我試圖尋找這個(gè)問題的答案，但意識(shí)到有多個(gè)相似但沒有一個(gè)與這個(gè)匹配。我有一個(gè)具有這種結(jié)構(gòu)的 JSON 對象{ "model": { "serie" : "123456", "id" : "abc123" /// many fields }, "externalModel": { "serie" : "123456", "fieldX" : "abcde" // many fields as well}我在我的代碼中這樣做： ObjectMapper mapper = new ObjectMapper(); MyObject object = mapper.readValue(hit.getSourceAsString(), MyObject.class);其中 MyObject 具有以下形式：@JsonInclude(value = JsonInclude.Include.NON_NULL)@JsonIgnoreProperties(ignoreUnknown = true)public class MyObject { @JsonProperty("serie") String serie; @JsonProperty("id") Long id; MyObject() {} }當(dāng)我轉(zhuǎn)換時(shí)，我沒有得到任何異常，而是我得到了所有值都設(shè)置為null的 myObject我不知道會(huì)出什么問題，因?yàn)闆]有返回任何異常，知道嗎？
2 回答
皈依舞
TA貢獻(xiàn)1851條經(jīng)驗(yàn) 獲得超3個(gè)贊
您需要使用根屬性model，
您可以重命名MyObject并MyModel創(chuàng)建一個(gè)MyObject
@JsonInclude(value = JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class MyObject{
@JsonProperty("model")
MyModel model;
}
然后檢查model
GCT1015
TA貢獻(xiàn)1827條經(jīng)驗(yàn) 獲得超4個(gè)贊
實(shí)際上，您需要 MyObject 中的兩個(gè)對象。
@JsonInclude(value = JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class MyModel {
@JsonProperty("id")
private String id;
@JsonProperty("serie")
private String serie;
//Generate getters and setters of these two
}
@JsonInclude(value = JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class ExternalObject {
@JsonProperty("serie")
private String serie;
@JsonProperty("fieldX")
private String fieldX;
//Generate getters and setters of these two
}
@JsonInclude(value = JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class MyObject{
@JsonProperty("model")
private MyModel model;
@JsonProperty("externalModel")
private ExternalObject externalModel;
//Generate getters and setters of these two
}
現(xiàn)在，當(dāng)您像下面那樣使用它時(shí)，它會(huì)正常工作。
ObjectMapper mapper = new ObjectMapper();
MyObject object = mapper.readValue(hit.getSourceAsString(), MyObject.class);
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