Question

我需要从项目列表中选择满足搜索条件的50个项目。我还需要销售时的当前数量。物品清单有超过100 000条记录

这就是我现在所做的：

SELECT 
  items.id,
  CONCAT(prod_codes.code,items.item) AS item,
  items.price,
  sold.count
FROM items
LEFT JOIN prod_codes ON items.prod_code_id=prod_codes.id
LEFT JOIN (
    SELECT COUNT(*) AS count,CONCAT(prod_code,part_num) AS part 
    FROM `sold_items` 
    WHERE 1 
    GROUP BY CONCAT(prod_code,part_num)
    ORDER BY count DESC
) AS sold ON sold.part=CONCAT(prod_codes.code,items.item)
WHERE active_status=1
AND CONCAT(items.prod_code,items.item) collate utf8_general_ci LIKE '%".$search_part."%' 
AND items.description collate utf8_general_ci LIKE '%".$search_desc."%'

ORDER BY items.item

LIMIT 50

但是这个查询需要20秒才能执行。我确信有更好的方法可以做到。

Answer 1

您可以尝试这样做，而不是嵌套子查询：

SELECT 
  items.id,
  CONCAT(prod_codes.code,items.item) AS item,
  items.price,
  COUNT(*) AS count
FROM items
LEFT JOIN prod_codes ON items.prod_code_id = prod_codes.id
LEFT JOIN sold_items sold ON sold.prod_code = prod_codes.code AND sold.part_num = items.item
WHERE active_status = 1
AND CONCAT(items.prod_code,items.item) collate utf8_general_ci LIKE '%".$search_part."%' 
AND items.description collate utf8_general_ci LIKE '%".$search_desc."%'
ORDER BY items.item
GROUP BY prod_codes.code, items.item, items.id, items.price
LIMIT 50

Answer 2

替换WHERE子句的这一部分

CONCAT(items.prod_code,items.item) collate utf8_general_ci LIKE '%".$search_part."%'

用这个：

(items.prod_code collate utf8_general_ci LIKE '%".$search_part."%' OR items.item collate utf8_general_ci LIKE '%".$search_part."%')

那个CONCAT（）调用几乎肯定会让它失效，因为它无法使用索引。

此外，您有SQL注入问题等 - 使用prepared statements。

最后，请参阅@ ypercube的评论;这也是一个好点，并将产生重大影响。

SELECT之后的Mysql Join

2 个答案: