英文:
How to Store values in specific column by using JSON from Postman (values are separated by underscore_)
问题
以下是翻译好的内容：
任何人都可以帮助我吗？使用@RequestBody将数据从Postman以下面默认的JSON格式进行发布是很容易的：
{"id": 1, "firstName": "abc", "lastName": "xyz"}
但是如何以这种格式发布数据呢？
{"EMPLOYEE_DETAILS": "1_abc_xyz"}
并且数据将被存储在特定的列中。
@Entity
@Table(name = "employee")
public class Employee {
@Id
private int id;
@Column(name = "first_name")
private String firstName;
@Column(name = "last_name")
private String lastName;
// 生成setter和getter方法
}
英文:
Anybody please can help me it's easy to post data from Postman in below default JSON format {"id":1, "firstName":"abc", "lastName":"xyz"} by using @RequestBody Employee emp
but how to post data in this format {"EMPLOYEE_DETAILS":"1_abc_xyz"} And data will be stored in specific column
@Entity
@Table(name="employee")
public class Employee{
@Id private int id;
@Column(name = "first_name")
private String firstName;
@Column(name = "last_name")
private String lastName;
//Generate setters and getters
}
答案1
得分: 1
你可以在控制器中以 Map<String, String> 的形式获取输入，或者创建一个 DTO，然后将值存储在 String 数据类型中，就像这样：
String data = map.get("EMPLOYEE_DETAILS");
或者
String data = employeeDto.getEmployesDetails();
首先，使用下划线（_）将数据拆分，就像这样：
String[] splitter = data.split("_");
在 Employee 实体中创建带有所有参数的构造函数，然后像这样传递 splitter：
Employee employee = new Employee(splitter[0], splitter[1], splitter[2]);
谢谢。
英文:
You can take input in controller either Map<String,String> or create a DTO and then store value in String datatype like this
String data=map.get("EMPLOYEE_DETAILS");
or
String data=employeeDto.getEmployesDetails();
First split data by using _(underscore) like this
String splitter=data.split("_");
create all argument constructor in Employee enitity and pass by splitter like this
Employee employee =new Employee(splitter[0],splitter[1],splitter[2])
Thanks
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