使用xstream将XMLl转换为Java对象

我有以下xml。如何使用xstream转换为java对象。我尝试了几种方法，但最终还是得到了conversionexception。 代码如下。我不知道如何转换为

BookDetails

对象

XML字符串：

    <?xml version="1.0" encoding="UTF-8" standalone="no"?>
    <Books>
    <bookdetails>
    <bookId>20</bookId>
    <bookName>AAAA</bookName>
    <amount>35</amount>
    </bookdetails>
    </Books>

@XStreamAlias("bookDetails")
public class BookDetails {
@XStreamAlias("bookId")
private int bookId;
@XStreamAlias("bookName")
private String bookName;
@XStreamAlias("amount")
private int amount;
//getters and setters
}

@XStreamAlias("Books")
public class Books{

@XStreamAlias("bookDetails")
private List<BookDetails> bookDetails=new ArrayList<BookDetails>();
}

--------------unmarshall class method----

public BookDetails convertXml(String xml){

            xstream.processAnnotations(Books.class);
            xstream.processAnnotations(BookDetails.class);
            xstream.addImplicitCollection(Books.class,"BookDetails");
            return (BookDetails)xstream.fromXML(processOrderXML);
}

图书类：

    <?xml version="1.0" encoding="UTF-8" standalone="no"?>
    <Books>
    <bookdetails>
    <bookId>20</bookId>
    <bookName>AAAA</bookName>
    <amount>35</amount>
    </bookdetails>
    </Books>

@XStreamAlias("bookDetails")
public class BookDetails {
@XStreamAlias("bookId")
private int bookId;
@XStreamAlias("bookName")
private String bookName;
@XStreamAlias("amount")
private int amount;
//getters and setters
}

@XStreamAlias("Books")
public class Books{

@XStreamAlias("bookDetails")
private List<BookDetails> bookDetails=new ArrayList<BookDetails>();
}

--------------unmarshall class method----

public BookDetails convertXml(String xml){

            xstream.processAnnotations(Books.class);
            xstream.processAnnotations(BookDetails.class);
            xstream.addImplicitCollection(Books.class,"BookDetails");
            return (BookDetails)xstream.fromXML(processOrderXML);
}

@XStreamAlias（“书籍”）
公共课用书{
@XStreamAlias（“账簿明细”）
private List bookDetails=new ArrayList（）；
}
--------------解组类方法----
public BookDetails convertXml（字符串xml）{
processAnnotations（Books.class）；
processAnnotations（BookDetails.class）；
xstream.addImplicitCollection（Books.class，“BookDetails”）；
return（BookDetails）xstream.fromXML（processOrderXML）；
}

您的代码有几个问题

首先，您为

BookDetails

类使用了错误的别名。您将其声明为

@XStreamAlias（“bookDetails”）

，并且您的XML标记都是小写的

bookDetails

然后，您在定义隐式集合时使用了错误的字段名：它应该是

bookDetails

-作为列表字段名，而不是

bookDetails

最后，您必须解析完整的XML，然后才能从中获取

BookDetails

数据。而不是创建

BookDetails

实例

XStream

将为您提供

Books

实例

作为补充观察，您不必为具有确切名称（如XML标记）的字段添加别名

更正代码为：

@XStreamAlias("bookdetails")
public class BookDetails
{
    private int bookId;
    private String bookName;
    private int amount;
}

@XStreamAlias("Books")
public class Books
{
    private List<BookDetails> bookDetails = new ArrayList<BookDetails>();
}

public BookDetails convertXml(String xml)
{
    xstream.processAnnotations(Books.class);
    xstream.processAnnotations(BookDetails.class);
    xstream.addImplicitCollection(Books.class, "bookDetails");
    Books b = (Books) xstream.fromXML(xml);
    // and now you can return your BookDetails element (in case you want only first one)
    return b.bookDetails.get(0);
}

@XStreamAlias（“书籍详细信息”）
公共类书籍详情
{
私人书号；
私有字符串书名；
私人整数金额；
}
@XStreamAlias（“书籍”）
公共课用书
{
private List bookDetails=new ArrayList（）；
}
public BookDetails convertXml（字符串xml）
{
processAnnotations（Books.class）；
processAnnotations（BookDetails.class）；
xstream.addImplicitCollection（Books.class，“bookDetails”）；
Books b=（Books）xstream.fromXML（xml）；
//现在您可以返回BookDetails元素（以防只需要第一个元素）
返回b.bookDetails.get（0）；
}

请编辑您的问题，并将不起作用的代码与您遇到的确切错误一起包含在内。修改问题谢谢！我对此有了基本的了解。