 10000               4 2066                0 16000               3

now have tried using query find of '0', problem i'm facing first row returns 4 instead of 3 searching whole row , not trailing. i've tried using find occurrence count

declare @longsequence int(max) declare @findsubstring int(max) set @longsequence = credit set @findsubstring = '0' select (len(@longsequence) - len(replace(@longsequence, @findsubstring, '')))     cntreplacedvals, (len(@longsequence) - len(replace(@longsequence, @findsubstring, '')))/len(@findsubstring) cntoccurancevals

is there way can find trailing 0's , not ones in middle of value?

edit: typo

you use reverse , patindex this:

declare @a table (i int) insert @a select 123000 union select 123001 union select 100000 union select 123001  select i, patindex('%[1-9]%',reverse(cast(i varchar(50)))) - 1   cntoccurancevals @a

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