Count Occurances Of Given Value

Hello All,

I have a question that has been vexing me for some time now. It keeps coming up when I'm trying to write queries for SSRS reports. Lets say I have a table that has 3 columns to keep track of people's gender in a annonomys survay (very basic example):

Month (varchar) | Year (smallint) | Gender (bool)

I want to return a dataset that is grouped by Month and Year and that contains a count of each Gender which would look something like this:

Month | Year | [Male Count] | [Female Count]

January | 2006 | 100 | 120
February | 2006 | 130 | 110
March | 2006 | 120 | 145
April | 2006 | 105 | 125

How would I acheive a dataset like this? Is it possible? Do I need to join the table to itself? If so do I use an Inner Join, an Outer Join, or a Left/Right Join? Any help would be extremely appreciated.

Thanks!

Tennyson

There are several possibilities. One option:

SELECT [Month], [Year],
COUNT(CASE WHEN Gender = 0 THEN 1 END) AS MaleCount,
COUNT(CASE WHEN Gender = 1 THEN 1 END) AS FemaleCount
FROM YourSurveyTable

GROUP BY [Month], [Year]

-Sue

|||

Thank you Sue! Your help was much appreciated.

Thanks,

Tennyson

Count Occurances Of Given Value

Hello All,

I have a question that has been vexing me for some time now. It keeps coming up when I'm trying to write queries for SSRS reports. Lets say I have a table that has 3 columns to keep track of people's gender in a annonomys survay (very basic example):

Month (varchar) | Year (smallint) | Gender (bool)

I want to return a dataset that is grouped by Month and Year and that contains a count of each Gender which would look something like this:

Month | Year | [Male Count] | [Female Count]

January | 2006 | 100 | 120
February | 2006 | 130 | 110
March | 2006 | 120 | 145
April | 2006 | 105 | 125

How would I acheive a dataset like this? Is it possible? Do I need to join the table to itself? If so do I use an Inner Join, an Outer Join, or a Left/Right Join? Any help would be extremely appreciated.

Thanks!

Tennyson

There are several possibilities. One option:

SELECT [Month], [Year],
COUNT(CASE WHEN Gender = 0 THEN 1 END) AS MaleCount,
COUNT(CASE WHEN Gender = 1 THEN 1 END) AS FemaleCount
FROM YourSurveyTable

GROUP BY [Month], [Year]

-Sue

|||

Thank you Sue! Your help was much appreciated.

Thanks,

Tennyson

count occurances in sql

Hi,

I am trying to count the times a machine has been scanned in to a table. what happens is everytime a product is read at a station it is entered in to the table with the following data

stationid
modelnumber
serialnumber

i want to establish a list of every machine that has been read at station id 15 but not station id 2 or 5, does this make sense? basically to establish what machines are not being read at points a and b but end up at point c.

obviously to do this i need to check the model and serials throughout the table to see if there is another occurance of it against station 2 or 5.

i have done this before but cannot recreate it for the lfie of me!

all help is greatly welcomed :)

Quote:

Originally Posted by 692909824

Hi,

I am trying to count the times a machine has been scanned in to a table. what happens is everytime a product is read at a station it is entered in to the table with the following data

stationid
modelnumber
serialnumber

i want to establish a list of every machine that has been read at station id 15 but not station id 2 or 5, does this make sense? basically to establish what machines are not being read at points a and b but end up at point c.

obviously to do this i need to check the model and serials throughout the table to see if there is another occurance of it against station 2 or 5.

i have done this before but cannot recreate it for the lfie of me!

all help is greatly welcomed :)

Your script should be:

SELECT *
FROM [TABLE]
WHERE serialno NOT IN
(SELECT serialno
FROM [TABLE]
WHERE stationid = 2 or stationid = 5)
ANDstationid =15

Count Occurances in a column

Is there a way to count the number of times a character or string
appears in a column in a table row/column?
Example, tblTable has column StringData varchar(1000)
StringData contains the value "Mary Had A Little Lamb Lamb and Then It
Died"
I want to know how many times "Lamb" appears in this string.
Any help is appreciated.
ThanksOne popular way to do this is:
SELECT ( LEN( stringdata ) -
LEN( REPLACE( stringdata, 'Lamb', '' ) ) ) / LEN( 'Lamb' )
FROM tbl ;
Alternatively, you can use a table of sequentially incrementing numbers and
construct a generic logic using SUBSTRING functions too.
Anith|||laurenq uantrell wrote:
> Is there a way to count the number of times a character or string
> appears in a column in a table row/column?
> Example, tblTable has column StringData varchar(1000)
> StringData contains the value "Mary Had A Little Lamb Lamb and Then It
> Died"
> I want to know how many times "Lamb" appears in this string.
> Any help is appreciated.
> Thanks
DECLARE @.str VARCHAR(1000)
SET @.str= 'Lamb'
SELECT LEN(REPLACE(stringdata,@.str,@.str+'_'))-LEN(stringdata)
FROM tbltable;
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx
--|||LEN has its problems, so here's one with DATALENGTH:
declare @.text nvarchar(4000)
declare @.string nvarchar(4000)
set @.text = N'Mary Had A Little Lamb Lamb and Then It Died'
set @.string = N'Lamb'
select (datalength(@.text) - datalength(replace(@.text, @.string, N''))) /
datalength(@.string)
ML
http://milambda.blogspot.com/|||Yup, that is right.
Anith|||DECLARE @.StringData varchar(1000)
DECLARE @.counter int
SET @.StringData = 'Mary Had A Little Lamb Lamb and Then It Died'
SET @.counter = 0
WHILE PATINDEX('%Lamb%',@.StringData) <> 0
BEGIN
SET @.counter = @.counter + 1
SET @.StringData = STUFF(@.StringData, PATINDEX('%Lamb%',@.StringData), 4, '')
END
PRINT @.counter
If you want to do this for all rows in a table, one way to accomplish this
is to put the above into a (gasp!) cursor:
DECLARE @.pkcol int --Change this to match that datatype of your PK column
DECLARE @.StringData varchar(1000)
DECLARE @.counter int
CREATE TABLE #results (pkcol int, StringCount int) --same here
DECLARE my_cursor CURSOR STATIC FORWARD_ONLY
FOR
SELECT pkcol, StringData FROM tblTable
FETCH NEXT FROM my_cursor INTO @.pkcol, @.StringData
WHILE (@.@.FETCH_STATUS = 0 )
BEGIN
SET @.counter = 0
WHILE PATINDEX('%Lamb%',@.StringData) <> 0
BEGIN
SET @.counter = @.counter + 1
SET @.StringData = STUFF(@.StringData, PATINDEX('%Lamb%',@.StringData), 4,
'')
END
INSERT INTO #results VALUES (@.pkcol, @.counter)
FETCH NEXT FROM my_cursor INTO @.pkcol, @.StringData
END
SELECT pkcol, StringCount FROM #results
"laurenq uantrell" wrote:

> Is there a way to count the number of times a character or string
> appears in a column in a table row/column?
> Example, tblTable has column StringData varchar(1000)
> StringData contains the value "Mary Had A Little Lamb Lamb and Then It
> Died"
> I want to know how many times "Lamb" appears in this string.
> Any help is appreciated.
> Thanks
>|||Mark,
A single SELECT with a table of numbers may do better than a cursor :
SELECT COUNT(*)
FROM Nbrs
WHERE SUBSTRING( @.stringdata, n, LEN( 'Lamb' ) ) = 'Lamb' ;
Anith|||There's only crap on TV, so...
http://milambda.blogspot.com/2006/0...th-strings.html
ML
http://milambda.blogspot.com/|||On 16 Feb 2006 14:11:22 -0800, laurenq uantrell wrote:

>Is there a way to count the number of times a character or string
>appears in a column in a table row/column?
>Example, tblTable has column StringData varchar(1000)
>StringData contains the value "Mary Had A Little Lamb Lamb and Then It
>Died"
>I want to know how many times "Lamb" appears in this string.
>Any help is appreciated.
>Thanks
Hi laurenq,
SELECT ( DATALENGTH(StringData)
- DATALENGTH(REPLACE(StringData, 'Lamb', '')) )
/ DATALENGTH('Lamb')
FROM tblTable
Hugo Kornelis, SQL Server MVP|||Thanks David and all other posters for these solutions.
lq