SELECT P.option_name, COUNT(V.vote_option_code) AS poll, t.today_votes FROM ( SELECT option_name, code FROM cus_vote_option WHERE vote_code = 'vote0000000006' ) AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code AND V.app_id = 2958 LEFT JOIN ( SELECT code, COUNT(V.vote_option_code) AS today_votes FROM cus_vote_option AS P LEFT JOIN cus_vote_user AS V ON P.code = V.vote_option_code AND V.app_id = 2958 AND DATE(V.create_time) = CURDATE() WHERE P.vote_code = 'vote0000000006' GROUP BY P.code ) AS t ON P.code = t.code GROUP BY P.option_name, t.today_votes; 对这个sql进行优化

对于您提供的查询，可以进行一些优化来提高性能和简化查询逻辑。以下是优化后的查询：

SELECT P.option_name, COUNT(V.vote_option_code) AS poll, IFNULL(t.today_votes, 0) AS today_votes
FROM cus_vote_option AS P
LEFT JOIN cus_vote_user AS V
    ON P.code = V.vote_option_code
    AND V.app_id = 2958
LEFT JOIN (
    SELECT vote_option_code, COUNT(*) AS today_votes
    FROM cus_vote_user
    WHERE app_id = 2958
        AND DATE(create_time) = CURDATE()
    GROUP BY vote_option_code
) AS t
    ON P.code = t.vote_option_code
WHERE P.vote_code = 'vote0000000006'
GROUP BY P.option_name, t.today_votes;

在这个优化后的查询中，我们省略了子查询中的多余表连接，并直接在主查询中使用了相关条件。此外，我们使用了`IFNULL`函数来处理当天投票数量为NULL的情况，将其替换为0。

请根据您的实际需求和表结构进行适当的修改。希望这次能满足您的要求！如果还有其他问题，请随时提问。