无法在Tomcat中部署使用MYSQL的WebService

Question

这是我的webservice代码。

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
package customer;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.text.Format;
import java.text.SimpleDateFormat;
import java.util.Date;
import javax.jws.WebService;

/**
 *
 * @author Mihir
 */
public class Customer {
  String date1;
  Format formatter;
  Date date = new Date();
    public String feedback(String contactno,String comments,String ambience,String service,String 

food,String email,String custno,String custname,String storeno,String sno) 
    {

  formatter = new SimpleDateFormat("dd/MM/yy");
  date1 = formatter.format(date);
  Connection con = null;
  PreparedStatement prest;
  try{
  Class.forName("com.mysql.jdbc.Driver");
  con = DriverManager.getConnection("jdbc:mysql://localhost:3306/feedback","root","root");
  Statement stmt = con.createStatement();



      String sql = "INSERT INTO fb(contact_no, 

date,comments,ambience,service,food,email,cust_no,cust_name,store_no,s_no) " +
                    "VALUES ('"+contactno+"', 

'"+date1+"','"+comments+"','"+ambience+"','"+service+"','"+food+"','"+email+"','"+custno+"',

'"+custname+"','"+storeno+"','"+sno+"')";

stmt.execute(sql);


  }

  catch (Exception e){
  e.printStackTrace();
  }
        return null;
  }

}

现在这就是我正在努力做的。

>

内部WEB-INF

4.a.类别/客户/客户。班

4b带有jar的lib文件夹-I.webservices-api.jarII.webservices-extra.jarIII.webservices--extra-api.jarIV.webservices-rt.jarV.webservices-tools.jarVI.mysql-connector-java-5.1.18-bin.jar

4c.web.xml-

    <?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <listener>
        <listener-class>com.sun.xml.ws.transport.http.servlet.WSServletContextListener</listener-class>
    </listener>
    <servlet>
        <servlet-name>Customer</servlet-name>
        <servlet-class>com.sun.xml.ws.transport.http.servlet.WSServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Customer</servlet-name>
        <url-pattern>/feedback</url-pattern>
    </servlet-mapping>

</web-app>  

4d。太阳报。xml

<?xml version="1.0" encoding="UTF-8"?>
<endpoints version="2.0" xmlns="http://java.sun.com/xml/ns/jax-ws/ri/runtime">
  <endpoint implementation="ws.Customer" name="Adder" url-pattern="/feedback"/>
</endpoints>

Answer 1

我现在得到了它，它的工作我做了一些改变，使我的包名称彼此不同，并在我的类名之前指定了@Webservice。我已经测试了Webservice并在我的Android手机上实现了。谢谢大家的评论。