MySQL：选择第n行，其中至少有一行具有属性集？

Question

I've a table in MySQL called words .

One column is called word (the actually word) and the other is called special

I've a query I'm trying to perform which - if possible - should do this:

SELECT word FROM words WHERE special = 1 AT LEAST ONCE

This is of cause a pseudo-like query but the thing I want is to get randomly nth records from my words table where AT LEAST ONE of the words has the attribute special set to 1.

It means that all nth records is allowed to be 1 if this is the case or all but one of the nth records can have special = 0 but there have to be at least one which has the special = 1.

I've tried something like: SELECT * FROM words HAVING COUNT(isNum = 1) > 1 ORDER BY RAND() LIMIT 10 or something like this.. It does not give me the result I want back, actually it only returns one result.

Can this be done by SQL??

Thanks

Answer 1

Here is an approach that should be relatively efficient, sort of. It selects one random "special" word, then n + 1 other words (which might contain the first word). It orders them so the random one is guaranteed to be first, and then select n of them.

select word
from ((select word, 0 as ordering
       from words
       where special = 1
       order by rand()
       limit 1
      ) union all
      (select word, rand() as ordering
       from words
       limit n
      )
     ) t
group by word
order by min(ordering)
limit n;

Answer 2

SELECT *
FROM words
HAVING COUNT(special = 1) > 1
ORDER BY RAND()
LIMIT 10

Using COUNT() here without GROUP simply counts the number of records and returns 1 row, then applies the HAVING clause to that one row, yielding one record in your result set.

If you add COUNT() to the SELECT clause, you'll kind of see your mistake:

SELECT COUNT(special = 1), words.*
FROM words
HAVING COUNT(special = 1) > 1
ORDER BY RAND()
LIMIT 10

The record that it returns after the count is indeterminate.

So, you must add GROUP BY to get multiple records back (one per word):

SELECT *
FROM words
GROUP BY word
HAVING COUNT(special = 1) > 1
ORDER BY RAND()
LIMIT 10

Now, you'll notice that it returns all words with more than one record, regardless of whether it has special = 1 .

That's because special = 1 is a Boolean expression that returns 0 or 1 . COUNT() increments on both COUNT(0) and COUNT(1) . Actually, it increments on anything but COUNT(NULL) . Now, you realize that you really want SUM() .

SELECT *
FROM words
GROUP BY word
HAVING SUM(special = 1) > 1
ORDER BY RAND()
LIMIT 10

Alternatively, and perhaps more straight-forward:

SELECT DISTINCT word
FROM words
WHERE special = 1
ORDER BY RAND()
LIMIT 10