CODEWITHSUNDEEP
sql
Rphani
Rphani

Reputation: 1

Display Count and distinct values for all columns in a table

I have a table with 700 columns. I am trying to get a list of distinct values for each column and their count. I am using the below query to get the result for 1 column

Select distinct col1, count(*) from MyTable group by 1.

Result:

col1    count(*)
a       10
b       20
c       40

How can I get the result for all columns using a single query in the most optimal way?

Upvotes: 0

Views: 6241

Answers (3)

Sekkou
Sekkou

Reputation: 1

Replace [table name] with the table you need counts for. 

DECLARE @table varchar(100) = '[table name]'
DECLARE @i INT = 1, @cntOUT int, @SQL nvarchar(500) = ''
DECLARE @ParmDef nvarchar(500) = N'@cnt int OUTPUT'; 

SELECT column_id, name, 0 as record_count
INTO #T1  
FROM sys.all_columns c 
WHERE c.object_id = (SELECT object_id FROM sys.objects WHERE name = @table AND type = 'U')

WHILE @i <= (SELECT MAX(column_id) FROM #T1)
BEGIN
    SELECT @SQL = 'SELECT @cnt = COUNT(DISTINCT ' + name + ') FROM ' + @table + ';' 
    FROM #T1 WHERE column_id = @i;

    EXECUTE sp_executesql @stmt = @SQL, @ParmDefinition = @ParmDef, @cnt = @cntOUT OUTPUT;
    UPDATE #T1 SET record_count = @cntOUT WHERE column_id = @i
    SET @i = @i + 1
END

SELECT * FROM #T1
--DROP TABLE #T1

Upvotes: 0

GerardV
GerardV

Reputation: 384

A list with distinct values for each column is impossible. What if column A has 5 distinct values and column B has 7. What would your list look like?

The other question is easier, but as @Gordon Linoff states, takes 2 steps. Elaborating on his answer, for MS SQL:

select replace(replace(' count(distinct([col])) as [col],',
                       '[tab]', table_name
                      ), '[col]', column_name
               )
from information_schema.columns
where table_name = 'your_table';

Copy the results and paste them in a new query window between.

SELECT
[[results query 1]]
FROM your_table

Remember to delete the last ',' from query 1 results.

Upvotes: 0

Gordon Linoff
Gordon Linoff

Reputation: 1270061

The basic query is:

select col001, count(*) from MyTable group by col001 union all
select col002, count(*) from MyTable group by col002 union all
. . . 
select col700, count(*) from MyTable group by col700 ;

Not pleasant, but that is basically the query you need to run. SQL doesn't really do multiple independent aggregations more efficiently than doing them separately (even using grouping sets, in my experience).

You can construct the query. One way is to run something like this:

select replace(replace('select [col], count(*) as cnt from [tab] group by [col] union all ',
                       '[tab]', table_name
                      ), '[col]', column_name
               )
from information_schema.columns
where table_name = 'mytable' and table_schema = ??;

You can then copy the generated SQL (removing the final union all) and run it.

Note: That above is generic; the exact code might differ by database.

Upvotes: 1

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