Count numeric chars in string - sql

Question

Using tsql I want to count a numeric chars in string. For example i've got 'kick0my234ass' string and i wanna count how many (4 in that example) numbers are in that string. I can't use regex, just plain tslq.

Answer 1

You COULD do this I suppose:
declare #c varchar(30)
set #c = 'kick0my234ass'
select #c, len(replace(#c,' ','')) - len(replace(replace(replace(replace(replace(replace(replace(replace(replace(replace(replace(#c,'0',''),'1',''),'2',''),'3',''),'4',''),'5',''),'6',''),'7',''),'8',''),'9',''),' ',''))

Answer 2

You'll first have to split the character string in its individual characters, evaluate which are numeric, and finally count those that are. This will do the trick:
DECLARE #test TABLE (Example NVARCHAR(255))
INSERT #test
VALUES ('kick0my234ass')
SELECT COUNT(1)
FROM #test AS T
INNER JOIN master..spt_values v
ON v.type = 'P'
AND v.number < len(T.Example)
WHERE SUBSTRING(T.Example, v.number + 1, 1) LIKE '[0-9]'

Answer 3

You could try this solution with regular expressions (if you'd allow them):
it uses recursive CTE, at every recursive step, one digit is removed from given string and the condition is to stop, when there are no digits in string. The rows are also numbered with consecutive ids, so the last id is the amount of removed digits from string.
declare #str varchar(100) = 'kick0my123ass';
with cte as (
select 1 [id], stuff(#str,PATINDEX('%[0-9]%', #str),1,'') [col]
union all
select [id] + 1, stuff([col],PATINDEX('%[0-9]%', [col]),1,'') from cte
where col like '%[0-9]%'
)
--this will give you number of digits in string
select top 1 id from cte order by id desc

Answer 4

Use a WHILE loop to each each character is a numeric or not.
Query
declare #text as varchar(max) = 'kick0my234ass';
declare #len as int;
select #len = len(#text);
if(#len > 0)
begin
declare #i as int = 1;
declare #count as int = 0;
while(#i <= #len)
begin
if(substring(#text, #i, 1) like '[0-9]')
set #count += 1;
set #i += 1;
end
print 'Count of Numerics in ' + #text + ' : ' + cast(#count as varchar(100));
end
else
print 'Empty string';

Answer 5

If simplicity & performance are important I suggest a purely set-based solution. Grab a copy of DigitsOnlyEE which will remove all non-numeric characters. Then use LEN against the output.
DECLARE #string varchar(100) = '123xxx45ff678';
SELECT string = #string, digitsOnly, DigitCount = LEN(digitsOnly)
FROM dbo.DigitsOnlyEE(#string);
Results
string digitsOnly DigitCount
------------------ ----------- ------------
123xxx45ff678 12345678 8

Answer 6

using a Tally Table created by an rCTE:
CREATE TABLE #Sample (S varchar(100));
INSERT INTO #Sample
VALUES ('kick0my234 ass');
GO
WITH Tally AS(
SELECT 1 AS N
UNION ALL
SELECT N + 1
FROM Tally
WHERE N + 1 <= 100)
SELECT S.S, SUM(CASE WHEN SUBSTRING(S,T.N, 1) LIKE '[0-9]' THEN 1 ELSE 0 END) AS Numbers
FROM #Sample S
JOIN Tally T ON LEN(S.S) >= T.N
GROUP BY S.S;
For future reference, also post your owns attempts please. We aren't here (really) to do your work for you.