将xpath表达式结果转换为json

时间:2014-12-17 10:21:19

标签: java xml json

我想将dom nodelist转换为json数组并将结果发送到rest客户端:

xml的每个节点代表以下内容:

<A NAME="x" COUNT="y">
  <B KEY="z1" VALUE="z2"/>
  <B KEY="z3" VALUE="z4"/>
</A>

我希望我能为输出提供一个对象数组,其中每个对象如下所示:

{"NAME":"x", 
 "COUNT":"y", 
 "B": [ {"KEY": "z1,  VALUE:"z2"},
        {"KEY":"z3", VALUE:"z4"} ]
}

我尝试使用GSON库:

package com.a;


import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.IOException;
import java.nio.charset.Charset;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;

import com.google.gson.Gson;
import com.google.gson.GsonBuilder;

public class Test {
    private static final String XPATH = "/A/B";

    public static void main(String[] args) throws ParserConfigurationException, SAXException, IOException, XPathExpressionException {
        File f = new File("C:/Users/abc/Desktop/a.xml");
        DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = null;
        builder = builderFactory.newDocumentBuilder();
        Document xmlDocument = builder.parse(f);
        XPath xPath =  XPathFactory.newInstance().newXPath();

        NodeList nodeList = (NodeList) xPath.compile(XPATH).evaluate(xmlDocument, XPathConstants.NODESET);
        Gson gson = new GsonBuilder().setPrettyPrinting().create();

        String jsonOutput = gson.toJson(nodeList);
        System.out.println(jsonOutput);




    }

}

但是我收到了错误

  

线程中的异常&#34; main&#34;

中的java.lang.StackOverflowError      

java.lang.StringBuffer.append(StringBuffer.java:224)

     

java.io.StringWriter.write(StringWriter.java:84)

     

go.google.gson.stream.JsonWriter.newline(JsonWriter.java:569)

     

com.google.gson.stream.JsonWriter.beforeName(JsonWriter.java:586)

如何修复此代码?

因为可以将整个xml转换为json(Quickest way to convert XML to JSON in Java

我假设可以将dom节点转换为json。这有什么不对?

1 个答案:

答案 0 :(得分:3)

出了什么问题?您只是不使用链接(Quickest way to convert XML to JSON in Java

中描述的lib

Gson将使用java反射从任何对象生成json字符串。从DOM Document(或节点),即使它没有以StackOverflowError结束,它也不会产生您期望的结果。这是您的XML的结果:

{"fNamespacesEnabled":false,"mutationEvents":false,"actualEncoding":"UTF-8","standalone":false,"fDocumentURI":"...a.xml","changes":0,"allowGrammarAccess":false,"errorChecking":true,"ancestorChecking":true,"xmlVersionChanged":false,"documentNumber":0,"nodeCounter":0,"xml11Version":false,"flags":6}

实际上,如果在DOM文档上调用了任何方法(例如:getDocumentElement),gson.toJson最终会出现StackOverflowError。

正如您在链接中所看到的,可以在此处找到可以完成工作的jar:http://mvnrepository.com/artifact/org.json/json

这意味着您将使用XPath提取的节点重新转换为字符串。 你可以这样做:

private static String toString(Node n) throws TransformerFactoryConfigurationError, TransformerException {
    Transformer transformer = TransformerFactory.newInstance().newTransformer();
    StreamResult result = new StreamResult(new StringWriter());
    DOMSource source = new DOMSource(n);
    transformer.transform(source, result);
    return result.getWriter().toString();
}

尽管如此,您所要做的就是遍历nodeList,将其转换为字符串,然后将其转换为json

for (int i = 0; i < nodeList.getLength(); i++) {
    Node n = nodeList.item(i);
    JSONObject xmlJSONObj = XML.toJSONObject(toString(n));
    String jsonPrettyPrintString = xmlJSONObj.toString(1);
    System.out.println(jsonPrettyPrintString);
}