如何通过数据库在TypeScript class中定义属性？

Question

This is my Error when I define properties in my employee.ts

click

export class Employee {
    id:number;
    fname: string;
    lastName:string;
    emailId: string; }

&我遇到了这种错误

属性 'id' 没有初始值设定项并且未明确赋值构造函数。
属性 'fname' 没有初始值设定项并且未明确赋值构造函数。
属性 'lastName' 没有初始化器，也没有明确赋值在构造函数中。
属性 'emailId' 没有初始化器，也没有明确赋值在构造函数中。

这也是我的数据库 click

mysql> SELECT *
    -> FROM employees;
+----+-----------------+------------+-----------+
| id | email_id        | first_name | last_name |
+----+-----------------+------------+-----------+
|  1 | mara@gmail.com  | Amara      | Perera    |
|  2 | sadun@gmail.com | Sadun      | Fernando  |
+----+-----------------+------------+-----------+
2 rows in set (0.01 sec)

Here is mY Employee.java file created on eclips IDE


import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "employees")
public class Employee {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;
    
    @Column(name = "first_name")//column annotation to provide columns
    private String firstName;
    
    @Column(name = "last_name")
    private String lastName;
    
    @Column(name = "email_id")
    private String emailId;
    
    public Employee() {
        
    }
    
    //generate constructor
    public Employee(String firstName, String lastName, String emailId) {
        super();
        this.firstName = firstName;
        this.lastName = lastName;
        this.emailId = emailId;
    }
    public long getId() {
        return id;
    }
    public void setId(long id) {
        this.id = id;
    }
    public String getFirstName() {
        return firstName;
    }
    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
    public String getLastName() {
        return lastName;
    }
    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
    public String getEmailId() {
        return emailId;
    }
    public void setEmailId(String emailId) {
        this.emailId = emailId;
    }
    
    
    

}

谁能帮我解决这个错误吗？？

Answer 1

编译器看到缺少初始化代码的显式类型。您应该将字段标记为可能未定义

class Employee {
    id?: number;
}

或在构造函数中初始化它们

export class Employee {
    constructor(public readonly id: number) {}  
}

Answer 2

TypeScript 正试图警告您这样的情况：

const employee = new Employee();

因为此时employee中的所有属性都未定义，但您将它们定义为数字和字符串。您可以使属性可选，在构造函数中初始化它们或将以下内容添加到 tsconfig.json:

"strictPropertyInitialization": false

在 Java 你没有这种类型的支票。

如何通过数据库在TypeScript class中定义属性？

How to define properties in TypeScript class through the database?

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