如何在SQL中合并两个表并对具有相同ID的值求和？_Sql_Sqlite

如何在SQL中合并两个表并对具有相同ID的值求和？

sql sqlite

如何在SQL中合并两个表并对具有相同ID的值求和？,sql,sqlite,Sql,Sqlite,我是新来的SQL抱歉，如果这个问题最终变得微不足道我有两张桌子：表1“客户”： CREATE TABLE customer (ID, Name, Surname); INSERT INTO customer (ID, Name, Surname) VALUES (1, 'Smith', 'Hulk'), (2, 'Jake', 'Brook'), (3, 'Vladimir', 'Computin'), (4, 'Joe', 'Door'); CREATE TABLE pay

我是新来的SQL抱歉，如果这个问题最终变得微不足道

我有两张桌子：

表1“客户”：

 CREATE TABLE customer (ID, Name, Surname);
 INSERT INTO customer (ID, Name, Surname)
 VALUES (1, 'Smith', 'Hulk'),
 (2, 'Jake', 'Brook'),
 (3, 'Vladimir', 'Computin'),
 (4, 'Joe', 'Door');

 CREATE TABLE payment (payment_id, customer_id, payment_date, payment_amount);
 INSERT INTO payment (payment_id, customer_id, payment_date, payment_amount)
 VALUES (1, 1, "11/09/2011", 100),
 (2, 2, "15/10/2013", 50),
 (3, 4, "2/1/2011", 30),
 (4, 3, "12/09/2011", 200),
 (5, 2, "15/10/2013", 200),
 (6, 3, "2/1/2011", 10),
 (7, 1, "11/09/2011", 120),
 (8, 4, "15/10/2013", 100);

CREATE TABLE payment (Name, Surname, total_payments);
 INSERT INTO payment (Name, Surname, total_payments)
 VALUES ('Smith', 'Hulk',220),
 ('Jake', 'Brook', 250),
 ('Vladimir', 'Computin', 210),
 ('Joe', 'Door', 130)

SELECT payments.ID, sum(payments.payment_amount)
FROM payments
GROUP BY payments.ID

表2“付款”：

 CREATE TABLE customer (ID, Name, Surname);
 INSERT INTO customer (ID, Name, Surname)
 VALUES (1, 'Smith', 'Hulk'),
 (2, 'Jake', 'Brook'),
 (3, 'Vladimir', 'Computin'),
 (4, 'Joe', 'Door');

 CREATE TABLE payment (payment_id, customer_id, payment_date, payment_amount);
 INSERT INTO payment (payment_id, customer_id, payment_date, payment_amount)
 VALUES (1, 1, "11/09/2011", 100),
 (2, 2, "15/10/2013", 50),
 (3, 4, "2/1/2011", 30),
 (4, 3, "12/09/2011", 200),
 (5, 2, "15/10/2013", 200),
 (6, 3, "2/1/2011", 10),
 (7, 1, "11/09/2011", 120),
 (8, 4, "15/10/2013", 100);

CREATE TABLE payment (Name, Surname, total_payments);
 INSERT INTO payment (Name, Surname, total_payments)
 VALUES ('Smith', 'Hulk',220),
 ('Jake', 'Brook', 250),
 ('Vladimir', 'Computin', 210),
 ('Joe', 'Door', 130)

SELECT payments.ID, sum(payments.payment_amount)
FROM payments
GROUP BY payments.ID

所需输出：

 CREATE TABLE customer (ID, Name, Surname);
 INSERT INTO customer (ID, Name, Surname)
 VALUES (1, 'Smith', 'Hulk'),
 (2, 'Jake', 'Brook'),
 (3, 'Vladimir', 'Computin'),
 (4, 'Joe', 'Door');

 CREATE TABLE payment (payment_id, customer_id, payment_date, payment_amount);
 INSERT INTO payment (payment_id, customer_id, payment_date, payment_amount)
 VALUES (1, 1, "11/09/2011", 100),
 (2, 2, "15/10/2013", 50),
 (3, 4, "2/1/2011", 30),
 (4, 3, "12/09/2011", 200),
 (5, 2, "15/10/2013", 200),
 (6, 3, "2/1/2011", 10),
 (7, 1, "11/09/2011", 120),
 (8, 4, "15/10/2013", 100);

CREATE TABLE payment (Name, Surname, total_payments);
 INSERT INTO payment (Name, Surname, total_payments)
 VALUES ('Smith', 'Hulk',220),
 ('Jake', 'Brook', 250),
 ('Vladimir', 'Computin', 210),
 ('Joe', 'Door', 130)

SELECT payments.ID, sum(payments.payment_amount)
FROM payments
GROUP BY payments.ID

以下是我尝试过的：

 CREATE TABLE customer (ID, Name, Surname);
 INSERT INTO customer (ID, Name, Surname)
 VALUES (1, 'Smith', 'Hulk'),
 (2, 'Jake', 'Brook'),
 (3, 'Vladimir', 'Computin'),
 (4, 'Joe', 'Door');

 CREATE TABLE payment (payment_id, customer_id, payment_date, payment_amount);
 INSERT INTO payment (payment_id, customer_id, payment_date, payment_amount)
 VALUES (1, 1, "11/09/2011", 100),
 (2, 2, "15/10/2013", 50),
 (3, 4, "2/1/2011", 30),
 (4, 3, "12/09/2011", 200),
 (5, 2, "15/10/2013", 200),
 (6, 3, "2/1/2011", 10),
 (7, 1, "11/09/2011", 120),
 (8, 4, "15/10/2013", 100);

CREATE TABLE payment (Name, Surname, total_payments);
 INSERT INTO payment (Name, Surname, total_payments)
 VALUES ('Smith', 'Hulk',220),
 ('Jake', 'Brook', 250),
 ('Vladimir', 'Computin', 210),
 ('Joe', 'Door', 130)

SELECT payments.ID, sum(payments.payment_amount)
FROM payments
GROUP BY payments.ID

这将对具有相同ID的表“付款”中的条目求和

但是，我不知道如何使用ID，以便在总付款前显示姓名（如所需输出）

任何帮助都将不胜感激

您需要将

客户

的

左侧

加入到

付款

和聚合：

SELECT c.Name, c.Surname, 
       COALESCE(SUM(p.payment_amount), 0) payment_amount
FROM customer c LEFT JOIN payment p
ON p.customer_id = c.id
GROUP BY c.ID, c.Name, c.Surname

SELECT c.ID, c.Name, c.SurName, sum(p.payment_amount)
FROM clients c LEFT JOIN
     payments p
     ON c.id = p.customer_id
GROUP BY c.ID, c.Name, c.SurName;

请参阅。
结果:

您需要将

customer

的

LEFT

加入到

payment

并聚合：

SELECT c.Name, c.Surname, 
       COALESCE(SUM(p.payment_amount), 0) payment_amount
FROM customer c LEFT JOIN payment p
ON p.customer_id = c.id
GROUP BY c.ID, c.Name, c.Surname

SELECT c.ID, c.Name, c.SurName, sum(p.payment_amount)
FROM clients c LEFT JOIN
     payments p
     ON c.id = p.customer_id
GROUP BY c.ID, c.Name, c.SurName;

请参阅。
结果:

您可能需要

JOIN

将它们引入查询，然后进行适当的聚合：

SELECT c.Name, c.Surname, 
       COALESCE(SUM(p.payment_amount), 0) payment_amount
FROM customer c LEFT JOIN payment p
ON p.customer_id = c.id
GROUP BY c.ID, c.Name, c.Surname

SELECT c.ID, c.Name, c.SurName, sum(p.payment_amount)
FROM clients c LEFT JOIN
     payments p
     ON c.id = p.customer_id
GROUP BY c.ID, c.Name, c.SurName;

注意：使用相关子查询通常更快：

SELECT c.*, 
       (SELECT SUM(p.payment_amount)
        FROM payments p
        WHERE c.id = p.customer_id
       ) as total_payments
FROM clients c ;

特别是，这可以有效地使用

付款（客户id、付款金额）的索引
您可以需要加入以将其纳入查询，然后进行适当的聚合：
SELECT c.Name, c.Surname, 
       COALESCE(SUM(p.payment_amount), 0) payment_amount
FROM customer c LEFT JOIN payment p
ON p.customer_id = c.id
GROUP BY c.ID, c.Name, c.Surname

SELECT c.ID, c.Name, c.SurName, sum(p.payment_amount)
FROM clients c LEFT JOIN
     payments p
     ON c.id = p.customer_id
GROUP BY c.ID, c.Name, c.SurName;

注意：使用相关子查询通常更快：
SELECT c.*, 
       (SELECT SUM(p.payment_amount)
        FROM payments p
        WHERE c.id = p.customer_id
       ) as total_payments
FROM clients c ;

特别是，这可以有效地使用付款索引（客户id、付款金额）
尝试：
select c.Name, c.Surname, SUM(p.payment_amount)
from Payments as p, Clients as c
where p.customerID = c.ID
group by c.Name, c.Surname

我想它会起作用的
试试：
select c.Name, c.Surname, SUM(p.payment_amount)
from Payments as p, Clients as c
where p.customerID = c.ID
group by c.Name, c.Surname

我想它会起作用的
你需要先把两张桌子连在一起。您需要找到表之间的“关系”，以便能够与付款的客户一起汇总付款。在开始将GROUPBY子句应用于查询之前，首先编写连接以关联数据。在这种情况下，您可以通过客户连接表。您需要首先将两个表连接在一起。您需要找到表之间的“关系”，以便能够与付款的客户一起汇总付款。在开始将GROUPBY子句应用于查询之前，首先编写连接以关联数据。在这种情况下，您可以通过客户加入您的表格。\u id感谢您的回复！从技术上讲，它是有效的，但由于某种原因，它改变了罗尔斯的顺序/谢谢你的回复！从技术上讲，它是有效的，但由于某种原因，它改变了罗尔斯的顺序/谢谢你的回复，它工作得很好！谢谢你让我知道这件事！！伟大的资源！很遗憾，我只能接受一个答案。@Geom你接受最新答案有什么原因吗？没有技术原因。这只是奥卡姆的剃须刀。不知何故，最简单的解决方案总是吸引人。不幸的是，我还不熟悉“COALESCE”，所以我更好地理解了其他解决方案中的情况。我希望有一种方法可以接受这两个答案，因为这两个答案都给了我很多帮助@Geom…不熟悉“合并”。。。那你应该查一下。您需要coalesce（），以防客户没有任何付款，因此您得到的结果是0，而不是NULL。我会的，谢谢！谢谢你的回复，它工作得很好！谢谢你让我知道这件事！！伟大的资源！很遗憾，我只能接受一个答案。@Geom你接受最新答案有什么原因吗？没有技术原因。这只是奥卡姆的剃须刀。不知何故，最简单的解决方案总是吸引人。不幸的是，我还不熟悉“COALESCE”，所以我更好地理解了其他解决方案中的情况。我希望有一种方法可以接受这两个答案，因为这两个答案都给了我很多帮助@Geom…不熟悉“合并”。。。那你应该查一下。您需要coalesce（），以防客户没有任何付款，因此您得到的结果是0，而不是NULL。我会的，谢谢！非常感谢。这正是我要找的！非常感谢。这正是我要找的！