我有表#include <stdio.h>
#include <stdlib.h>
#define BLOCKS 20
int *memory[BLOCKS];
int main( void )
{
int seg;
FILE *stream = fopen( “file.csv”, “r” );
if ( !stream )
// print error and exit
printf( “Please enter a number: “);
if ( scanf( “%d”, &seg ) != 1 )
// print error and exit
for ( size_t b = 0; b < BLOCKS; b++ )
{
/**
* Allocate memory for a block (row)
*/
memory[b] = malloc( sizeof *b * seg );
if ( !memory[b] )
// print error and exit
/**
* Read that row from the file - since it has
* a .csv extension, I am assuming it is
* comma-delimited. Note that malloc is not
* required to initialize memory to any specific
* value - the initial value of each memory[b][s]
* is indeterminate.
*/
for ( size_t s; s < seg; s++ )
if ( fscanf( stream, “%d%*c”, &memory[b][s] )) != 1 )
// print error and exit
}
fclose( stream );
/**
* Do stuff with memory here
*/
/**
* After you’re done with memory, free it
*/
for ( size_t b = 0; b < BLOCKS; b++ )
free( memory[b] );
return EXIT_SUCCESS;
}
和users。在orders中的每个UPDATE行之后。我要更新orders表中的DATA，即concat（OLD.DATA +已更新的ID）。
users例如：
Table 'users'.
ID NAME DATA
1 John 1|2
2 Michael 3|4
3 Someone 5
Table 'orders'.
ID USER CONTENT
1 1 ---
2 1 ---
3 2 ---
4 2 ---
5 3 ---
结果：
SELECT `data` from `users` where `id` = 2; // Result: 3|4
UPDATE `orders` SET '...' WHERE `id` > 0;
**NEXT LOOP**
UPDATE `users` SET `data` = concat(OLD.data, ID.rowUpdated) WHERE `user` = 1;
UPDATE `users` SET `data` = concat(OLD.data, ID.rowUpdated) WHERE `user` = 1;
UPDATE `users` SET `data` = concat(OLD.data, ID.rowUpdated) WHERE `user` = 2;
UPDATE `users` SET `data` = concat(OLD.data, ID.rowUpdated) WHERE `user` = 2;
UPDATE `users` SET `data` = concat(OLD.data, ID.rowUpdated) WHERE `user` = 3;
我该怎么办？
答案 0 :(得分：0)
我认为您犯的错误与不久前一样，即将数组/对象存储在列中。
在您的情况下，我建议使用以下表格：
"V"+-----------+-----------+
| id | user_name |
+-----------+-----------+
| 1 | John |
+-----------+-----------+
| 2 | Michael |
+-----------+-----------+
其中+-----------+-----------+------------+
| id | user_id |date_ordered|
+-----------+-----------+------------+
| 1 | 1 | 2019-03-05 |
+-----------+-----------+------------+
| 2 | 2 | 2019-03-05 |
+-----------+-----------+------------+
是user_id的外键
users其中+-----------+-----------+------------+------------+------------+
| id | order_id | item_sku | qty | price |
+-----------+-----------+------------+------------+------------+
| 1 | 1 | 1001 | 1 | 2.50 |
+-----------+-----------+------------+------------+------------+
| 2 | 1 | 1002 | 2 | 3.00 |
+-----------+-----------+------------+------------+------------+
| 3 | 2 | 1001 | 2 | 2.00 |
+-----------+-----------+------------+------------+------------+
是order_id的外键
现在让您困惑的部分。您将需要使用一系列orders来访问每个用户的相关数据。
JOIN这将返回：
SELECT
t3.id AS user_id,
t3.user_name,
t1.id AS order_id,
t1.date_ordered,
SUM((t2.price * t2.qty)) AS order_total
FROM orders t1
JOIN sales t2 ON (t2.order_id = t1.id)
LEFT JOIN users t3 ON (t1.user_id = t3.id)
WHERE user_id=1
GROUP BY order_id;
在任何使用关系数据库的项目中（也就是说，如果您正确地设计了数据库），这些+-----------+--------------+------------+------------+--------------+
| user_id | user_name | order_id |date_ordered| order_total |
+-----------+--------------+------------+------------+--------------+
| 1 | John | 1 | 2019-03-05 | 8.50 |
+-----------+--------------+------------+------------+--------------+
语句的类型基本上都应该出现。通常，我为每个复杂的查询创建一个JOIN，然后可以通过简单的view
例如：
SELECT * FROM orders_view然后可以通过以下方式访问此
：CREATE
ALGORITHM = UNDEFINED
DEFINER = `root`@`localhost`
SQL SECURITY DEFINER
VIEW orders_view AS (
SELECT
t3.id AS user_id,
t3.user_name,
t1.id AS order_id,
t1.date_ordered,
SUM((t2.price * t2.qty)) AS order_total
FROM orders t1
JOIN sales t2 ON (t2.order_id = t1.id)
LEFT JOIN users t3 ON (t1.user_id = t3.id)
GROUP BY order_id
)
将返回与上面的查询相同的结果。
根据您的需要，您可能需要再添加一些表（SELECT * FROM orders_view WHERE user_id=1;
，addresses等），并向每个表添加更多行。很多时候，您会发现需要products 5个以上的表进入一个视图，有时您可能需要两次JOIN同一张表。
我希望这会有所帮助，尽管它不能完全回答您的问题！
答案 1 :(得分：0)
在插入（或更新）ORDERS表之后更新USERS表可能是一个坏主意。避免两次存储数据。在您的情况下：您始终可以通过查询ORDERS表获得用户的所有“订单ID”。因此，您无需再次将它们存储在USERS表中。示例（已通过MySQL 8.0测试，请参见dbfiddle）：
表格和数据
create table users( id integer primary key, name varchar(30) ) ;
insert into users( id, name ) values
(1, 'John'),(2, 'Michael'),(3, 'Someone') ;
create table orders(
id integer primary key
, userid integer
, content varchar(3) references users (id)
);
insert into orders ( id, userid, content ) values
(101, 1, '---'),(102, 1, '---')
,(103, 2, '---'),(104, 2, '---'),(105, 3, '---') ;
也许可以实现 VIEW （类似于以下内容）来解决问题。 （优点：您不需要其他的列或表。）
-- View
-- Inner SELECT: group order ids per user (table ORDERS).
-- Outer SELECT: fetch the user name (table USERS)
create or replace view userorders (
userid, username, userdata
)
as
select
U.id, U.name, O.orders_
from (
select
userid
, group_concat( id order by id separator '|' ) as orders_
from orders
group by userid
) O join users U on O.userid = U.id ;
该视图到位后，您只需从中进行选择，就可以始终获取当前的“用户数据”，例如
select * from userorders ;
-- result
userid username userdata
1 John 101|102
2 Michael 103|104
3 Someone 105
-- add some more orders
insert into orders ( id, userid, content ) values
(1000, 1, '***'),(4000, 1, '***'),(7000, 1, '***')
,(2000, 2, ':::'),(5000, 2, ':::'),(8000, 2, ':::')
,(3000, 3, '@@@'),(6000, 3, '@@@'),(9000, 3, '@@@') ;
select * from userorders ;
-- result
userid username userdata
1 John 101|102|1000|4000|7000
2 Michael 103|104|2000|5000|8000
3 Someone 105|3000|6000|9000