计算重复行sql的结果数量

Question

我有一张类似的表：

 customer         employee
------        ---------
  A              Adam
  A              Adam
  A              Adam
  B              Bob
  C              Adam
  C              Adam
  B              Bob
  D              Dan

我试图弄清楚如何显示亚当有多少次重复客户以及鲍勃有多少次。 所以在上面的例子中它会返回：

Adam =＆gt; 2（他有2个回头客，客户A和C）

Bob =＆gt; 1（他有1个重复客户，客户b）

Dan =＆gt; 0（他只有一个顾客，没有重复）

到目前为止我所拥有的是：

SELECT customer, COUNT(*) as cnt
FROM table
GROUP BY employee
HAVING cnt > 1

但我似乎无法完成剩下的工作。

Answer 1

您可以尝试以下查询：

SELECT 
    t.employee, COUNT(*) AS cnt
FROM
    (SELECT 
        employee, customer, COUNT(*) AS actcnt
    FROM
        customers
    GROUP BY customer , employee HAVING actcnt > 1) AS t
GROUP BY t.employee
ORDER BY t.employee;

DEMO

Answer 2

首先需要找到重复客户：

SELECT employee,customer FROM yourtable GROUP BY employee, customer having COUNT(*) > 1;

之后，您可以找到每个员工有多少回头客：

SELECT employee, COUNT(*) from (
 SELECT employee,customer 
  FROM yourtable 
  GROUP BY employee, customer 
  HAVING COUNT(*) > 1
 ) t
GROUP BY employee;

Answer 3

select employee, count(*)-1 as cnt
from t1
group by employee,customer
having cnt > 0;

Answer 4

  

如果你COUNT重复员工：

SELECT id,customer,employee,COUNT(*) FROM USER  GROUP BY employee HAVING COUNT(*)>0

  

如果你COUNT重复客户：

SELECT id,customer,employee,COUNT(*) FROM USER  GROUP BY customer HAVING COUNT(*)>0

  

//注意：我总计数，你改了）COUNT（）＆gt; 0到COUNT（）＆gt; 1    ＆安培;＆安培;表=用户为

CREATE TABLE `user` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `customer` varchar(200) DEFAULT NULL,
  `employee` varchar(200) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=latin1;

/ *表user * /

的数据

insert  into `user`(`id`,`customer`,`employee`) 
values (1,'A','Adam'),(2,'A','Adam'),(3,'A','Adam'),(4,'B','Bob'),(5,'C','Adam'),(6,'C','Adam'),(7,'B','Bob'),(8,'D','Dan');

Answer 5

选择客户，员工   从表 GROUP BY客户，员工   有计数（*）＆gt; 1