试试这样的。

select u.user_name, u.user_email, u.user_phone, count(distinct b.bid),
       count(distinct c.cid) from user_table u 
   left outer join business_list b 
       on  u.user_id=b.user_id 
   left outer join classified_list c 
       on c.user_id=u.user_id 
group by 
    u.user_name,u.user_email,u.user_phone

此查询将提供所需的结果。它加入了 user 表到a UNION 的 business_list 以及 classified_list 表，并使用条件聚合对与每个用户关联的业务和分类的数量求和：

SELECT u.user_name AS `User Name`,
       u.user_email AS `User Email`,
       u.user_phone AS `User Phone`,
       SUM(CASE WHEN bc.type = 'business' THEN 1 ELSE 0 END) AS `No Of Business`,
       SUM(CASE WHEN bc.type = 'classified' THEN 1 ELSE 0 END) AS `No Of Classified`
FROM user u
JOIN (SELECT 'business' AS type, user_id 
      FROM business_list
      UNION ALL
      SELECT 'classified' AS type, user_id
      FROM classified_list) bc
ON bc.user_id = u.user_id
GROUP BY u.user_id;

或者，您可以使用 LEFT JOIN 每一张table， COUNT 惯性导航与制导 DISTINCT 每个表中的值：

SELECT u.user_name AS `User Name`,
       u.user_email AS `User Email`,
       u.user_phone AS `User Phone`,
       COUNT(DISTINCT b.bid) AS `No Of Business`,
       COUNT(DISTINCT c.cid) AS `No Of Classified`
FROM user u
LEFT JOIN business_list b ON b.user_id = u.user_id
LEFT JOIN classified_list c ON c.user_id = u.user_id
GROUP BY u.user_id

两个查询的输出相同：

User Name   User Email      User Phone  No Of Business  No Of Classified
Jose        jose@yahoo.in   457855654   2               1
Tom         tom@yahoo.in    5464644654  0               1
Nick        nick@yahoo.in   4545645644  2               1
Rose        rose@yahoo.in   554545441   1               2

sqlfiddle演示