java - Spring REST - 从 JSON 输入将实体添加到数据库 - 数据库外键错误

Question

我有两个实体:A 和 B:

在 Controller 类中我有这些方法:

@RequestMapping(value="/json/get", method=RequestMethod.GET)
    public @ResponseBody List<B> getAll() {
        return BManager.findAll();
    }

@RequestMapping(value="/json/add", method=RequestMethod.POST)
    public @ResponseBody void addB(@RequestBody B b) {      
        B.setA(AManager.findByAddress(123456)); // I must be from JSON, not static!
        BManager.save(b);           
    }

URL /json/get 返回给我:

{  
   "value":1,
   "title":"hello"
}

A 没有值。但没关系。问题是当我想从 JSON 输入创建新的 B 实体时。

我发送

curl -X POST -H "Content-Type: application/json" -H "Accept: application/json" -d '{"value":2,"title":"world"}' http://localhost:8080/test/json/add

如果我在方法 addB 行 B.setA(AManager.findByAddress(123456)); 中，它将实体保存到数据库。但我需要像这样在输入 JSON 中指定地址(我可以更改 json 语法):

curl -X POST -H "Content-Type: application/json" -H "Accept: application/json" -d '{"address":123456,"value":2,"title":"world"}' http://localhost:8080/test/json/add

它崩溃了:

SEVERE: Servlet.service() for servlet [spring] in context with path [/test] threw exception [Request processing failed; nested exception is org.springframework.dao.DataIntegrityViolationException: could not execute statement; SQL [n/a]; constraint [null]; nested exception is org.hibernate.exception.ConstraintViolationException: could not execute statement] with root cause
com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException: Column 'A_id' cannot be null

问题:

1) 如何在输入 JSON 中指定 B 的地址(A 类)？

2) 如果我想要从方法 getAll 输出 JSON(B 类)中的地址(来自 A 类)，我该怎么做？

我想获取/json/get:

{  
      "addrress":123456,
   "value":1,
   "title":"hello"
}

Answer 1

您的错误是由于 Address 类的主键在保存时为空，请尝试如下操作

@RequestMapping(value="/json/add", method=RequestMethod.POST)
    public @ResponseBody void addB(@RequestBody B b) { 

        A a = AManager.findByAddress(123456);

        if(a == null)
        {
          // you need to create a new obj in here

           a = new A();
           a.setId(anyId);
            //remain code
        }

         //then set in here  
        B.setA(a); // I must be from JSON, not static!
        BManager.save(b);           
    }

对于问题 #2，请发布您的 POJO

已更新

您需要将实体 B 作为

 Class B {
         String id;
         String name;
         Address address;        
        //getter setters

        }



 Class Address
 {
   String address1;
   String address2;
   String code;

   //getter setters
 }

因此，在这种关系中，您需要将 json 发送为

{
  name:"bla bla",
  id:"12345"
  address:{
            address1:"any dummy address",
            address2:"any dummy address",
            code:"12345"

  }

}

现在你的 Controller 地址对象将被 spring 自动映射到“b”对象

@RequestMapping(value="/json/add", method=RequestMethod.POST)
    public @ResponseBody void addB(@RequestBody B b) { 

        // you can get your address object here or can save it directly
       Address address = b.getAddress();

        BManager.save(b);           
    }