MySQL - Select from a list of numbers those without a counterpart in the id field of a table

Solution 1:

This is a problem that is pretty common: generating a relation on the fly without creating a table. SQL solutions for this problem are pretty awkward. One example using a derived table:

SELECT n.id
FROM
  (SELECT 2 AS id 
   UNION SELECT 3 
   UNION SELECT 4 
   UNION SELECT 5 
   UNION SELECT 6 
   UNION SELECT 7) AS n
  LEFT OUTER JOIN foos USING (id)
WHERE foos.id IS NULL;

But this doesn't scale very well, because you might have many values instead of just six. It can become tiresome to construct a long list with one UNION needed per value.

Another solution is to keep a general-purpose table of ten digits on hand, and use it repeatedly for multiple purposes.

CREATE TABLE num (i int);
INSERT INTO num (i) VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9);

SELECT n.id
FROM 
  (SELECT n1.i + n10.i*10 AS id
   FROM num AS n1 CROSS JOIN num AS n10
   WHERE n1.i + n10.i*10 IN (2, 3, 4, 5, 6, 7)) AS n
  LEFT OUTER JOIN foos USING (id)
WHERE foos.id IS NULL;

I show the inner query generating values from 0..99 even though this isn't necessary for this case. But you might have values greater than 10 in your list. The point is that with one table num, you can generate large numbers without having to resort to very long chains with one UNION per value. Also, you can specify the list of desired values in one place, which is more convenient and readable.

Solution 2:

I can't find a solution to your precise problem that doesn't use a temporary table, but an alternate way of doing your query using a sub-select instead of a join is:

SELECT bars.* FROM bars WHERE bars.ID NOT IN (SELECT ID FROM foos)

Like the other posters I originally wrote:

SELECT * FROM foos WHERE foos.ID NOT IN (2, 4, 5, 6, 7)

but then I realised that this is producing the opposite to what you want.

Solution 3:

If you use PHP, you can make this work without creating any temporary tables.

SELECT ID FROM foos WHERE foos.ID IN (2, 4, 5, 6, 7)

You can use PHP's array_diff() function to convert this to the desired result. If your list (2,4,5,6,7) is in an array called $list and the result of the query above is in an array $result, then

$no_counterparts = array_diff($list, $result);

...will return all the numbers in your list with no counterpart in your database table. While this solution doesn't perform the entire operation within the query, the post-processing you need to do in PHP is minimal to get what you want, and it may be worthwhile to avoid having to create a temporary table.

Solution 4:

I had a similar problem. I had a range where the auto-incrementing primary key had some missing values, so first I found how many there were: select count(*) from node where nid > 1962. Comparing this number against the highest value, I got the number missing. Then I ran this query: select n2.nid from node n1 right join node n2 on n1.nid = (n2.nid - 1) where n1.nid is null and n2.nid > 1962 This will find the number of non-consecutive missing records. It won't show consecutive ones, and I'm not entirely certain how to do that, beyond changing the ON clause to allow greater latitude (which would make the JOIN table substantially larger). In any case, this gave me five results out of the total seven missing, and the other two were guaranteed to be next to at least one of the five. If you have a larger number missing, you'll probably need some other way of finding the remaining missing.