Question

我在我的选择查询中使用子查询，但正如您在示例中所看到的，我使用type='jump'执行相同的子查询两次以与不同的值进行比较。优化此查询的最佳方法是什么？感谢

SELECT * FROM tableone WHERE ONE > 10 AND PAS != '' AND CMD = 'xxx' 
AND (SELECT COUNT() FROM tabletwo WHERE tabletwo.id = tableone.id AND TYPE='walk' AND TIMESTAMP > $last24hours) <= $walklimit 
AND (SELECT COUNT() FROM tabletwo WHERE tabletwo.id = tableone.id AND TYPE='jump' AND TIMESTAMP > $last24hours) <= $jumpslimit
AND (SELECT COUNT() FROM tabletwo WHERE tabletwo.id = tableone.id AND TYPE='jump' AND TIMESTAMP > $last24hours) <= $jumpsHOURlimit
LIMIT 1;

更新：使用JOIN或使用min（）简单的解决方案更好？

Answer 1

您的两个子查询完全相同。唯一的区别在于极限。因此，请使用min()：

SELECT t1.*
FROM tableone t1
WHERE t1.ONE > 10 AND t1.PAS <> '' AND t1.CMD = 'xxx'  AND
      (SELECT COUNT(*)
       FROM tabletwo
       WHERE tabletwo.id = tableone.id AND TYPE = 'walk' AND TIMESTAMP > $last24hours
      ) <= $walklimit AND
      (SELECT COUNT(*)
       FROM tabletwo
       WHERE tabletwo.id = tableone.id AND TYPE='jump' AND TIMESTAMP > $last24hours
      ) <= min($jumpslimit, $jumpsHOURlimit)
LIMIT 1;

Answer 2

如果我正确理解您的问题，您还可以使用JOIN将它们合并为一个查询，然后使用COUNT和CASE来过滤结果：

SELECT t1.*
FROM tableone t1
   JOIN tabletwo t2 ON tabletwo.id = tableone.id AND t2.TIMESTAMP > $last24hours 
WHERE t1.ONE > 10 AND t1.PAS <> '' AND t1.CMD = 'xxx' 
GROUP BY t1.id
HAVING COUNT(CASE WHEN t2.TYPE='walk' THEN 1 END) <= $walklimit
   AND COUNT(CASE WHEN t2.TYPE='jump' THEN 1 END) <= min($jumpslimit, $jumpsHOURlimit)
LIMIT 1;

ГЗsqlite子查询记住结果

2 个答案: