LEN function not including trailing spaces in SQL Server

Question

I have the following test table in SQL Server 2005:

CREATE TABLE [dbo].[TestTable]
(
 [ID] [int] NOT NULL,
 [TestField] [varchar](100) NOT NULL
) 

Populated with:

INSERT INTO TestTable (ID, TestField) VALUES (1, 'A value');   -- Len = 7
INSERT INTO TestTable (ID, TestField) VALUES (2, 'Another value      '); -- Len = 13 + 6 spaces

When I try to find the length of TestField with the SQL Server LEN() function it does not count the trailing spaces - e.g.:

-- Note: Also results the grid view of TestField do not show trailing spaces (SQL Server 2005).
SELECT 
 ID, 
 TestField, 
 LEN(TestField) As LenOfTestField, -- Does not include trailing spaces
FROM 
 TestTable

How do I include the trailing spaces in the length result?

Answer 1

This is clearly documented by Microsoft in MSDN at http://msdn.microsoft.com/en-us/library/ms190329(SQL.90).aspx, which states LEN "returns the number of characters of the specified string expression, excluding trailing blanks". It is, however, an easy detail on to miss if you're not wary.

You need to instead use the DATALENGTH function - see http://msdn.microsoft.com/en-us/library/ms173486(SQL.90).aspx - which "returns the number of bytes used to represent any expression".

Example:

SELECT 
    ID, 
    TestField, 
    LEN(TestField) As LenOfTestField,           -- Does not include trailing spaces
    DATALENGTH(TestField) As DataLengthOfTestField      -- Shows the true length of data, including trailing spaces.
FROM 
    TestTable

Answer 2

You can use this trick:

LEN(Str + 'x') - 1

Answer 3

I use this method:

LEN(REPLACE(TestField, ' ', '.'))

I prefer this over DATALENGTH because this works with different data types, and I prefer it over adding a character to the end because you don't have to worry about the edge case where your string is already at the max length.

Note: I would test the performance before using it against a very large data set; though I just tested it against 2M rows and it was no slower than LEN without the REPLACE...

Answer 4

"How do I include the trailing spaces in the length result?"

You get someone to file a SQL Server enhancement request/bug report because nearly all the listed workarounds to this amazingly simple issue here have some deficiency or are inefficient. This still appears to be true in SQL Server 2012. The auto trimming feature may stem from ANSI/ISO SQL-92 but there seems to be some holes (or lack of counting them).

Please vote up "Add setting so LEN counts trailing whitespace" here:

https://feedback.azure.com/forums/908035-sql-server/suggestions/34673914-add-setting-so-len-counts-trailing-whitespace

Retired Connect link: https://connect.microsoft.com/SQLServer/feedback/details/801381

Answer 5

There are problems with the two top voted answers. The answer recommending DATALENGTH is prone to programmer errors. The result of DATALENGTH must be divided by the 2 for NVARCHAR types, but not for VARCHAR types. This requires knowledge of the type you're getting the length of, and if that type changes, you have to diligently change the places you used DATALENGTH.

There is also a problem with the most upvoted answer (which I admit was my preferred way to do it until this problem bit me). If the thing you are getting the length of is of type NVARCHAR(4000), and it actually contains a string of 4000 characters, SQL will ignore the appended character rather than implicitly cast the result to NVARCHAR(MAX). The end result is an incorrect length. The same thing will happen with VARCHAR(8000).

What I've found works, is nearly as fast as plain old LEN, is faster than LEN(@s + 'x') - 1 for large strings, and does not assume the underlying character width is the following:

DATALENGTH(@s) / DATALENGTH(LEFT(LEFT(@s, 1) + 'x', 1))

This gets the datalength, and then divides by the datalength of a single character from the string. The append of 'x' covers the case where the string is empty (which would give a divide by zero in that case). This works whether @s is VARCHAR or NVARCHAR. Doing the LEFT of 1 character before the append shaves some time when the string is large. The problem with this though, is that it does not work correctly with strings containing surrogate pairs.

There is another way mentioned in a comment to the accepted answer, using REPLACE(@s,' ','x'). That technique gives the correct answer, but is a couple orders of magnitude slower than the other techniques when the string is large.

Given the problems introduced by surrogate pairs on any technique that uses DATALENGTH, I think the safest method that gives correct answers that I know of is the following:

LEN(CONVERT(NVARCHAR(MAX), @s) + 'x') - 1

This is faster than the REPLACE technique, and much faster with longer strings. Basically this technique is the LEN(@s + 'x') - 1 technique, but with protection for the edge case where the string has a length of 4000 (for nvarchar) or 8000 (for varchar), so that the correct answer is given even for that. It also should handle strings with surrogate pairs correctly.

Answer 6

LEN cuts trailing spaces by default, so I found this worked as you move them to the front

(LEN(REVERSE(TestField))

So if you wanted to, you could say

SELECT
t.TestField,
LEN(REVERSE(t.TestField)) AS [Reverse],
LEN(t.TestField) AS [Count]
FROM TestTable t
WHERE LEN(REVERSE(t.TestField)) <> LEN(t.TestField)

Don't use this for leading spaces of course.

Answer 7

You need also to ensure that your data is actually saved with the trailing blanks. When ANSI PADDING is OFF (non-default):

Trailing blanks in character values inserted into a varchar column are trimmed.

Answer 8

This is the best algorithm I've come up with which copes with the maximum length and variable byte count per character issues:

ISNULL(LEN(STUFF(@Input, 1, 1, '') + '.'), 0)

This is a variant of the LEN(@Input + '.') - 1 algorithm but by using STUFF to remove the first character we ensure that the modified string doesn't exceed maximum length and remove the need to subtract 1.

ISNULL(..., 0) is added to deal with the case where @Input = '' which causes STUFF to return NULL.

This does have the side effect that the result is also 0 when @Input is NULL which is inconsistent with LEN(NULL) which returns NULL, but this could be dealt with by logic outside this function if need be

Here are the results using LEN(@Input), LEN(@Input + '.') - 1, LEN(REPLACE(@Input, ' ', '.')) and the above STUFF variant, using a sample of @Input = CAST(' S' + SPACE(3998) AS NVARCHAR(4000)) over 1000 iterations

Algorithm	DataLength	ExpectedResult	Result	ms
LEN	8000	4000	2	14
+DOT-1	8000	4000	1	13
REPLACE	8000	4000	4000	514
STUFF+DOT	8000	4000	4000	0

In this case the STUFF algorithm is actually faster than LEN()!

I can only assume that internally SQL looks at the last character and if it is not a space then optimizes the calculation But that's a good result eh?

Don't use the REPLACE option unless you know your strings are small - it's hugely inefficient

Answer 9

You should define a CLR function that returns the String's Length field, if you dislike string concatination. I use LEN('x' + @string + 'x') - 2 in my production use-cases.

Answer 10

If you dislike the DATALENGTH because of of n/varchar concerns, how about:

select DATALENGTH(@var)/isnull(nullif(DATALENGTH(left(@var,1)),0),1)

which is just

select DATALENGTH(@var)/DATALENGTH(left(@var,1))

wrapped with divide-by-zero protection.

By dividing by the DATALENGTH of a single char, we get the length normalised.

(Of course, still issues with surrogate-pairs if that's a concern.)

Answer 11

use SELECT DATALENGTH('string ')