How to count instances of character in SQL Column

Question

I have an sql column that is a string of 100 \'Y\' or \'N\' characters. For example:

YYNYNYYNNNYYNY...

What is the easiest way

Answer 1

Try This. It determines the no. of single character occurrences as well as the sub-string occurrences in main string.

SELECT COUNT(DECODE(SUBSTR(UPPER(:main_string),rownum,LENGTH(:search_char)),UPPER(:search_char),1)) search_char_count
FROM DUAL
connect by rownum <= length(:main_string);

Answer 2

The second answer provided by nickf is very clever. However, it only works for a character length of the target sub-string of 1 and ignores spaces. Specifically, there were two leading spaces in my data, which SQL helpfully removes (I didn't know this) when all the characters on the right-hand-side are removed. Which meant that

" John Smith"

generated 12 using Nickf's method, whereas:

" Joe Bloggs, John Smith"

generated 10, and

" Joe Bloggs, John Smith, John Smith"

Generated 20.

I've therefore modified the solution slightly to the following, which works for me:

Select (len(replace(Sales_Reps,' ',''))- len(replace((replace(Sales_Reps, ' ','')),'JohnSmith','')))/9 as Count_JS

I'm sure someone can think of a better way of doing it!

Answer 3

You can also Try This

-- DECLARE field because your table type may be text
DECLARE @mmRxClaim nvarchar(MAX) 

-- Getting Value from table
SELECT top (1) @mmRxClaim = mRxClaim FROM RxClaim WHERE rxclaimid_PK =362

-- Main String Value
SELECT @mmRxClaim AS MainStringValue

-- Count Multiple Character for this number of space will be number of character
SELECT LEN(@mmRxClaim) - LEN(REPLACE(@mmRxClaim, 'GS', ' ')) AS CountMultipleCharacter

-- Count Single Character for this number of space will be one
SELECT LEN(@mmRxClaim) - LEN(REPLACE(@mmRxClaim, 'G', '')) AS CountSingleCharacter

Output:

Answer 4

In SQL Server:

SELECT LEN(REPLACE(myColumn, 'N', '')) 
FROM ...

Answer 5

This snippet works in the specific situation where you have a boolean: it answers "how many non-Ns are there?".

SELECT LEN(REPLACE(col, 'N', ''))

If, in a different situation, you were actually trying to count the occurrences of a certain character (for example 'Y') in any given string, use this:

SELECT LEN(col) - LEN(REPLACE(col, 'Y', ''))

Answer 6

try this

declare @v varchar(250) = 'test.a,1  ;hheuw-20;'
-- LF   ;
select len(replace(@v,';','11'))-len(@v)