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ks_on_v
Asked:2020-02-12 15:52:32 +0000 UTC2020-02-12 15:52:32 +0000 UTC

错误：不应该在 setterless 属性上调用“set”

  • 772

尝试反序列化 Jackson 中的响应时出错

org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Can not construct instance of org.proj.model.profile.Profile, problem: Should never call 'set' on setterless property
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@2d881384; line: 10, column: 1] (through reference chain: java.util.ArrayList[0]); nested exception is com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of org.proj.model.profile.Profile, problem: Should never call 'set' on setterless property
at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@2d881384; line: 10, column: 1] (through reference chain: java.util.ArrayList[0])
at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.readJavaType(AbstractJackson2HttpMessageConverter.java:208)
at org.springframework.http.converter.json.AbstractJackson2HttpMessageConverter.read(AbstractJackson2HttpMessageConverter.java:200)

此时，我在 Profile 构造函数中添加了另一个变量

@JsonProperty("childrens") List<String> childrens

调用后出现错误：

return restClient.exchange(PROFILES_URL + "/name" + path, HttpMethod.GET, restClient.makeHttpEntityWithAuth(), new ParameterizedTypeReference<Collection<Profile>>() {
            });

我在网上搜索了错误的名称，但没有找到。

简介类：

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
public class Profile{

private final String id;
private final List<String> childrens;

public Profile(
        @JsonProperty("id") String id,
        @JsonProperty("childrens") List<String> childrens) {
    this.id = id;
    this.childrens = childrens;
}

public String getId() {
    return id;
}

public List<String> getChildrens() {
    return childrens;
}

有什么问题？

java

1 个回答

  1. Best Answer

    你忘记@JsonCreator了构造函数。这就是杰克逊试图使用 Setter 而不是构造函数的原因，而您只是没有 setter。

    @JsonCreator //here
public Profile(
         @JsonProperty("id") String id,
         @JsonProperty("childrens") List<String> childrens) {
     this.id = id;
     this.childrens = childrens;
}

    这是关于这个主题的更多内容

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