In SQL how can I remove (from displaying on my report no deleting from database) the first 3 characters (CN=) and everything after the comma that is followed by "OU" so that I am left with the name and last name in the same column? for example:
CN=Tom Chess,OU=records,DC=1234564786_data for testing, 1234567
CN=Jack Bauer,OU=records,DC=1234564786_data for testing, 1234567
CN=John Snow,OU=records,DC=1234564786_data for testing, 1234567
CN=Anna Rodriguez,OU=records,DC=1234564786_data for testing, 1234567
Desired display:
Tom Chess
Jack Bauer
John Snow
Anna Rodriguez
I tried playing with TRIM but I don't know how to do it without declaring the position and with names and last names having different lengths I really don't know how to handle that.
Thank you in advance
Update: I wonder about an approach of using Locate to match the position of the comma and then feed that to a sub-string. Not sure if a approach like would work and not sure how to put the syntax together. What do you think? will it be a feasible approach?
You can try this one SUBSTRING(ColumnName, 4, CHARINDEX(',', ColumnName) - 4)
In Postgres, you could use split_part() assuming no name contains a ,
select substr(split_part(the_column, ',', 1), 4)
from ...
Db2 11.x for LUW:
with tab (str) as (values
' CN = Tom Chess , OU = records,DC=1234564786_data for testing, 1234567'
, 'CN=Jack Bauer,OU=records,DC=1234564786_data for testing, 1234567'
, 'CN=John Snow,OU=records,DC=1234564786_data for testing, 1234567'
, 'CN=Anna Rodriguez,OU=records,DC=1234564786_data for testing, 1234567'
)
select REGEXP_REPLACE(str, '^\s*CN\s*=\s*(.*)\s*,\s*OU\s*=.*', '\1')
from tab;
Note, that such a regex pattern allows an arbitrary number of spaces as in the 1-st record of example above.
In Oracle 11g, it might work.
REGEXP_SUBSTR(REGEXP_SUBSTR(COLUMN_NAME, '[^CN=]+',1,1),'[^,OU]+',1,1)
I think there has to be a loop to handle this. Here's SQL Server function that will parse this out. (I know the question didn't specify SQL Server, but it's an example of how it can be done.)
select dbo.ScrubFieldValue(value) from table will return what you're looking for
CREATE FUNCTION ScrubFieldValue
(
#Input varchar(8000)
)
RETURNS varchar(8000)
AS
BEGIN
DECLARE #retval varchar(8000)
DECLARE #charidx int
DECLARE #remaining varchar(8000)
DECLARE #current varchar(8000)
DECLARE #currentLength int
select #retval = ''
select #remaining = #Input
select #charidx = CHARINDEX('CN=', #remaining,2)
while(LEN(#remaining) > 0)
BEGIN
--strip current row from remaining
if (#charidx > 0)
BEGIN
select #current = SUBSTRING(#remaining, 1, #charidx - 1)
END
else
BEGIN
select #current = #remaining
END
select #currentLength = LEN(#current)
-- get current name
select #current = SUBSTRING(#current, 4, CHARINDEX(',OU', #current)-4)
select #retval = #retval + #current + ' '
-- strip off current from remaining
select #remaining =substring(#remaining,#currentLength + 1,
LEN(#remaining) - #currentLength)
select #charidx = CHARINDEX('CN=', #remaining,2)
END
RETURN #retval
END
On my version of DB2 for Z/OS CHARINDEX throws a syntax error. Here are two ways to work around that.
SUBSTRING(ColumnName, 4, INSTR(ColumnName,',',1) - 4)
SUBSTRING(ColumnName, 4, LOCATE_IN_STRING(ColumnName,',') - 4)
I should add that the version is V12R1
If input str is wellformed (i.e. looks like your sample data without any additional tokens such as space), you could use something like:
substr(str,locate('CN=', str)+length('CN='), locate(',', str)-length('CN=')-1)
If your Db2 version support REGEXP, that's a better choice.
Related
I have a requirement to run a function over certain fields to identify and redact any numbers which are 5 digits or longer, ensuring all but the last 4 digits are replaced with *
For example: "Some text with 12345 and 1234 and 12345678" would become "Some text with *2345 and 1234 and ****5678"
I've used PATINDEX to identify the the starting character of the pattern:
PATINDEX('%[0-9][0-9][0-9][0-9][0-9]%', TEST_TEXT)
I can recursively call that to get the starting character of all the occurrences, but I'm struggling with the actual redaction.
Does anyone have any pointers on how this can be done? I know to use REPLACE to insert the *s where they need to be, it's just the identification of what I should actually be replacing I'm struggling with.
Could do it on a program, but I need it to be T-SQL (can be a function if needed).
Any tips greatly appreciated!
You can do this using the built in functions of SQL Server. All of which used in this example are present in SQL Server 2008 and higher.
DECLARE #String VARCHAR(500) = 'Example Input: 1234567890, 1234, 12345, 123456, 1234567, 123asd456'
DECLARE #StartPos INT = 1, #EndPos INT = 1;
DECLARE #Input VARCHAR(500) = ISNULL(#String, '') + ' '; --Sets input field and adds a control character at the end to make the loop easier.
DECLARE #OutputString VARCHAR(500) = ''; --Initalize an empty string to avoid string null errors
WHILE (#StartPOS <> 0)
BEGIN
SET #StartPOS = PATINDEX('%[0-9][0-9][0-9][0-9][0-9]%', #Input);
IF #StartPOS <> 0
BEGIN
SET #OutputString += SUBSTRING(#Input, 1, #StartPOS - 1); --Seperate all contents before the first occurance of our filter
SET #Input = SUBSTRING(#Input, #StartPOS, 500); --Cut the entire string to the end. Last value must be greater than the original string length to simply cut it all.
SET #EndPos = (PATINDEX('%[0-9][0-9][0-9][0-9][^0-9]%', #Input)); --First occurance of 4 numbers with a not number behind it.
SET #Input = STUFF(#Input, 1, (#EndPos - 1), REPLICATE('*', (#EndPos - 1))); --#EndPos - 1 gives us the amount of chars we want to replace.
END
END
SET #OutputString += #Input; --Append the last element
SET #OutputString = LEFT(#OutputString, LEN(#OutputString))
SELECT #OutputString;
Which outputs the following:
Example Input: ******7890, 1234, *2345, **3456, ***4567, 123asd456
This entire code could also be made as a function since it only requires an input text.
A dirty solution with recursive CTE
DECLARE
#tags nvarchar(max) = N'Some text with 12345 and 1234 and 12345678',
#c nchar(1) = N' ';
;
WITH Process (s, i)
as
(
SELECT #tags, PATINDEX('%[0-9][0-9][0-9][0-9][0-9]%', #tags)
UNION ALL
SELECT value, PATINDEX('%[0-9][0-9][0-9][0-9][0-9]%', value)
FROM
(SELECT SUBSTRING(s,0,i)+'*'+SUBSTRING(s,i+4,len(s)) value
FROM Process
WHERE i >0) calc
-- we surround the value and the string with leading/trailing ,
-- so that cloth isn't a false positive for clothing
)
SELECT * FROM Process
WHERE i=0
I think a better solution it's to add clr function in Ms SQL Server to manage regexp.
sql-clr/RegEx
Here is an option using the DelimitedSplit8K_LEAD which can be found here. https://www.sqlservercentral.com/articles/reaping-the-benefits-of-the-window-functions-in-t-sql-2 This is an extension of Jeff Moden's splitter that is even a little bit faster than the original. The big advantage this splitter has over most of the others is that it returns the ordinal position of each element. One caveat to this is that I am using a space to split on based on your sample data. If you had numbers crammed in the middle of other characters this will ignore them. That may be good or bad depending on you specific requirements.
declare #Something varchar(100) = 'Some text with 12345 and 1234 and 12345678';
with MyCTE as
(
select x.ItemNumber
, Result = isnull(case when TRY_CONVERT(bigint, x.Item) is not null then isnull(replicate('*', len(convert(varchar(20), TRY_CONVERT(bigint, x.Item))) - 4), '') + right(convert(varchar(20), TRY_CONVERT(bigint, x.Item)), 4) end, x.Item)
from dbo.DelimitedSplit8K_LEAD(#Something, ' ') x
)
select Output = stuff((select ' ' + Result
from MyCTE
order by ItemNumber
FOR XML PATH('')), 1, 1, '')
This produces: Some text with *2345 and 1234 and ****5678
I have a string:
DECLARE #UserComment AS VARCHAR(1000) = 'bjones marked inspection on system UP for site COL01545 as Refused to COD won''t pay upfront :Routeid: 12 :Inspectionid: 55274'
Is there a way for me to extract everything from the string after 'Inspectionid: ' leaving me just the InspectionID to save into a variable?
Your example doesn't quite work correctly. You defined your variable as varchar(100) but there are more characters in your string than that.
This should work based on your sample data.
DECLARE #UserComment AS VARCHAR(1000) = 'bjones marked inspection on system UP for site COL01545 as Refused to COD won''t pay upfront :Routeid: 12 :Inspectionid: 55274'
select right(#UserComment, case when charindex('Inspectionid: ', #UserComment, 0) > 0 then len(#UserComment) - charindex('Inspectionid: ', #UserComment, 0) - 13 else len(#UserComment) end)
I would do this as:
select stuff(#UserComment, 1, charindex(':Inspectionid: ', #UserComment) + 14, '')
This works even if the string is not found -- although it will return the whole string. To get an empty string in this case:
select stuff(#UserComment, 1, charindex(':Inspectionid: ', #UserComment + ':Inspectionid: ') + 14, '')
Firstly, let me say that your #UserComment variable is not long enough to contain the text you're putting into it. Increase the size of that first.
The SQL below will extract the value:
DECLARE #UserComment AS VARCHAR(1000); SET #UserComment = 'bjones marked inspection on system UP for site COL01545 as Refused to COD won''t pay upfront :Routeid: 12 :Inspectionid: 55274'
DECLARE #pos int
DECLARE #InspectionId int
DECLARE #IdToFind varchar(100)
SET #IdToFind = 'Inspectionid: '
SET #pos = CHARINDEX(#IdToFind, #UserComment)
IF #pos > 0
BEGIN
SET #InspectionId = CAST(SUBSTRING(#UserComment, #pos+LEN(#IdToFind)+1, (LEN(#UserComment) - #pos) + 1) AS INT)
PRINT #InspectionId
END
You could make the above code into a SQL function if necessary.
If the Inspection ID is always 5 digits then the last argument for the Substring function (length) can be 5, i.e.
SELECT SUBSTRING(#UserComment,PATINDEX('%Inspectionid:%',#UserComment)+14,5)
If the Inspection ID varies (but is always at the end - which your question slightly implies), then the last argument can be derived by subtracting the position of 'InspectionID:' from the overall length of the string. Like this:
SELECT SUBSTRING(#UserComment,PATINDEX('%Inspectionid:%',#UserComment)+14,LEN(#usercomment)-(PATINDEX('%Inspectionid:%',#UserComment)+13))
Table FOO has a column FILEPATH of type VARCHAR(512). Its entries are absolute paths:
FILEPATH
------------------------------------------------------------
file://very/long/file/path/with/many/slashes/in/it/foo.xml
file://even/longer/file/path/with/more/slashes/in/it/baz.xml
file://something/completely/different/foo.xml
file://short/path/foobar.xml
There's ~50k records in this table and I want to know all distinct filenames, not the file paths:
foo.xml
baz.xml
foobar.xml
This looks easy, but I couldn't find a DB2 scalar function that allows me to search for the last occurrence of a character in a string. Am I overseeing something?
I could do this with a recursive query, but this appears to be overkill for such a simple task and (oh wonder) is extremely slow:
WITH PATHFRAGMENTS (POS, PATHFRAGMENT) AS (
SELECT
1,
FILEPATH
FROM FOO
UNION ALL
SELECT
POSITION('/', PATHFRAGMENT, OCTETS) AS POS,
SUBSTR(PATHFRAGMENT, POSITION('/', PATHFRAGMENT, OCTETS)+1) AS PATHFRAGMENT
FROM PATHFRAGMENTS
)
SELECT DISTINCT PATHFRAGMENT FROM PATHFRAGMENTS WHERE POS = 0
I think what you're looking for is the LOCATE_IN_STRING() scalar function. This is what Info Center has to say if you use a negative start value:
If the value of the integer is less than zero, the search begins at
LENGTH(source-string) + start + 1 and continues for each position to
the beginning of the string.
Combine that with the LENGTH() and RIGHT() scalar functions, and you can get what you want:
SELECT
RIGHT(
FILEPATH
,LENGTH(FILEPATH) - LOCATE_IN_STRING(FILEPATH,'/',-1)
)
FROM FOO
One way to do this is by taking advantage of the power of DB2s XQuery engine. The following worked for me (and fast):
SELECT DISTINCT XMLCAST(
XMLQuery('tokenize($P, ''/'')[last()]' PASSING FILEPATH AS "P")
AS VARCHAR(512) )
FROM FOO
Here I use tokenize to split the file path into a sequence of tokens and then select the last of these tokens. The rest is only conversion from SQL to XML types and back again.
I know that the problem from the OP was already solved but I decided to post the following anyway to hopefully help others like me that land here.
I came across this thread while searching for a solution to my similar problem which had the exact same requirement but was for a different kind of database that was also lacking the REVERSE function.
In my case this was for a OpenEdge (Progress) database, which has a slightly different syntax. This made the INSTR function available to me that most Oracle typed databases offer.
So I came up with the following code:
SELECT
SUBSTRING(
foo.filepath,
INSTR(foo.filepath, '/',1, LENGTH(foo.filepath) - LENGTH( REPLACE( foo.filepath, '/', '')))+1,
LENGTH(foo.filepath))
FROM foo
However, for my specific situation (being the OpenEdge (Progress) database) this did not result into the desired behaviour because replacing the character with an empty char gave the same length as the original string. This doesn't make much sense to me but I was able to bypass the problem with the code below:
SELECT
SUBSTRING(
foo.filepath,
INSTR(foo.filepath, '/',1, LENGTH( REPLACE( foo.filepath, '/', 'XX')) - LENGTH(foo.filepath))+1,
LENGTH(foo.filepath))
FROM foo
Now I understand that this code won't solve the problem for T-SQL because there is no alternative to the INSTR function that offers the Occurence property.
Just to be thorough I'll add the code needed to create this scalar function so it can be used the same way like I did in the above examples.
-- Drop the function if it already exists
IF OBJECT_ID('INSTR', 'FN') IS NOT NULL
DROP FUNCTION INSTR
GO
-- User-defined function to implement Oracle INSTR in SQL Server
CREATE FUNCTION INSTR (#str VARCHAR(8000), #substr VARCHAR(255), #start INT, #occurrence INT)
RETURNS INT
AS
BEGIN
DECLARE #found INT = #occurrence,
#pos INT = #start;
WHILE 1=1
BEGIN
-- Find the next occurrence
SET #pos = CHARINDEX(#substr, #str, #pos);
-- Nothing found
IF #pos IS NULL OR #pos = 0
RETURN #pos;
-- The required occurrence found
IF #found = 1
BREAK;
-- Prepare to find another one occurrence
SET #found = #found - 1;
SET #pos = #pos + 1;
END
RETURN #pos;
END
GO
To avoid the obvious, when the REVERSE function is available you do not need to create this scalar function and you can just get the required result like this:
SELECT
SUBSTRING(
foo.filepath,
LEN(foo.filepath) - CHARINDEX('\', REVERSE(foo.filepath))+2,
LEN(foo.filepath))
FROM foo
You could just do it in a single statement:
select distinct reverse(substring(reverse(FILEPATH), 1, charindex('/', reverse(FILEPATH))-1))
from filetable
I currently have a table Telephone it has entries like the following:
9073456789101
+773456789101
0773456789101
What I want to do is remove only the 9 from the start of all the entries that have a 9 there but leave the others as they are.
any help would be greatly appreciated.
While all other answer are probably also working, I'd suggest to try and use STUFF function to easily replace a part of the string.
UPDATE Telephone
SET number = STUFF(number,1,1,'')
WHERE number LIKE '9%'
SQLFiddle DEMO
Here is the code and a SQLFiddle
SELECT CASE
WHEN substring(telephone_number, 1, 1) <> '9'
THEN telephone_number
ELSE substring(telephone_number, 2, LEN(telephone_number))
END
FROM Telephone
Update Telephone set number = RIGHT(number,LEN(number)-1) WHERE number LIKE '9%';
I recently solved a similar problem with a combination of RIGHT(), LEN() & PATINDEX(). PATINDEX will return the integer 1 when it finds a 9 as the first character and 0 otherwise. This method allows all records to be returned at once without a CASE WHEN statement.
SELECT
RIGHT(number, LEN(number) - PATINDEX('9%', number))
FROM Telephone
UPDATE dbo.Telephone
SET column_name = SUBSTRING(column_name, 2, 255)
WHERE column_name LIKE '9%';
Stuff is a great function for this. However, using it with an update statement with a where clause is great, but what if I was doing an insert, and I needed all of the rows inserted in one pass. The below will remove the first character if it is a period, does not use the slower case statement, and converts nulls to an empty string.
DECLARE #Attachment varchar(6) = '.GIF',
#Attachment2 varchar(6)
SELECT
#Attachment2 = ISNULL(ISNULL(NULLIF(LEFT(#Attachment, 1), '.'), '') + STUFF(#Attachment, 1, 1, ''), '')
SELECT
#Attachment2
DECLARE #STR nvarchar(200) = 'TEST'
SET #STR = STUFF(#STR,1,1,'')
PRINT #STR
Result will be "EST"
You can use replace in select statement instead of where or update
SELECT REPLACE(REPLACE('_'+number,'_9',''),'_','') FROM #tbl
I have an sql column that is a string of 100 'Y' or 'N' characters. For example:
YYNYNYYNNNYYNY...
What is the easiest way to get the count of all 'Y' symbols in each row.
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
Maybe something like this...
SELECT
LEN(REPLACE(ColumnName, 'N', '')) as NumberOfYs
FROM
SomeTable
Below solution help to find out no of character present from a string with a limitation:
1) using SELECT LEN(REPLACE(myColumn, 'N', '')), but limitation and
wrong output in below condition:
SELECT LEN(REPLACE('YYNYNYYNNNYYNY', 'N', ''));
--8 --Correct
SELECT LEN(REPLACE('123a123a12', 'a', ''));
--8 --Wrong
SELECT LEN(REPLACE('123a123a12', '1', ''));
--7 --Wrong
2) Try with below solution for correct output:
Create a function and also modify as per requirement.
And call function as per below
select dbo.vj_count_char_from_string('123a123a12','2');
--2 --Correct
select dbo.vj_count_char_from_string('123a123a12','a');
--2 --Correct
-- ================================================
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: VIKRAM JAIN
-- Create date: 20 MARCH 2019
-- Description: Count char from string
-- =============================================
create FUNCTION vj_count_char_from_string
(
#string nvarchar(500),
#find_char char(1)
)
RETURNS integer
AS
BEGIN
-- Declare the return variable here
DECLARE #total_char int; DECLARE #position INT;
SET #total_char=0; set #position = 1;
-- Add the T-SQL statements to compute the return value here
if LEN(#string)>0
BEGIN
WHILE #position <= LEN(#string) -1
BEGIN
if SUBSTRING(#string, #position, 1) = #find_char
BEGIN
SET #total_char+= 1;
END
SET #position+= 1;
END
END;
-- Return the result of the function
RETURN #total_char;
END
GO
try this
declare #v varchar(250) = 'test.a,1 ;hheuw-20;'
-- LF ;
select len(replace(#v,';','11'))-len(#v)
If you want to count the number of instances of strings with more than a single character, you can either use the previous solution with regex, or this solution uses STRING_SPLIT, which I believe was introduced in SQL Server 2016. Also you’ll need compatibility level 130 and higher.
ALTER DATABASE [database_name] SET COMPATIBILITY_LEVEL = 130
.
--some data
DECLARE #table TABLE (col varchar(500))
INSERT INTO #table SELECT 'whaCHAR(10)teverCHAR(10)whateverCHAR(10)'
INSERT INTO #table SELECT 'whaCHAR(10)teverwhateverCHAR(10)'
INSERT INTO #table SELECT 'whaCHAR(10)teverCHAR(10)whateverCHAR(10)~'
--string to find
DECLARE #string varchar(100) = 'CHAR(10)'
--select
SELECT
col
, (SELECT COUNT(*) - 1 FROM STRING_SPLIT (REPLACE(REPLACE(col, '~', ''), 'CHAR(10)', '~'), '~')) AS 'NumberOfBreaks'
FROM #table
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!
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:
If you need to count the char in a string with more then 2 kinds of chars, you can use instead of 'n' - some operator or regex of the chars accept the char you need.
SELECT LEN(REPLACE(col, 'N', ''))
Try this:
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);
It determines the number of single character occurrences as well as the sub-string occurrences in main string.
Here's what I used in Oracle SQL to see if someone was passing a correctly formatted phone number:
WHERE REPLACE(TRANSLATE('555-555-1212','0123456789-','00000000000'),'0','') IS NULL AND
LENGTH(REPLACE(TRANSLATE('555-555-1212','0123456789','0000000000'),'0','')) = 2
The first part checks to see if the phone number has only numbers and the hyphen and the second part checks to see that the phone number has only two hyphens.
for example to calculate the count instances of character (a) in SQL Column ->name is column name
'' ( and in doblequote's is empty i am replace a with nocharecter #'')
select len(name)- len(replace(name,'a','')) from TESTING
select len('YYNYNYYNNNYYNY')- len(replace('YYNYNYYNNNYYNY','y',''))
DECLARE #char NVARCHAR(50);
DECLARE #counter INT = 0;
DECLARE #i INT = 1;
DECLARE #search NVARCHAR(10) = 'Y'
SET #char = N'YYNYNYYNNNYYNY';
WHILE #i <= LEN(#char)
BEGIN
IF SUBSTRING(#char, #i, 1) = #search
SET #counter += 1;
SET #i += 1;
END;
SELECT #counter;