Count for a list of items with zero for does not exist

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If I have a table t1 with:

my_col
------
foo
foo
bar

And I have a list with foo and hello

How can I get:

my_col | count
-------|-------
foo    | 2
hello  | 0

If I just do

SELECT my_col, COUNT(*) 
FROM t1
WHERE my_col in ('foo', 'hello')
GROUP BY my_col

I get

my_col | count
-------|------
foo    | 2

without any value for hello.

I'm specifically wanting this to be in reference to a list of items because this will be called in a program where the list is a variable.

Ideally you should maintain a separate table with all the possible column values which you want to appear in your report. In the absence of that, we can try using a CTE here:

WITH cte AS (
    SELECT 'foo' AS my_col UNION ALL
    SELECT 'bar' UNION ALL
    SELECT 'hello'
)

SELECT
    a.my_col,
    COUNT(b.my_col) AS count
FROM cte a
LEFT JOIN t1 b
    ON a.my_col = b.my_col
WHERE
    a.my_col IN ('foo', 'hello')
GROUP BY
    a.my_col;

Demo

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Here's yet another way, using values:

select
  t2.my_col, count (t1.my_col)
from
  (values ('foo'), ('hello')) as t2 (my_col)
  left join t1 on t1.my_col = t2.my_col
group by
  t2.my_col

Note that count (t1.my_col) returns a 0 for "hello" since nulls are not counted. count (*) by contast would have returned 1 for "hello" because it was counting the row.

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You can turn your list into a set of rows and use a LEFT JOIN, like :

SELECT x.val, COUNT(t.my_col) 
FROM 
    (SELECT 'foo' val UNION SELECT 'hello') x
    LEFT JOIN t ON t.my_col = x.val
GROUP BY x.val

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Postgres solution:

One way is to place the 'list' into an ARRAY, and then convert the ARRAY into a column using unnest. Then perform a left join on that column with the other table and perform a count.

WITH t1 AS (
     SELECT 'foo' AS my_col UNION ALL
     SELECT 'foo' UNION ALL
     SELECT 'bar'
)

SELECT
    a.my_col,
    COUNT(b.my_col) AS count
FROM unnest(ARRAY['foo', 'hello']) a (my_col)
LEFT JOIN t1 b
    ON a.my_col = b.my_col
GROUP BY
    a.my_col;

The issue I had with the other answers is that (while they they helped me get to the solution) they did not provide a solution where the items of interest were in a single list (which isn't an actual sql term, so the fault is on me).

However, my real use case is to perform a native query using java and hibernate, and unfortunately the above does not work because the typing cannot be determined. Instead I converted my list into a single string and used string_to_array in place of the ARRAY function.

So the solution that worked best for my use case is below (but at this point, the other answers would be just as correct since I'm now having to do manual string manipulation, but I'm leaving this here for the sake of posterity)

WITH t1 AS (
    SELECT 'foo' AS my_col UNION ALL
    SELECT 'foo' UNION ALL
    SELECT 'bar'
)

SELECT
    a.my_col,
    COUNT(b.my_col) AS count
FROM unnest(string_to_array('foo, hello', ',')) a (my_col)
LEFT JOIN t1 b
    ON a.my_col = b.my_col
GROUP BY
    a.my_col;

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Comments
  • where is the list coming from? is it stored in a table?
  • The values are not guaranteed to be in a table. They will come from an xml file.
  • I think count (*) will return a 1 for hello. If you change it to count (t1.*) that should yield the desired results.
  • @Hambone : thanks for checking and commenting ! I updated the answer accordingly
  • Your explanation make a lot of sense; I see why the posted solutions won't exactly meet the need. I don't know about Java/JDBC, but in ADO.net using Npgsql, you can pass an array as a parameter. For example cmd.Parameters[0].Value = new string[3] { "foo", "bar", "baz" }. If your database adapter supports this natively, I'd recommend using that in lieu of passing a joined string and then splitting it.
  • I agree passing the array directly would be preferred. It seems there is support for passing arrays, but I kept getting postgres type errors. So something was getting lost in the interface. I'm guessing there's some way to do it, but in reality I'm guessing someone above me will change their mind about what they want, and then all bets are off anyways :)