使用Retrofit反序列化带有文本和子标签的XML标记

我正在使用Retrofit和SimpleXML来解析来自某些公共API的XML响应.我一直在很好地处理所有内容,直到我偶然发现包含自由文本和子标记的XML标记 - 如以下示例所示:

   Some free-style text
   Even more text!

为了尝试使用Simple-XML注释进行反序列化,我已经采用了两种方式.请记住,基本上'a'是一个条目标签列表:

首先:

@ElementList(entry = "a", inline = true, required = false) List aList;

public static class A {
    @Text(required = false) protected String a;
}

@ElementList(entry = "a", inline = true, required = false) List aList;

// First, declare the 'a' tags in the root class hosting them (this is pretty standard):
@ElementList(entry = "a", inline = true) List aList;

// Second, create and apply a custom converter as described:
@Root
@Convert(CustomConverter.class)
public class A {
    String content = "";

    public String getContent() {
        return content;
    }
}

public class CustomConverter implements Converter {

    @Override
    public A read(InputNode node) throws Exception {
        A a = new A();

        String value = node.getValue();
        if (value != null) {
            a.content = value;
        }

        InputNode nodeB = node.getNext();
        if (nodeB != null) {
            value = nodeB.getValue();
            if (value != null) {
                a.content += value;
            }
        }

        return a;
    }

    @Override
    public void write(OutputNode node, A value) throws Exception {
        // N/A
    }
}

If you can't explain it 
simply
, you don't 

   understand it 
   well enough.

 -- Albert Einstein

HTML标签与整体解析XML无关:我真正需要的是将'a'下的内容反序列化为名为'A'的类下的纯文本.所以这是我的(递归)转换器:

@Override
public A read(InputNode node) throws Exception {
    final StringBuilder sb = new StringBuilder(1024);
    concatNodesTree(sb, node);

    A a = new A();
    a.content = sb.toString();
    return a;
}

private void concatNodesTree(StringBuilder sb, InputNode root) throws Exception {

    if (root.isElement()) {
        sb.append("<").append(root.getName()).append(">");
    }

    String value = root.getValue();
    if (value != null) {
        sb.append(value);
    }

    InputNode node = root.getNext();
    while (node != null) {
        concatNodesTree(sb, node);

        // Each time a sub-tree is 'over', getValue() will deserialize the potentially upcoming free-text
        value = root.getValue();
        if (value != null) {
            sb.append(value);
        }
        node = root.getNext();
    }

    if (root.isElement()) {
        sb.append("");
    }
}

注意:在此解决方案中,'a'标记也将被解析为最终字符串.为避免这样做,可以为根节点发出特殊情况concatNodesTree()方法.