MySQL查询返回错误的COUNT值

时间:2018-01-01 18:13:32

标签: mysql

我在这里分享了我的查询。所有总值都给出了错误的结果,因为非常大的数值不是所需的COUNT值。我的查询有什么问题?

           SELECT bd.business_id,
                  bd.business_name,
                  bd.business_address,
                  bd.created_at,
                  bd.updated_at AS extent_expiry_date,
                  bd.manager_id as user_id,
                  COUNT(m.manager_id) AS total_managers,
                  COUNT(a.agent_id) AS total_agents,
                  COUNT(c.customer_id) AS total_customers,
                  COUNT(t.task_id) AS total_tasks,
                  COUNT(t.order_id) AS total_orders
            FROM business_details bd
            LEFT JOIN managers m ON m.business_id = bd.business_id
            LEFT JOIN agents a ON a.business_id = bd.business_id
            LEFT JOIN customers c ON c.business_id = bd.business_id
            LEFT JOIN tasks t ON t.business_id = bd.business_id
            GROUP BY bd.business_id

1 个答案:

答案 0 :(得分:2)

由于你的问题中没有设置样本数据,我的猜测是你得到的是相同的manager_id,agent_id ...由于左连接多次获得正确的计数使用不同

SELECT bd.business_id,
      bd.business_name,
      bd.business_address,
      bd.created_at,
      bd.updated_at AS extent_expiry_date,
      bd.manager_id as user_id,
      COUNT(distinct m.manager_id) AS total_managers,
      COUNT(distinct a.agent_id) AS total_agents,
      COUNT(distinct c.customer_id) AS total_customers,
      COUNT(distinct t.task_id) AS total_tasks,
      COUNT(distinct t.order_id) AS total_orders
FROM business_details bd
LEFT JOIN managers m ON m.business_id = bd.business_id
LEFT JOIN agents a ON a.business_id = bd.business_id
LEFT JOIN customers c ON c.business_id = bd.business_id
LEFT JOIN tasks t ON t.business_id = bd.business_id
GROUP BY bd.business_id

还根据新版本包含所有非聚合列,并且大多数RDBMS拒绝这些类型的查询

GROUP BY 
bd.business_id,
bd.business_name,
bd.business_address,
bd.created_at,
bd.updated_at,
bd.manager_id