Question

我的classfieds网站有九个类别。当用户搜索关键字时，我想在每个类别中显示搜索结果和搜索结果的数量。

如何优化sql查询？

我尝试了什么？为每个类别运行一个循环：

select * from ads where title like '%keyword%';
select count(*) from ads where title like '%keyword%' and category_id = 1;
select count(*) from ads where title like '%keyword%' and category_id = 2;
select count(*) from ads where title like '%keyword%' and category_id = 3;
select count(*) from ads where title like '%keyword%' and category_id = 4;
.....

有什么更好的建议可以让sql查询更快？

Answer 1

使用GROUP BY子句

SELECT
   COUNT(*)
   ,category_id 
FROM 
   ads 
WHERE 
   title like '%keyword%' 
GROUP BY category_id

Answer 2

nickles80的查询将有效。但如果您想在一个查询中同时执行这两项操作，请尝试以下操作：

SELECT ads.*, cnt.CountResult
FROM ads
INNER JOIN (
    SELECT category_id, COUNT(1) AS CountResult
    FROM ads 
    where title like '%keyword%' 
    GROUP BY category_id
) cnt ON cnt.category_id = ads.category_id
where title like '%keyword%'

Answer 3

你可以通过IF（）（如果是mysql）来计算（）计数，或者更多通用的情况/

select 
      count(*) as TotalAds,
      sum( case when category_id = 1 then 1 else 0 end ) as TotalCategory1,
      sum( case when category_id = 2 then 1 else 0 end ) as TotalCategory2,
      sum( case when category_id = 3 then 1 else 0 end ) as TotalCategory3,
      sum( case when category_id = 4 then 1 else 0 end ) as TotalCategory4,
      sum( case when category_id = 5 then 1 else 0 end ) as TotalCategory5,
      sum( case when category_id = 6 then 1 else 0 end ) as TotalCategory6,
      sum( case when category_id = 7 then 1 else 0 end ) as TotalCategory7,
      sum( case when category_id = 8 then 1 else 0 end ) as TotalCategory8,
      sum( case when category_id = 9 then 1 else 0 end ) as TotalCategory9
   from
      ads 
   where 
      title like '%keyword%';

如何针对搜索结果优化每个类别的计数查询？

3 个答案: