Put the property Files in the classpath ,which contains key value pairs.
If the files are putin any other location thean classpath , give the path in tomcat startup.sh or classpath.sh next to the bootstrap jar listing.
Now access the files as accessing any other Bundle
ResourceBundle myResources = ResourceBundle.getBundle("propertfilename");
Then u can do a bundle.getValue() to the required key value pair.
Wednesday, November 12, 2008
XML -RPC
Apache XML-RPC is a Java implementation of XML-RPC, a popular protocol that uses XML over HTTP to implement remote procedure calls.
Steps
1.Client class
XmlRpcClientConfigImpl config = new XmlRpcClientConfigImpl();
config.setServerURL(new URL("http://127.0.0.1:8080/xmlrpc"));
XmlRpcClient client = new XmlRpcClient();
client.setConfig(config);
Object[] params = new Object[]{new Integer(33), new Integer(9)};
Integer result = (Integer) client.execute("Calculator.add", params);
2.Server Class
public class Calculator {
public int add(int i1, int i2) {
return i1 + i2;
}
public int subtract(int i1, int i2) {
return i1 - i2;
}
}
3. Create Propert File
Create a property file, which contains at least one property. The property name is arbitrary, and the property value is the fully qualified name of the Calculator class. For example, like that:
Calculator=org.apache.xmlrpc.demo.Calculator
The property file must be called XmlRpcServlet.properties, and it must be located in the package org.apache.xmlrpc.webserver. In other words, you would typically put it into the directory org/apache/xmlrpc/webserver and add it to your jar file.
4. Add Entry to Web.xml
<servlet>
<servlet-name> XmlRpcServlet</servlet-name>
<servlet-class>org.apache.xmlrpc.webserver.XmlRpcServlet</servlet-class>
<init-param>
<param-name>
enabledForExtensions</param-name>
<param-value>true</param-value>
<description> Sets, whether the servlet supports vendor extensions for XML-RPC. </description>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>XmlRpcServlet</servlet-name>
<url-pattern>/xmlrpc</url-pattern>
</servlet-mapping>
Reference http://ws.apache.org/xmlrpc/
Steps
1.Client class
XmlRpcClientConfigImpl config = new XmlRpcClientConfigImpl();
config.setServerURL(new URL("http://127.0.0.1:8080/xmlrpc"));
XmlRpcClient client = new XmlRpcClient();
client.setConfig(config);
Object[] params = new Object[]{new Integer(33), new Integer(9)};
Integer result = (Integer) client.execute("Calculator.add", params);
2.Server Class
public class Calculator {
public int add(int i1, int i2) {
return i1 + i2;
}
public int subtract(int i1, int i2) {
return i1 - i2;
}
}
3. Create Propert File
Create a property file, which contains at least one property. The property name is arbitrary, and the property value is the fully qualified name of the Calculator class. For example, like that:
Calculator=org.apache.xmlrpc.demo.Calculator
The property file must be called XmlRpcServlet.properties, and it must be located in the package org.apache.xmlrpc.webserver. In other words, you would typically put it into the directory org/apache/xmlrpc/webserver and add it to your jar file.
4. Add Entry to Web.xml
<servlet>
<servlet-name> XmlRpcServlet</servlet-name>
<servlet-class>org.apache.xmlrpc.webserver.XmlRpcServlet</servlet-class>
<init-param>
<param-name>
enabledForExtensions</param-name>
<param-value>true</param-value>
<description> Sets, whether the servlet supports vendor extensions for XML-RPC. </description>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>XmlRpcServlet</servlet-name>
<url-pattern>/xmlrpc</url-pattern>
</servlet-mapping>
Reference http://ws.apache.org/xmlrpc/
XStream - Convert XMl to Java & Vice Versa
XStream is a simple library to serialize objects to XML and back again.
XStream xstream = new XStream();
Person joe = new Person("Joe", "Walnes");
String xml = xstream.toXML(joe)
The resulting XML looks like this:
<code>
<person>
<firstname>Joe</firstname>
<lastname>Walnes</lastname>
<phone>
<code>123</code>
<number>1234-456</number>
</phone>
<fax>
<code>123</code>
<number>9999-999</number>
</fax>
</person></code>
</person></code>
Deserializing an object back from XML
Person newJoe = (Person)xstream.fromXML(xml);
Reference :http://xstream.codehaus.org/