SQL*Plus查询帮助从不同的表中求和和计数，并获得正确答案

这是我的问题，给了我错误的金额。我怎样才能修好它？它应该给167700，但它给我251550的贷款总额。计数正常=15。我不能改变桌子上的任何东西

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

简单规则：切勿在

FROM

子句中使用逗号。始终使用明确、正确的

JOIN

语法和

ON

子句中的条件。那你就不会忘记他们了

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

因此：

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

---

当然，我编造了用于连接的列的名称，因为您的问题没有描述表的信息。

Duh！两个表中没有公用键。怎么样？您如何跟踪哪些客户接受了哪些贷款

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

为什么要对来自两个不相关表的数据运行一个查询？您创建的交叉联接（对两个表的枚举没有任何条件）只是将第一个表中的每一行联接到第二个表中的每一行

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

customer表似乎有15行，所有15个名称都是不同的。当您计算DISTINCT时，您得到了正确的数字，即使在交叉联接中，每个customer_名称出现了很多次

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

另一方面，每个贷款金额重复15次。167700 x 15=2515500

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

如果需要在一行中同时显示贷款总额和（不同的）客户数，则需要

    create table loan
   (loan_number     varchar(15) not null,
    branch_name varchar(15) not null,
    amount      number not null,
    primary key(loan_number));

    create table customer
   (customer_name   varchar(15) not null,
    customer_street     varchar(12) not null,
    customer_city   varchar(15) not null,
    primary key(customer_name));


select SUM(amount),
COUNT( distinct customer_name)
from loan,customer;

select (select sum(amount) from loan) as total_amount, 
       (select count (distinct customer_name) from customer) as distinct_customers
from   dual
;

---

我知道如何加入一个表，但我没有一个共同的关键在他们。我将编辑我的问题。@d_PI。嗯，贷款是给某人的。所以，我想象在某个地方有一个表，在贷款和客户之间有一个链接。有另一个表用来跟踪客户和贷款。它不起作用-可能我在错误的空格中有一个括号。不是每个来这里问问题的人都是懒惰的，在他们提问之前不尝试。你不应该假设。我可能不擅长提问和描述问题，但我知道如何使用谷歌。@d_PI-你说的“它不工作”是什么意思？你收到错误消息了吗？您是否完全按照编写的方式使用查询？剩下的，我不知道你指的是什么（用“懒惰”和“谷歌”之类的词）-那是关于什么的？只是现在尝试你的查询-它确实有效。就像我说的“也许我把括号放错地方了”。我在上找到了相同的示例，但我只是缺少正确的括号顺序。-不管怎么说，谢谢，其他的事情只是我感到沮丧，因为我觉得你们认为我一点都不知道我在说什么，但实际上只是因为我提出了一个不充分的问题。