已解决

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/"
}

那是错误消息

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> {
}

UsersRepository.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;
    }
}

Users.java

摘要；

在地址（localhost：8080 / rest / users / all）上，Get操作正在顺利运行。但是，当我尝试使用（localhost：8080 / rest / users / load）上的帖子创建新用户时，出现错误：

“ message”：“ JSON解析错误：无法从START_ARRAY令牌中反序列化com.springboot.sprinboot.model.Users的实例；嵌套的异常是com.fasterxml.jackson.databind.exc.MismatchedInputException：无法反序列化com.springboot.sprinboot.model.Users的实例不在[来源：（PushbackInputStream）；第1行，第1列]中的START_ARRAY令牌中\ n“，

示例json

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

已解决

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

所有人，谢谢，我不需要添加，因为id是主键。

Answer 1

您应该发送与此类似的JSON

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

}

很可能您使用的是[而不是{或两者都使用

Spring Boot JSON解析错误：无法反序列化错误

已解决

1 个答案: