Count spaces (ASCII code 32) at the beginning and end of string - sql

Question

I'm struggling with the following problem. I have a column with this kind of data:
'abbb ccc '
' aaa abbb ccc'
'abbb ccc '
' aaa abbb ccc '
' ccc'
'aaa abbb'
I want to count the number of spaces on the left and the number of spaces on the right of each string.

Answer 1

Try using a combination of LEN, LTRIM, and REVERSE:
SELECT
LEN(col) - LEN(LTRIM(col)) AS num_left,
LEN(REVERSE(col)) - LEN(LTRIM(REVERSE(col))) AS num_right
FROM yourTable;
Demo
As #AaronDietz mentioned in the comment below, LEN actually also trims whitespace on the right. But LEN does not affect leading whitespace. To compensate for this we can reverse the string and then do the calculation using LTRIM.

Answer 2

DECLARE #s NVARCHAR(50)='abbb ccc '
DECLARE #t NVARCHAR(50)=' aaa abbb ccc'
SELECT
RightSpaces = LEN(RTRIM(REVERSE(#s))) - LEN(#s),
LeftSpaces = LEN(#t) - LEN(LTRIM(REVERSE(#t)))
output
RightSpaces LeftSpaces
3 4
or you can use DATALENGTH and remove the need for REVERSE
SELECT
LeftSpaces = (DATALENGTH(#s)- DATALENGTH(RTRIM(#s)))/2,
RightSpaces = (DATALENGTH(#t)- DATALENGTH(LTRIM(#t)))/2
output
RightSpaces LeftSpaces
3 4

Answer 3

You can ditch LEN and TRIM functions and use PATINDEX instead:
SELECT
str,
PATINDEX('%[^ ]%', str) - 1 AS leading_spaces,
PATINDEX('%[^ ]%', REVERSE(str)) - 1 AS trailing_spaces
FROM testdata
Output:
| str | leading_spaces | trailing_spaces |
+--------------------+----------------+-----------------+
| abbb·ccc··· | 0 | 3 |
| ····aaa·abbb·ccc | 4 | 0 |
| abbb·ccc··· | 0 | 3 |
| ···aaa·abbb·ccc··· | 3 | 3 |
| ···ccc | 3 | 0 |
| aaa·abbb | 0 | 0 |
SQL Fiddle

Answer 4

Use datalength to get the full length
len will rtrim
declare #t varchar(100) = ' abbb ccc ';
select len(#t) as 'len', datalength(#t) as 'datalength',
datalength(#t) - datalength(ltrim(#t)) as 'left',
datalength(#t) - datalength(rtrim(#t)) as 'right'

Answer 5

You can do this...
SELECT Len(col + 'End') - Len(Rtrim(col)) - 3 AS Rno,
Len(col) - Len(Ltrim(col)) AS Lno
Since LEN function does not include trailing spaces, you could add extra text at the end of the string and remove the length of that text later.
Refer this example (4 white spaces)
SELECT LEN('test ') = 4
SELECT LEN('test ' + 'End') = 11
SELECT LEN('test ' + 'End') - 3 = 8