Spring @RequestMapping方法無@RequestBody

Question

在Spring Boot Controller方法中，如何獲取POST的主體？我見過的所有例子都使用@RequestBody。如何在不使用@RequestBody的情況下獲取身體？

我在寫一個方法來處理Slack事件。當Slack POST一個事件時，主體是JSON並且通常包含一個「用戶」鍵。根據事件的類型，「user」的值可以是字符串或對象。正因爲如此，我不能創建一個類，寫

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

答：實現由@ChiDov建議的方法，解決的辦法是保持@RequestBody，進口

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

定義用戶字段（和一個新的字段來存儲「用戶」，如果它是一個簡單的字符串值）作爲

@OneToOne 
private SlackEventUser user; 
private String slackUserId; 

並定義它的設置器方法

@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; 
    } 
} 

Answer 1

更新：我會讓你的DTO喜歡：

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

和控制器：

@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 
    } 
} 

獲得啓示：How to deserialize dynamic field using Jackson?