使用Simple-XML-Framework时的ValueRequired异常

Siv*_*gam 5 android simple-framework

我正在尝试使用Simple-XML-framework.但我得到ValueRequiredException

org.simpleframework.xml.core.ValueRequiredException: Unable to satisfy @org.simpleframework.xml.ElementList(data=false, empty=true, entry=, inline=false, name=, required=true, type=void) on field 'Employees' public java.util.List com.example.xmlparsing.Company.Employees for class com.example.xmlparsing.Company at line 2
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我的XML文件包含以下内容:

<?xml version="1.0" encoding="utf-8"?>
<Company>
<Emp>
    <Name>Venkat</Name>
    <ID>661511</ID>  
</Emp>
<Emp>
    <Name>Shiv</Name>
    <ID>661311</ID> 
</Emp>
</Company>
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我的注释类如下:

Company.java:

package com.example.xmlparsing;

import java.util.List;

import org.simpleframework.xml.ElementList;
import org.simpleframework.xml.Root;

@Root
public class Company {

@ElementList
public List<Emp> Employees;

}
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Emp.java:

package com.example.xmlparsing;

import org.simpleframework.xml.Element;
import org.simpleframework.xml.Root;
import org.simpleframework.xml.Text;

@Element(name="Emp")
public class Emp {

@Element
public String Name;

@Element
public String ID;


}
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可能是什么问题呢?怎么解决?

oll*_*llo 9

您的代码中有两个错误:

首先,你需要@Root()而不是@ElementEmp课堂上:

@Root(name="Emp")
public class Emp {

@Element
public String Name;

@Element
public String ID;


}
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其次,如果你使用这样的列表:

@Root
public class Company {

@ElementList
public List<Emp> Employees;

}
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它包装成另一个xml标签.
您可以更改xml或只是使用@ElementList(inline = true).