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2.8.7
2.8.3
2.8.2
2.8.7
2.8.5
2.8.7
2.8.7
2.8.5
2.6.0
2.8.3
2.6.4
2.6.3
2.8.4
2.8.0
2.6.3
2.8.5
2.8.5
2.8.5
2.6.0
2.8.2
How do I bring a unique value version sorted by the number of these versions?
At the exit I want to get the following:
2.8.5 5
2.8.7 4
2.6.0 2
2.6.3 2
2.8.2 2
2.8.3 2
2.8.4 2
2.6.4 1
2.8.0 1
ORDER BY count unique versions))
Sorry for my bad english
341
There is no way this query can be executed reasonably. Either DISTINCT doesn't work (because the added extended sort key column changes its semantics), or ORDER BY doesn't work (because after DISTINCT we can no longer access the extended sort key column).
Yes, you can use COUNT() and DISTINCT together to display the count of only distinct rows.
To count distinct values, you can use distinct in aggregate function count(). The result i3 tells that we have 3 distinct values in the table.
To count the number of different values that are stored in a given column, you simply need to designate the column you pass in to the COUNT function as DISTINCT . When given a column, COUNT returns the number of values in that column. Combining this with DISTINCT returns only the number of unique (and non-NULL) values.
SELECT
version,
COUNT(*) AS num
FROM
my_table
GROUP BY
version
ORDER BY
COUNT(*) DESC
SELECT version, COUNT(*) FROM tablename
GROUP BY version
ORDER BY COUNT(*) DESC;
or, alternate syntax
SELECT version, COUNT(*) FROM tablename
GROUP BY 1
ORDER BY 2 DESC;
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