Java，简单XML解析列表

Question

如何在 Java 中解析此列表？我有 List<Image>，其中 returns 来自服务器，但我无法获得单个项目。

<images>
   <image>
       uploads/posts/2008-10/1225141003_1-21.jpg
   </image>
    <image>
       uploads/posts/2008-10/1225141003_1-22.jpg
   </image>
</images>

@Root(name = "Images") public class Images {
@ElementList(required=false, inline = true)
private List<Image> imageList;

public List<Image> getImageList() {
    return imageList;
} 
}

@Root(name = "image") public class Image {

//Some code.......

}

Answer 1

试试这个：

String inputStreamToString(InputStream is) {
    String line = "";
    String total = "";
    BufferedReader rd = new BufferedReader(new InputStreamReader(is));

    try {
        while ((line = rd.readLine()) != null) {
            total +=line;
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
    return total;
}

编辑

如果你能得到 xml:

String responseXML = inputStreamToString(yourXMLResponseFromServer.getEntity().getContent());

Answer 2

为了解释 XML 文件，我一直使用 org.w3c.dom.Document 界面，它提供了一个类似于 Javascript 的文档修改。查看oracle网站上的Documentation！

Answer 3

圣诞快乐

使用 DOM 和 Xpath

1 解析你的字符串

String xml="<my_xml/>";
DocumentBuilderFactory builderFactory =DocumentBuilderFactory.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document document = builder.parse(new InputSource(new StringReader(xml)));

2 使用 xpath

XPath xPath = XPathFactory.newInstance().newXPath();
String expression="/images/image";
XPathExpression expr = xpath.compile(expression) ; 
NodeList nodes  = (NodeList) expr.evaluate(document,  XPathConstants.NODESET);

3次迭代

for (int k = 0; k < nodes.getLength(); k++)
    {
    Node nodeSegment = nodes.item(k);
    if (nodeSegment.getNodeType() == Node.ELEMENT_NODE)
        {
        Element eElement = (Element) nodeSegment;
        System.out.println("TEXT CONTENT="+eElement.getTextContent());

备选方案：

如果您知道自己有 1 或 2 张图像：

expression="/images/image[1]"; // first one
String value = xPath.evaluate(expression, document);
System.out.println("EVALUATE:"+value);

Answer 4

我是这样解决这个问题的：

@Root(name = "images") 
public class Images {

     @ElementList(entry = "image", required=false, inline = true)
     private List<String> imageList;

     public List<String> getImageList() {
         return imageList;
     }
}

Java，简单XML解析列表

Java, Simple XML parsing list

java

xml

android

xml-parsing

simple-framework