Question

我想将员工体验数据存储在数据库中，员工可以一次添加多种体验。

数组：

Array
(
    [start_date] => Array
        (
            [0] => 2016-07-12
            [1] => 2016-09-16
        )

    [end_date] => Array
        (
            [0] => 2016-09-09
            [1] => 2017-01-20
        )

    [total_month] => Array
        (
            [0] => 2
            [1] => 4
        )

    [firm_name] => Array
        (
            [0] => sadsadasd
            [1] => 34343
        )

    [turnover] => Array
        (
            [0] => 343443434
            [1] => 443434
        )

    [student_experiece] => Array
        (
            [0] => 343434
            [1] => 343434
        )

    [orderof_supply_food] => Array
        (
            [0] => 343443
            [1] => 434343434
            [2] => 34334
            [3] => 34343
        )

    [payment_against_bill] => Array
        (
            [0] => 3434343
            [1] => 434343
        )

    [vat] => Array
        (
            [0] => 3434343
            [1] => 3434343434
        )

）

我所做的是将数组转换为字符串，然后存储在数据库中，如下所示：

[start_date] = 2016-07-12,2016-09-16,
[end_date] =2016-09-09,2016-01-20

数据库结构：

我认为这不是正确的方法。请提出任何建议吗？

Answer 1

在数据库中存储以逗号分隔的值通常会违反First Normal Form。有关详细信息，请参阅this answer。

我建议使用两个表：一个用于员工，一个用于体验。然后，您可以在＆＃34;体验＆＃34;中使用外键。表引用＆＃34;雇员＆＃34;中的主键。表

为了示例，我简化了您的表格结构：

CREATE TABLE IF NOT EXISTS `employees` (
  `id` mediumint(8) UNSIGNED NOT NULL AUTO_INCREMENT,
  `name_first` varchar(50) NOT NULL,
  `name_last` varchar(50) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=6 DEFAULT CHARSET=latin1;

CREATE TABLE IF NOT EXISTS `experiences` (
  `id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT,
  `employee_id` int(11) NOT NULL,
  `date` date NOT NULL,
  `description` text NOT NULL,
  PRIMARY KEY (`id`),
  KEY `employee_id` (`employee_id`)
) ENGINE=MyISAM AUTO_INCREMENT=7 DEFAULT CHARSET=latin1;


INSERT INTO `employees` (`id`, `name_first`, `name_last`) VALUES
(1, 'Jane', 'Doe'),
(2, 'John', 'Doe'),
(3, 'Elmer', 'Fudd'),
(4, 'Tester', 'McTesterson'),
(5, 'Sally', 'Jones');

INSERT INTO `experiences` (`id`, `employee_id`, `date`, `description`) VALUES
(1, 2, '2016-07-01', 'Amazing Experience'),
(2, 4, '2016-07-02', 'Testing'),
(3, 2, '2016-07-03', 'Another Experience'),
(4, 3, '2016-07-05', 'Test Experience'),
(5, 1, '2016-07-07', 'Sample Experience'),
(6, 2, '2016-07-15', 'An experience');

请注意，experiences表有一个名为＆＃34; employee_id＆＃34;的外键。此值与employees表中员工的主键相关。

使用此结构，您无需为每次体验重复员工数据。可以轻松添加新体验并与现有员工相关联。您可以使用JOIN获取关联数据，具体取决于您所需的输出。

按日期升序获取所有体验：

SELECT emp.`name_first`,emp.`name_last`,exp.`date`,exp.`description`
FROM `experiences` exp
LEFT JOIN `employees` emp ON (emp.`id`=exp.`employee_id`)
WHERE 1
ORDER BY exp.`date` ASC;

name_first  name_last   date        description 
John        Doe         2016-07-01  Amazing Experience  
Tester      McTesterson 2016-07-02  Testing 
John        Doe         2016-07-03  Another Experience  
Elmer       Fudd        2016-07-05  Test Experience 
Jane        Doe         2016-07-07  Sample Experience   
John        Doe         2016-07-15  An experience

Working Example

计算每位员工的经验：

SELECT emp.`name_last`,emp.`name_first`,COUNT(exp.`id`) as `experience_count`
FROM `employees` emp
LEFT JOIN `experiences` exp ON (exp.`employee_id`=emp.`id`)
GROUP BY emp.`id`
ORDER BY emp.`name_last` ASC;

name_last   name_first  experience_count
Doe         John        3
Doe         Jane        1
Fudd        Elmer       1
Jones       Sally       0
McTesterson Tester      1

Working Example

以下是您可能会发现有用的信息资源：
Database Normalization - Explained with Examples
Normalization in Database (With Example)
Normalization of Database

Answer 2

您可以使用json_encode()将PHP-Array编码为JSON-String。当您从数据库中获取JSON-String时，可以使用json_decode()将其解码为PHP数组。然后，您可以修改它并将其存储回数据库。

如何在数据库中存储员工体验数据

2 个答案: