Question

CREATE TABLE customer (   id_customer NUMBER PRIMARY KEY,   name
   VARCHAR2(10),   surname VARCHAR2(20));

CREATE TABLE customer_wer (
  id_customer NUMBER,
  name VARCHAR2(10),
  surname VARCHAR2(20),
  data_from DATE,
  data_to DATE NOT NULL,
  CONSTRAINT customer_wer_pk PRIMARY KEY (id_customer, data_from),
  CONSTRAINT customer_wer_fk FOREIGN KEY (id_customer)
  REFERENCES customer (id_customer));

 SELECT COUNT(customer.id_customer) 
   FROM customer JOIN customer_wer
     ON customer.id_customer = customer_wer.id_customer
  WHERE (SELECT customer_wer.id_customer 
           FROM customer_wer JOIN customer
             ON customer_wer.id_customer = customer.id_customer
          GROUP BY customer.id_customer) >= 8;

在＆＃39;客户＆＃39;表有5个客户和＆＃39; customer_wer＆＃39;有48个职位。来自＆＃39;客户＆＃39;的每位客户table在＆＃39; customer_wer＆＃39;中有8个相关位置。表。我想知道有多少客户在＆＃39; customer_wer＆＃39;中有8个或更多相关职位。我如何使用select来做到这一点？

提前谢谢你。

Answer 1

例如，您可以使用MAX：

SELECT COUNT(customer.id_customer) 
  FROM customer JOIN customer_wer
    ON customer.id_customer = customer_wer.id_customer
 WHERE (SELECT MAX (customer_wer.id_customer) 
          FROM customer_wer JOIN customer
            ON customer_wer.id_customer = customer.id_customer
         GROUP BY customer.id_customer) >= 8;

或 - 如果可能的话 - 甚至更好 - 为您的子选择添加一些条件，以便它返回唯一的结果。

另一个选项，如果要检查子查询返回的行之一是否为8，则使用IN运算符：

SELECT COUNT(customer.id_customer) 
  FROM customer JOIN customer_wer
    ON customer.id_customer = customer_wer.id_customer
 WHERE 8 IN (SELECT customer_wer.id_customer
               FROM customer_wer JOIN customer
                 ON customer_wer.id_customer = customer.id_customer
              GROUP BY customer.id_customer);

或使用EXISTS并将条件>= 8移至子查询中：

SELECT COUNT(customer.id_customer) 
  FROM customer JOIN customer_wer
    ON customer.id_customer = customer_wer.id_customer
 WHERE EXISTS (SELECT customer_wer.id_customer
                 FROM customer_wer JOIN customer
                   ON customer_wer.id_customer = customer.id_customer
                WHERE customer_wer.id_customer >= 8
                GROUP BY customer.id_customer);

p.s。 :(根据最新评论添加）

SELECT COUNT(c1.id_customer) 
  FROM customer c1
 WHERE 2 <= (SELECT COUNT (cw1.id_customer)
               FROM customer_wer cw1
              WHERE c1.id_customer = cw1.id_customer);

如何修复ORA-01427：单行子查询返回多行？

1 个答案: