I can perform the following SQL Server selection of distinct (or non-repeating names) from a column in one table like so:
SELECT COUNT(DISTINCT [Name]) FROM [MyT
In case you have different amounts of columns per table, like:
And you want to count the amount of distinct values of different column names, what it was useful to me in AthenaSQL was to use CROSS JOIN
since your output would be only one row, it would be just 1 combination:
SELECT * FROM (
SELECT COUNT(DISTINCT name1) as amt_name1,
COUNT(DISTINCT name2) as amt_name2,
COUNT(DISTINCT name3) as amt_name3,
FROM table1 ) t1
CROSS JOIN
(SELECT COUNT(DISTINCT name4) as amt_name4,
COUNT(DISTINCT name5) as amt_name5,
MAX(t3.amt_name6) as amt_name6
FROM table2
CROSS JOIN
(SELECT COUNT(DISTINCT name6) as amt_name6
FROM table3) t3) t2
Would return a table with one row and their counts:
amt_name1 | amt_name2 | amt_name3 | amt_name4 | amt_name5 | amt_name6
4123 | 675 | 564 | 2346 | 18667 | 74567
After the clarification, use:
SELECT x.name, COUNT(x.[name])
FROM (SELECT [name]
FROM [MyTable]
UNION ALL
SELECT [name]
FROM [MyTable2]
UNION ALL
SELECT [name]
FROM [MyTable3]) x
GROUP BY x.name
If I understand correctly, use:
SELECT x.name, COUNT(DISTINCT x.[name])
FROM (SELECT [name]
FROM [MyTable]
UNION ALL
SELECT [name]
FROM [MyTable2]
UNION ALL
SELECT [name]
FROM [MyTable3]) x
GROUP BY x.name
UNION
will remove duplicates; UNION ALL
will not, and is faster for it.
EDIT: Had to change after seeing recent comment.
Does this give you what you want? This gives a count for each person after combining the rows from all tables.
SELECT [NAME], COUNT(*) as TheCount
FROM
(
SELECT [Name] FROM [MyTable1]
UNION ALL
SELECT [Name] FROM [MyTable2]
UNION ALL
SELECT [Name] FROM [MyTable3]
) AS [TheNames]
GROUP BY [NAME]
Here's another way:
SELECT x.name, SUM(x.cnt)
FROM ( SELECT [name], COUNT(*) AS cnt
FROM [MyTable]
GROUP BY [name]
UNION ALL
SELECT [name], COUNT(*) AS cnt
FROM [MyTable2]
GROUP BY [name]
UNION ALL
SELECT [name], COUNT(*) AS cnt
FROM [MyTable3]
GROUP BY [name]
) AS x
GROUP BY x.name