## Include a column to count records with a specific value

I want to return all data in a table and append a column that counts the number of records in a subset (say, the number of houses in a neighborhood).

I tried

CASE WHEN EXISTS (SELECT 1 as [parcels] FROM dbo.parcels p2 WHERE p2.Neighborhood = p.Neighborhood) THEN COUNT([parcels]) END -- can't count outside subquery as [TotalProps]

The subquery itself returns a value of 1 for each property record in any given neighborhood, but I can't count/sum the [parcels] outside of the subquery in a THEN statement.

Input Table:

dbo.parcels ID Address Neighborhood == ======= ============ 1 123 Main St MITO 2 124 Main St MITO 3 200 2nd St MITO 4 201 2nd St MITO 5 5 Park Ave FAIRWIND 6 1600 Baker St GALLERY 7 1601 Baker St GALLERY 8 1602 Baker St GALLERY

SELECT *, <<<COUNT(neighborhood props)>>> as [TotalProps] FROM dbo.parcels p

Expected Output:

ID Address Neighborhood TotalProps == ======= ============ ========== 1 123 Main St MITO 4 2 124 Main St MITO 4 3 200 2nd St MITO 4 4 201 2nd St MITO 4 5 5 Park Ave FAIRWIND 1 6 1600 Baker St GALLERY 3 7 1601 Baker St GALLERY 3 8 1602 Baker St GALLERY 3

You can use COUNT OVER PARTITION aggregate:

SELECT p.*, COUNT(ID) OVER(PARTITION BY Neighborhood) AS TotalProps FROM dbo.parcels p

**Excel formula: Count rows that contain specific values,** To count rows that contain specific values, you can use an array formula based on the MMULT, TRANSPOSE, COLUMN, and SUM functions. How to count cells with certain text (partial match) The formula discussed in the previous example matches the criteria exactly. If there is at least one different character in a cell, for instance an extra space in the end, that won't be an exact match and such a cell won't be counted.

Use window functions:

select p.*, count(*) over (partition by neighborhood) from dbo.parcels p;

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Keeping things simple - a basic subselect will give you what you need ...

SELECT p.*, ( select count(*) FROM dbo.parcels p2 WHERE p2.neighborhood = p1.neighborhood ) AS hoodcount FROM dbo.parcels p

**How to count how many cells contain certain text or value in Excel?,** You can count number of cells which contain certain text or value with formula as Reuse: Quickly insert complex formulas, charts and anything that you have without losing Data; Split Cells Content; Combine Duplicate Rows/Columns. The SQL COUNT() function returns the number of rows in a table satisfying the criteria specified in the WHERE clause. It sets the number of rows or non NULL column values. COUNT() returns 0 if there were no matching rows. Syntax: COUNT(*) COUNT( [ALL|DISTINCT] expression ) The above syntax is the general SQL 2003 ANSI standard syntax.

**How to Count in Excel Functions,** The arguments (e.g. value1) can be cell references, or values typed the COUNT function will include any cells that contain dates. In cell A7, enter a COUNTA formula, to count the numbers in column A: =COUNTA(A1:A5). I thought I could use "=Sum[late_count]" in the footer (if the text box is named "late_count"), but that doesn't work--when I try to open the query, it asks for late_count. I'm using "=Count(*)" to get the total number of records--is there a way to use Count but just against certain field values?

**Count number of times value appears in particular column in MySQL?,** Insert records in the table using insert command. The query is as follows. mysql> insert into CountSameValue values(1,'Sam',67); Query OK,� To count cells that contain either one value or another, you can either use a helper column then tally up the count, or a more complex single cell formula. Background When you count cells with "OR" criteria, you need to be careful not to double

I want to show the count of specific value of a column from a table as my KPI - I have used below function - CountValues = CALCULATE ( COUNTROWS ( Sheet1 ), Sheet1[XXX_place] = " open_XXX " ) but the count is showing as 1 instead of actual count. Please let me know hat is incorrect in it.

##### Comments

- Pretty sure you meant to partition by neighborhood. :)
- @SeanLange . . . Thank you.