在xml下面阅读的最简单，最好的方法是什么？

Question

I am new to XML.

Please let me know easy and best way to read xml below in java. In my xml, I have queries as root and query as child element in it.

<queries>
    <query id="getUserByName">
        select * from users where name=?
    </query>
    <query id="getUserByEmail">
        select * from users where email=?
    </query>
</queries>

I will pass query id, based on that we need to fetch corresponding query. Please help me with code for better understanding.

Answer 1

With XPath, it's simple.

import java.io.ByteArrayInputStream;
import java.io.InputStream;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathFactory;
import org.xml.sax.InputSource;

public class Test {

  public static final String xml = 
    "<queries>"
    + "  <query id=\"getUserByName\">"
    + "    select * from users where name=?"
    + "  </query>"
    + "  <query id=\"getUserByEmail\">"
    + "    select * from users where email=?" 
    + "  </query>"
    + "</queries>"; 


  public static void main(String[] args) throws Exception {
    System.out.println(getQuery("getUserByName"));
    System.out.println(getQuery("getUserByEmail"));

  }

  public static String getQuery (String id) throws Exception {
    InputStream is = new ByteArrayInputStream(xml.getBytes("UTF8"));
    InputSource inputSource = new InputSource(is);
    XPath xpath = XPathFactory.newInstance().newXPath();
    return xpath.evaluate("/queries/query[@id='" + id +"']", inputSource);
  }
}

Answer 2

A very easy code to implement would be JAXB parser. Personally I love this one as it establishes everything using simple annotations.

Steps.

1. Create a couple of bean classes with the structure of your xml. In your case Queries class containing List<Query> . Define Query to contain a string variable. If you take the time to go through the annotations, I'm sure you can do this even with a single bean class but with multiple annotations.

2. Pass your string of XML to a JAXB context of Queries class and you are done.

3. You'll get one Java object for each Query tag. Once you get the bean class, manipulation becomes easy.

Ref:

JAXB Hello World Example

JAXB Tutorial

Answer 3

A good solution would be to load these queries to a map first and access it later based on the map. To load queries to a map you could do something like:

Map<String, String> queriesMap = new HashMap<String, String>();
DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance();

DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder();
ByteArrayInputStream inputStream = new ByteArrayInputStream("<queries>    <query id=\"getUserByName\">        select * from users where name=?    </query>    <query id=\"getUserByEmail\">        select * from users where email=?    </query></queries>".getBytes());
// you could use something like: new FileInputStream("queries.xml");

Document doc = documentBuilder.parse(inputStream);

// get queries elements
NodeList queriesNodes = doc.getElementsByTagName("queries");

// iterate over it
for (int i = 0; i < queriesNodes.getLength(); i++) {

    // get queries element
    Node node = queriesNodes.item(i);

    // get query elements (theoretically)
    NodeList queryNodes = node.getChildNodes();
    for (int j = 0; j < queryNodes.getLength(); j++) {
        Node queryNode = queryNodes.item(j);

        // if not element just skip to next one (in case of text nodes for the white spaces)
        if (!(queryNode.getNodeType() == Node.ELEMENT_NODE)) {
            continue;
        }
        // get query
        Node idAttr = queryNode.getAttributes().getNamedItem("id");

        if (idAttr != null) {
            queriesMap.put(idAttr.getTextContent(), StringUtils.trim(queryNode.getTextContent()));
        }
    }
}

System.out.println(queriesMap);