Spring 没有@RequestBody 的@RequestMapping 方法

Spring @RequestMapping method without @RequestBody

在 Spring 引导控制器方法中,如何获取 POST 的主体?我见过的所有示例都使用@RequestBody。如何在不使用@RequestBody 的情况下获取正文?

我正在编写一种方法来处理 Slack 事件。当 Slack POST 是一个事件时,正文在 JSON 中并且通常包含一个 "user" 键。根据事件的类型,"user" 的值可以是字符串或对象。因此,我无法创建单个 Class 并写入

@RequestMapping(path = "/slackRequest", method = RequestMethod.POST)
    public String handleSlackRequest(@RequestBody final SlackRequest slackRequest)

答案:实现@ChiDov建议的方法,解决方案是保留@RequestBody,import

import com.fasterxml.jackson.annotation.JsonSetter;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;

定义用户字段(如果它是一个简单的字符串值,则定义一个新字段来存储 'user')为

@OneToOne
private SlackEventUser user;
private String slackUserId;

并将其 Setter 方法定义为

@JsonSetter("user")
public void setUser(JsonNode userNode) {
    ObjectMapper mapper = new ObjectMapper();
    mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
    if (userNode.isObject()) {
        SlackEventUser slackEventUser = mapper.convertValue(userNode, SlackEventUser.class);
        this.user = slackEventUser;
    } else {
        String userString = mapper.convertValue(userNode, String.class);
        this.slackUserId = userString;
        this.user = null;
    }
}

已更新:我会让你的 DTO 像 :

Class SlackRequest{
    ...
    private String eventType;
    private JsonNode user;
    ...
    public String getUser(){
        return user.asText();
    }
}

在控制器中:

@RequestMapping(path = "/slackRequest", method = RequestMethod.POST)
    public String handleSlackRequest(@RequestBody final SlackRequest slackRequest){
    if(slackRequest.getEventType == "objectEvent"){
      SomeObject user = mapper.convertValue(slackRequest.getUser(), SomeObject.class);
      // do something with the object
    }else{
     // do something with the user string
   }
}

从中获取灵感:How to deserialize dynamic field using Jackson?