How to count instances of character in SQL Column

Question

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', ''))

In SQL Server:

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

This gave me accurate results every time...

This is in my Stripes field...

Yellow, Yellow, Yellow, Yellow, Yellow, Yellow, Black, Yellow, Yellow, Red, Yellow, Yellow, Yellow, Black

• 11 Yellows
• 2 Black
• 1 Red

SELECT (LEN(Stripes) - LEN(REPLACE(Stripes, 'Red', ''))) / LEN('Red') 
  FROM t_Contacts

DECLARE @StringToFind VARCHAR(100) = "Text To Count"

SELECT (LEN([Field To Search]) - LEN(REPLACE([Field To Search],@StringToFind,'')))/COALESCE(NULLIF(LEN(@StringToFind), 0), 1) --protect division from zero
FROM [Table To Search]

This will return number of occurance of N

select ColumnName, LEN(ColumnName)- LEN(REPLACE(ColumnName, 'N', '')) from Table

The easiest way is by using Oracle function:

SELECT REGEXP_COUNT(COLUMN_NAME,'CONDITION') FROM TABLE_NAME