This question already has answers here:
Number of times a particular character appears in a string
(9 answers)
Closed 4 years ago.
If i have string 'ABCDAAARSTRLAMA ' how can i count 'A' this char from the given string in sql server 2008
use len and Replace function
declare #myvar varchar(20)
set #myvar = 'ABCDAAARSTRLAMA'
select len(#myvar) - len(replace(#myvar,'A',''))
http://sqlfiddle.com/#!18/063d8/1
SELECT LENGTH( 'ABCDAAARSTRLAMA' ) - LENGTH( replace( 'ABCDAAARSTRLAMA', 'A', '' ) );
Length Function
The Length function in SQL is used to get the length of a string. This function has a different name for different databases:
MySQL: LENGTH( )
Oracle: LENGTH( )
SQL Server: LEN( )
Syntax
The syntax for the Length function is as follows:
LENGTH(str)
Try this one
SELECT (LENGTH('ABCDAAARSTRLAMA') - LENGTH(REPLACE('ABCDAAARSTRLAMA', 'A', '')));
select len('ABCDAAARSTRLAMA') - len(replace('ABCDAAARSTRLAMA', 'A', ''))
This would give the count of 'A' from the string
It's been a while...I'd like to get your advice on the most efficient way to join on only the number part of a field that may be prefixed and/or suffixed with up to 2 letters. Here's a simplified snippet of what I'm trying to do:
SELECT a, b, c
FROM table 1 t1
LEFT JOIN table 2 t2 ON t1.PolicyCode = t2.sPolicyID,
Where t2.sPolicyID could begin and/or end with up to 2 letters. Some examples: TG73100, S7286674, 2344506R, etc. We only want to join to just its numeric part in between the letters, i.e. 73100, 7286674 or 2344506 from the examples.
Could someone please advise on a simple way of doing this?
Here is one way:
LEFT JOIN table 2 t2 ON t1.PolicyCode =
LEFT(SUBSTRING(t2.sPolicyID, PATINDEX('%[0-9]%', t2.sPolicyID), 50),
PATINDEX('%[^0-9]%',
SUBSTRING(t2.sPolicyID, PATINDEX('%[0-9]%', t2.sPolicyID), 50)
+ 'a') -1)
To break this down, there are 4 main parts.
1: Find the position of the first number with PATINDEX:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT PATINDEX('%[0-9]%', #spolicyID)
--Returns 3
2: Use SUBSTRING() to cut off everything before the first letter:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50)
--Returns 123123xx
If we hardcoded the 3 that we know is returned from the first part, it would look like this:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT SUBSTRING(#spolicyID, 3), 50)
--50 is the number of characters to extract, set to something
--higher than the max string length to be safe
Of course, we don't want to hardcode it since it can change, but that makes seeing the different functions a bit easier.
3: Find the position of the next letter using PATINDEX again:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT PATINDEX('%[^0-9]%', SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50) + 'a')
--Returns 7 since it is looking at 123123xx
--The first x is in the 7th position
Note that we added an a onto the string. This is because if we had a string with no letters at the end, it would throw an error as the length 0 would be returned to SUBSTRING. You could add any letter or letters to the end and it would work, we are just making sure there is at least one. Try removing the + 'a' and using a string like xx123123 to see the error.
If we hardcoded the 123123xx from step 2 it would look like this (again just for easy example):
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT PATINDEX('%[^0-9]%', '123123xx' + 'a')
4: Use LEFT() to return everything before the trailing letters, leaving us with only the numbers in between:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
LEFT(SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50),PATINDEX('%[^0-9]%', SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50) + 'a') -1)
--Need to add `-1` because step 3 PATINDEX returns 7
--as the position of first trailing letter, and
--we want the 6 characters before that
And again hardcoded from step 2 and 3 for easy viewing:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
LEFT('123123xx', 7-1)
I have a string like this 'sdf,11-df,12-asd,sadfsdf'. But I need only numbers from this string in multiple rows like this.
11
12
I need numbers between (,) and (-) that's the actual requirement. String can be like 'sd47f,11-df,12-asd,sadfsdf,12-ds,32-fsdfsd' anything, But numbers will always between (,) and (-)
or find numbers between (,) and (-) it will also help me.
Thanks in advance.
I need any solution guys please help,
Following will work with SQL Server 2016 or higher:
DECLARE #str varchar(50) = 'sd47f,11-df,12-asd,sadfsdf,12-ds,32-fsdfsd'
SELECT
LEFT(Value, CHARINDEX('-', Value)-1)
FROM STRING_SPLIT(#str, ',')
WHERE PATINDEX('%[0-9]-%',Value) > 0
try this
DECLARE #data nvarchar(max) = 'sdf,11-df,12-asd,sadfsdf'
select substring(#data,PATINDEX('%[0-9]%',#data),2)
union all
select substring(#data,PATINDEX('%[-]%',REVERSE(#data))-1,2)
Update
You can also use a function which returns as a table
CREATE FUNCTION dbo.udf_GetNumeric (#strAlphaNumeric VARCHAR(256))
RETURNS TABLE AS RETURN ( select LEFT(value,2) as Value from
string_split(#strAlphaNumeric,',') where PATINDEX('%[0-9]-%',value)
> 0 )
END GO
SELECT * from
dbo.udf_GetNumeric('sd47f,11-df,12-asd,sadfsdf,12-ds,32-fsdfsd')
This question already has answers here:
Formatting Numbers by padding with leading zeros in SQL Server
(14 answers)
Closed 7 years ago.
I have been working on a query (in sql Server TSQL) which fills left of a number with 0's so output is always 5 digit.
So:
Select MuNumber From Mytable
for data 11,011,2132,1111
Creates output like
00011
02134
01111
I tried Lpad Function but numer of 0's can be different.
if Munumber is 1 we need 0000 and If MyNumber is 34 we need 000
Assuming that MuNumber is VARCHAR simply use RIGHT
SELECT RIGHT('00000' + MuNumber, 5)
FROM Mytable
Otherwise you need to convert it first
SELECT RIGHT('00000' + CONVERT(VARCHAR(5), MuNumber), 5)
FROM Mytable
And in general you can use this pattern:
DECLARE #num INT = 10;
SELECT RIGHT(REPLICATE('0', #num) + CONVERT(VARCHAR(5), MuNumber), #num)
FROM Mytable
Try this
select right('00000'+cast(col as varchar(5)),5) from table
You can use the user defined function udfLeftSQLPadding where you can find the source codes at SQL Pad Leading Zeros
After you create the function on your database, you can use it as follows
select
dbo.udfLeftSQLPadding(MuNumber,5,'0')
from dbo.Mytable
Another option:
declare #n int = 6
select stuff(replicate('0', #n), 6-len(n), len(n), n)
from (values('123'), ('2493'), ('35')) as x(n)
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;