如何使用SpringBoot從POST方法回傳物件？

我正在嘗試使用 Spring Boot 在兩個服務之間創建通信，但無法從 POST 回傳自定義物件。

我試過類似的東西

第一服務：

@RestController
@RequestMapping("/broker")
class HostController{
    @PostMapping(value = "/bid")
    public Bid bid(@RequestBody Auction auction){
        return new Bid(new URI("http://url:8080"));
    }
}

第二服務：

ResponseEntity<Bid> response = rest.postForEntity(hostURL   "/bid", auction, Bid.class);
response.getBody();

問題是我收到一條錯誤訊息，指出 Bid 類的“無法從物件值反序列化”，這讓我認為 Auction 正在發送，但 Bid 沒有發回。

我也不確定拍賣的序列化是如何發生的，因為我只輸入了“@RequestBody”，它就開始作業了。Auction 類內部甚至有一個 Bid 物件，但這似乎不是問題。

class Bid{
    private URI host;
    
    public Bid(URI host){ this.host = host; }

    public URI getHost() { return host; }
}

class Auction{
    URI host;

    private Bid winner; //Not defined when the problem happens

    public Auction(URI host){ this.host = host; }
    
    public URI getHost(){ return host; }
}

完整的堆疊跟蹤是巨大的，但我認為相關的部分是：

com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.gustavovbs.microservicesoffloading.Bid` (no Creators, like default constructor, exist): cannot deserialize from Object value (no delegate- or property-based Creator)\n at [Source: (PushbackInputStream); line: 1, column: 2]\n\tat
 
com.fasterxml.jackson.databind.exc.InvalidDefinitionException.from(InvalidDefinitionException.java:67)\n\tat
 com.fasterxml.jackson.databind.DeserializationContext.reportBadDefinition(DeserializationContext.java:1764)\n\tat
 com.fasterxml.jackson.databind.DatabindContext.reportBadDefinition(DatabindContext.java:400)\n\tat

com.fasterxml.jackson.databind.DeserializationContext.handleMissingInstantiator(DeserializationContext.java:1209)\n\tat
 com.fasterxml.jackson.databind.deser.BeanDeserializerBase.deserializeFromObjectUsingNonDefault(BeanDeserializerBase.java:1415)\n\tat
 com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:362)\n\tat
 com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:195)\n\tat
 com.fasterxml.jackson.databind.deser.DefaultDeserializationContext.readRootValue(DefaultDeserializationContext.java:322)\n\tat
 com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4593)\n\tat
 com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3601)\n\tat
 org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.readJavaType(AbstractJackson2HttpMessageConverter.java:378)\n\t... 63 more\n","message":"Type definition error: [simple type, class com.gustavovbs.microservicesoffloading.Bid]; 

nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.gustavovbs.microservicesoffloading.Bid` (no Creators, like default constructor, exist): cannot deserialize from Object value (no delegate- or property-based Creator)\n at [Source: (PushbackInputStream); line: 1, column: 2]","path":"/broker/broadcast"}"

uj5u.com熱心網友回復：

當你傳輸一些東西時，確保它是可序列化的，并且不要忘記添加任何 arg 建構式。這應該有效。

   class Bid implements Serializable {
    
    private URI host;

    public Bid();

    public URI getHost() {
        return host;
    }

    public void setHost(URI host) {
        this.host = host;
    }
}

uj5u.com熱心網友回復：

正如評論中所討論的，我設法通過將默認建構式添加到 Bid 類來使其作業。

class Bid{
    private URI host;

    public Bid(){}
    
    public Bid(URI host){ this.host = host; }

    public URI getHost() { return host; }
}

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