在Spring REST控制器中将JSON映射到Hibernate模型 [英] Mapping JSON to Hibernate models in Spring REST Controllers

问题描述

假设我有一个Spring 4 Web应用程序和一个基于Hibernate的持久层。我想创建一个支持我的模型的基本CRUD操作的 RestController 。创建一个获取记录的方法并不困难：

  @RequestMapping（value =/ stuff / list，method = RequestMethod .GET）
 public List< Stuff> getStuff（）{
 return stuffService.findAll（）; 
} 
  

杰克逊处理对象序列化，没有问题。但是如果我想添加一个通过POST请求创建新记录的方法呢？有没有简单的方法来支持这样的简单方法？

  @RequestMapping（value =/ stuff / new，method = RequestMethod.POST）
 public Integer getStuff（@RequestParam（stuff）Stuff stuff）{
 return stuffService.save（stuff）; 
} 
  

是这样的可能吗？或者我需要手动将发布的表单数据映射到新对象？

解决方案

以下是我解决问题的方法，脚步。首先，我的最终控制器方法：

  @RequestMapping（value =/ stuff / new，method = RequestMethod.POST）
 public Integer getStuff（@RequestBody Stuff stuff）{
 return stuffService.save（stuff）; 
} 
  

我已经有一个过滤器应用于应用程序API的所有请求，跨源数据共享，但需要进行修改以允许请求指定内容类型： b
$ b @Override public void doFilter（ServletRequest servletRequest，ServletResponse servletResponse，
FilterChain filterChain）throws IOException，ServletException {

HttpServletResponse response =（HttpServletResponse）servletResponse;
response.setHeader（Access-Control-Allow-Origin，*）;
response.setHeader（Access-Control-Allow-Methods，POST，GET，OPTIONS，DELETE）;
response.setHeader（Access-Control-Max-Age，3600）;
response.setHeader（Access-Control-Allow-Headers，x-requested-with，Content-Type）;
filterChain.doFilter（servletRequest，servletResponse）;


$ b @Override public void init（FilterConfig filterConfig）throws ServletException {}
$ b $ @Override public void destroy（）{}
}

在我的 web.xml中注册 file：

 < filter> 
< filter-name> cors< / filter-name> 
< filter-class> com.company.app.util.SimpleCORSFilter< / filter-class> 
< / filter> 
 
< filter-mapping> 
< filter-name> cors< / filter-name> 
< url-pattern> / api / *< / url-pattern> 
< / filter-mapping> 
  

现在，当我提出请求时，例如下面的请求，它会正确地将我提交的JSON映射到我的模型，并坚持一个新的实例：

$ $ p $ var stuff = {
name：Some stuff，
描述：这是一些东西。
$ .ajax（{
url：url，
method：post，
dataType：json，
data：JSON.stringify（stuff），
contentType：application / json
}）。done（function（data）{
console.log（data）;
} ）.fail（function（x，status，e）{
console.log（e）;
}）;


解决方案

为了告诉Spring你想让它应用一个解串器到内容而不是尝试标准的HTML表单绑定你使用 @RequestBody 。

 <$ c $ （@RequestBody的东西）{
返回stuffService.save（东西）; / /。 
} 
  

@RequestParam 是告诉它查找具有该名称的单个参数并应用标准数据绑定，而不是将POST的全部内容反序列化到对象中。

Let's say that I have a Spring 4 web application with a Hibernate-based persistence layer. I'd like to create a RestController that supports basic CRUD operations for my models. Creating a method for fetching records works without a hitch:

@RequestMapping(value = "/stuff/list", method = RequestMethod.GET)
public List<Stuff> getStuff(){
    return stuffService.findAll();
}

Jackson handles the object serialization, no problem. But what if I want to add a method for creating new records via a POST request? Is there any easy way to support a simple method like this?

@RequestMapping(value = "/stuff/new", method = RequestMethod.POST)
public Integer getStuff(@RequestParam("stuff") Stuff stuff){
    return stuffService.save(stuff);
}

Is something like this possible? Or do I need to manually map posted form data to a new object?

SOLUTION

Here is how I solved my problem, there were a couple steps. First, my final controller method:

@RequestMapping(value = "/stuff/new", method = RequestMethod.POST)
public Integer getStuff(@RequestBody Stuff stuff){
    return stuffService.save(stuff);
}

I already had a filter applied to all requests to the application API, to allow cross origin resource sharing, but modifications to this were needed to allow requests to specify content type:

public class SimpleCORSFilter implements Filter{

    @Override public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse,
            FilterChain filterChain) throws IOException, ServletException {

        HttpServletResponse response = (HttpServletResponse) servletResponse;
        response.setHeader("Access-Control-Allow-Origin", "*");
        response.setHeader("Access-Control-Allow-Methods", "POST, GET, OPTIONS, DELETE");
        response.setHeader("Access-Control-Max-Age", "3600");
        response.setHeader("Access-Control-Allow-Headers", "x-requested-with, Content-Type");
        filterChain.doFilter(servletRequest, servletResponse);

    }

    @Override public void init(FilterConfig filterConfig) throws ServletException { }

    @Override public void destroy() { }
}

Which is registered in my web.xml file:

<filter>
    <filter-name>cors</filter-name>
    <filter-class>com.company.app.util.SimpleCORSFilter</filter-class>
</filter>

<filter-mapping>
    <filter-name>cors</filter-name>
    <url-pattern>/api/*</url-pattern>
</filter-mapping>

Now when I make a request, such as the one below, it will correctly map my submitted JSON to my model and persist a new instance:

var stuff = {
    name: "Some stuff",
    description: "This is some stuff."
}

$.ajax({
    url: url,
    method: "post",
    dataType: "json",
    data: JSON.stringify(stuff),
    contentType: "application/json"
  }).done(function(data){
    console.log(data);
  }).fail(function(x, status, e){
    console.log(e);
  });

解决方案

To tell Spring that you want it to apply a deserializer to the content instead of attempt standard HTML form binding you use @RequestBody.

@RequestMapping(value = "/stuff/new", method = RequestMethod.POST)
public Integer getStuff(@RequestBody Stuff stuff){
    return stuffService.save(stuff);
}

@RequestParam is telling it to look for an individual parameter with that name and apply standard databinding, not to deserialize the entire content of the POST into an object.

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