1

我有一个看起来像这样的查询：

Select 
    tradesmen.id,
    (SELECT COUNT(quotes.id) FROM quotes WHERE quotes.tradesman_id = tradesmen.id) AS quoted
From
    tradesmen;

所以基本上数据库中的每一行都有一个子查询（50,000+）。现在，对于每个交易者来说，可能会有大约 1,000 - 2,000 条报价。

所以我可以使用这个子查询来计算它们。

或者，

我可以查询所有的商人。

select tradesman.id from tradesmen;

执行一次查询以获取所有报价计数

select tradesman_id as id, count(quotes.id) as quotes from quotes group by tradesman_id;

然后遍历每个商人并从数组中取出每个商人的计数。

mysql有多快？第二种方法会带来显着的好处还是两种方法都可以接受？

作为一般参考，我的实际查询是：

SELECT 
    tradesmen.*, 
    regions.name AS region_name, 
    GROUP_CONCAT(ptypes_tradesmen.ptype_id SEPARATOR '|') AS ptype_ids, 
    (SELECT 
         COUNT(quotes.id) 
     FROM 
         quotes 
     WHERE 
         quotes.tradesman_id = tradesmen.id
    ) AS quoted, 
    (SELECT 
         COUNT(quote_intentions.id) 
     FROM 
         quote_intentions 
     WHERE 
         quote_intentions.tradesman_id = tradesmen.id
    ) AS intended, 
    (SELECT 
         COUNT(quotes.id) FROM quotes 
     WHERE 
         quotes.tradesman_id = tradesmen.id 
         AND quotes.accepted = 1
    ) AS awarded
FROM 
    (`tradesmen`)
LEFT JOIN `regions` ON `regions`.`id` = `tradesmen`.`region_id`
LEFT JOIN `ptypes_tradesmen` ON `ptypes_tradesmen`.`tradesman_id` = `tradesmen`.`id`
GROUP BY `tradesmen`.`id`

更新

使用 ctrahey 的答案，我更改了查询。

所以，我们现在有三个版本的查询..

关注：

SELECT 
    tradesmen.*, 
    regions.name AS region_name, 
    GROUP_CONCAT(ptypes_tradesmen.ptype_id SEPARATOR '|') AS ptype_ids, 
    COUNT(quotes.id) AS quoted,
    COUNT(quote_intentions.id) AS intended,
    COUNT(NULLIF(quotes.accepted, 0)) AS awarded
FROM (`tradesmen`)
LEFT JOIN `regions` ON `regions`.`id` = `tradesmen`.`region_id`
LEFT JOIN `ptypes_tradesmen` ON `ptypes_tradesmen`.`tradesman_id` = `tradesmen`.`id`
LEFT JOIN quotes ON quotes.tradesman_id = tradesmen.id
LEFT JOIN quote_intentions ON quote_intentions.tradesman_id = tradesmen.id
GROUP BY `tradesmen`.`id`

我的修改版本：

SELECT 
    t.*, 
    r.name AS region_name, 
    GROUP_CONCAT(p.ptype_id SEPARATOR "|") AS ptype_ids, 
    COUNT(q.id) as quoted, 
    COUNT(i.id) as intended, 
    COUNT(NULLIF(q.accepted, 0)) as awarded
FROM (tradesmen t)
LEFT JOIN regions r ON r.id = t.region_id
LEFT JOIN quotes q ON t.id = q.tradesman_id
LEFT JOIN quote_intentions i ON t.id = i.tradesman_id
LEFT JOIN ptypes_tradesmen p ON p.tradesman_id = t.id
GROUP BY t.id

原本的：

SELECT 
    tradesmen.*, 
    regions.name AS region_name, 
    GROUP_CONCAT(ptypes_tradesmen.ptype_id SEPARATOR '|') AS ptype_ids, 
    (SELECT 
         COUNT(quotes.id) 
     FROM 
         quotes 
     WHERE 
         quotes.tradesman_id = tradesmen.id
    ) AS quoted, 
    (SELECT 
         COUNT(quote_intentions.id) 
     FROM 
         quote_intentions 
     WHERE 
         quote_intentions.tradesman_id = tradesmen.id
    ) AS intended, 
    (SELECT 
         COUNT(quotes.id) FROM quotes 
     WHERE 
         quotes.tradesman_id = tradesmen.id 
         AND quotes.accepted = 1
    ) AS awarded
FROM 
    (`tradesmen`)
LEFT JOIN `regions` ON `regions`.`id` = `tradesmen`.`region_id`
LEFT JOIN `ptypes_tradesmen` ON `ptypes_tradesmen`.`tradesman_id` = `tradesmen`.`id`
GROUP BY `tradesmen`.`id`

虽然它们都返回几乎相同的结果，但最后四个字段存在差异（因此我在这里从结果数组中删除所有其他行）。

从原始查询（正确结果）：

array('id' => '53',
  'ptype_ids' => '58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68',
  'quoted' => '6',
  'intended' => '14',
  'awarded' => '3'),

从 ctrahey 的查询：

array('id' => '53',
  'ptype_ids' => '58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|58|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|2|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|7|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|17|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|1|40|40|40|40|40|4',
  'quoted' => '2016',
  'intended' => '2016',
  'awarded' => '1008'),

从我修改后的查询中：

array('id' => '53',
  'ptype_ids' => '58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46|13|67|8|75|59|23|9|31|71|24|68|58|2|7|17|1|40|52|4|74|66|19|15|46',
  'quoted' => '2016',
  'intended' => '2016',
  'awarded' => '1008'),
4

1 回答 1

5

两者都不！执行正确的 JOIN（RDBMS 中的真正价值在哪里）：

SELECT 
  tradesmen.id,
  COUNT(quotes.id) as quoted
FROM 
  tradesmen
  LEFT JOIN quotes ON quotes.tradesman_id = tradesmen.id
GROUP BY tradesmen.id

此查询将正是您所需要的，而且速度很快！

编辑您的真实查询

实际查询中唯一需要注意的是接受引号计数中的 NULLIF 位，因为 (IIRC) COUNT()将计数 false/0，但不计数为 NULL。

SELECT 
    tradesmen.*, 
    regions.name AS region_name, 
    GROUP_CONCAT(ptypes_tradesmen.ptype_id SEPARATOR '|') AS ptype_ids, 
    COUNT(quotes.id) AS quoted
    COUNT(quote_intentions.id) AS intended
    COUNT(NULLIF(quotes.accepted, 0)) AS awarded
FROM (`tradesmen`)
LEFT JOIN `regions` ON `regions`.`id` = `tradesmen`.`region_id`
LEFT JOIN `ptypes_tradesmen` ON `ptypes_tradesmen`.`tradesman_id` = `tradesmen`.`id`
LEFT JOIN quotes ON quotes.tradesman_id = tradesmen.id
LEFT JOIN quote_intentions ON quote_intentions.tradesman_id = tradesmen.id
GROUP BY `tradesmen`.`id`
于 2012-08-27T04:55:18.943 回答