java - Spring Boot JSON解析错误: Cannot deserialize error

Question

{
    "timestamp": "2018-07-18T11:02:29.789+0000",
    "status": 400,
    "error": "Bad Request",
    "message": "JSON parse error: Cannot deserialize instance of `com.springboot.sprinboot.model.Users` out of START_ARRAY token; nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `com.springboot.sprinboot.model.Users` out of START_ARRAY token\n at [Source: (PushbackInputStream); line: 1, column: 1]",
    "path": "/rest/users/"
}

这是错误信息

package com.springboot.sprinboot.resource;

import com.springboot.sprinboot.model.Users;
import com.springboot.sprinboot.repository.UsersRepository;
import org.apache.tomcat.util.http.parser.MediaType;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.*;
import org.springframework.web.bind.annotation.RestController;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.GetMapping;

import java.awt.*;
import java.util.List;

@RestController
@RequestMapping(value = "/rest/users")
public class UsersResource {

    @Autowired
    UsersRepository usersRepository;

    @GetMapping(value = "/all")
    public List<Users> getAll(){
        return usersRepository.findAll();

    }

    
     @PostMapping (value = "/load")
public List<Users> persist(@RequestBody final Users users){
    usersRepository.save(users);
    return usersRepository.findAll();
    }
}

UsersResource.java

package com.springboot.sprinboot.repository;

import com.springboot.sprinboot.model.Users;
import org.springframework.data.jpa.repository.JpaRepository;

public interface UsersRepository extends JpaRepository<Users, Integer> {
}

用户库.java

package com.springboot.sprinboot.model;

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

@Entity
public class Users {

    @Id
    @GeneratedValue
    @Column(name = "id")
    private Integer id;
    @Column(name = "name")
    private String name;
    @Column(name = "team_name")
    private String teamName;
    @Column (name = "salary")
    private Integer salary;

    public Users() {
    }

    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getTeamName() {
        return teamName;
    }

    public void setTeamName(String teamName) {
        this.teamName = teamName;
    }

    public Integer getSalary() {
        return salary;
    }

    public void setSalary(Integer salary) {
        this.salary = salary;
    }
}

用户.java

总结；

在地址(localhost:8080/rest/users/all)，Get操作运行顺利。但是，当我尝试在 (localhost:8080/rest/users/load) 上创建一个新用户时，出现错误:

"message": "JSON parse error: Cannot deserialize instance of com.springboot.sprinboot.model.Users out of START_ARRAY token; nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of com.springboot.sprinboot.model.Users out of START_ARRAY token\n at [Source: (PushbackInputStream); line: 1, column: 1]",

示例 json

[
    {
        "id": 2,
        "name": "omer",
        "teamName": "omr",
        "salary": 200
    }
]

---

已解决

{      
        "name": "omer",
        "teamName": "omr",
        "salary": 200
}

谢谢大家，我是因为id是主键所以不能加

Answer 1

你应该发送一个类似这样的 JSON

{
     "id": 1,
     "name": "omer"
    ........ 

}

很可能你正在使用 [ 而不是 { 或者两者都用