Question

我需要使用jackson-dataformat-xml将一些XML文件反序列化为常规java对象。所以我在做：

import com.fasterxml.jackson.dataformat.xml.XmlMapper;

XmlMapper mapper = new XmlMapper();
return mapper.readValue(xmlString, Certificate.class);

xmlString 有外观：

    <?xml version="1.0" encoding="UTF-8"?>
    <doc>
        <r  key="0">
            <ATT_SEARCH DM="dm1" DS="ds1" DocType="1"/>
            <ATT_SEARCH DM="dm2" DS="ds2" DocType="2"/>
        </r>
    </doc>

班级证书：

package ua.max;

import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlElementWrapper;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlProperty;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlRootElement;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import java.util.List;


@JacksonXmlRootElement(localName = "doc")
@XmlAccessorType(XmlAccessType.FIELD)
public class Certificate {

    @JacksonXmlProperty(localName = "r")
    private R r;

    public R getR() {
        return r;
    }

    public void setR(R r) {
        this.r = r;
    }


    public class R {

        @JacksonXmlProperty(localName = "ATT_SEARCH")
        @JacksonXmlElementWrapper(useWrapping = false)
        private List<AttSearch> attSearch;

        public List<AttSearch> getAttSearch() {
            return attSearch;
        }

        public void setAttSearch(List<AttSearch> attSearch) {
            this.attSearch = attSearch;
        }

        @JacksonXmlProperty(isAttribute = true, localName = "key")
        private String key;

        public String getKey() {
            return key;
        }

        public void setKey(String key) {
            this.key = key;
        }


        public class AttSearch {

            @JacksonXmlProperty(isAttribute = true, localName = "DM")
            private String dm;

            @JacksonXmlProperty(isAttribute = true, localName = "DS")
            private String ds;

            @JacksonXmlProperty(isAttribute = true, localName = "DocType")
            private String docType;


            public String getDm() {
                return dm;
            }

            public void setDm(String dm) {
                this.dm = dm;
            }

            public String getDs() {
                return ds;
            }

            public void setDs(String ds) {
                this.ds = ds;
            }

            public String getDocType() {
                return docType;
            }

            public void setDocType(String docType) {
                this.docType = docType;
            }


        }


    }


}

在尝试去实现XML之后，我得到了例外：
“找不到类型[simple type，class ua.max.Certificate $ R]的合适构造函数：无法从JSON对象实例化”

我的尝试：
1.如果我为我的内部类添加修饰符“static”它正在工作，我得到java对象，但除了2个对象的列表“ATT-SEARCH”我得到的第一个是null 2.添加不同的构造函数没有任何影响

Answer 1

R和AttSearch应该是静态的：

 public static class R {
   // other stuff

 public static class AttSearch {
   // other stuff

否则编译器使用外部类引用作为参数创建默认构造函数，因此fastxml无法找到没有参数的构造函数并创建pojo。

异常“通过fastxml将XML文件解析为POJO时，找不到类型[simple type ...”的合适构造函数

1 个答案: