Question

我正在尝试创建一个查询我的oracle数据库的Web应用程序，并从表中返回一个菜单标题列表，但是当我导航到该页面时，它会抛出下面显示的区域。

type Exception report

message

description The server encountered an internal error () that prevented it from fulfilling this request.

exception

org.apache.jasper.JasperException: java.lang.NullPointerException
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:502)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:430)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260)
javax.servlet.http.HttpServlet.service(HttpServlet.java:717)

root cause

java.lang.NullPointerException
oracle.DatabaseConnector.selectAllMenus(DatabaseConnector.java:63)
org.apache.jsp.menus_jsp._jspService(menus_jsp.java:113)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:388)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260)
javax.servlet.http.HttpServlet.service(HttpServlet.java:717)

note The full stack trace of the root cause is available in the Apache Tomcat/6.0.35 logs.

以下是我的java类中查询数据库的代码中令人讨厌的部分。但我不确定为什么会导致上述错误？有人可以看看是否有问题吗？

public MenuList selectAllMenus(String ordering) {
    MenuList list = null;

    try {
        String strQuery = "SELECT title"
                + " FROM menus"
                + " ORDER BY " + ordering;
        PreparedStatement stmt = conn.prepareStatement(strQuery);

        results = stmt.executeQuery();

        list = new MenuList(results);
    } catch (SQLException e) {
        e.printStackTrace();
    }

    return list;
}

这是日志文件中的内容：

28-Feb-2012 14:40:57 org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet jsp threw exception
java.lang.NullPointerException
at oracle.DatabaseConnector.selectAllMenus(DatabaseConnector.java:63)
at org.apache.jsp.menus_jsp._jspService(menus_jsp.java:113)
at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:388)
at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313)
at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:293)
at org.apache.coyote.http11.Http11AprProcessor.process(Http11AprProcessor.java:877)
at org.apache.coyote.http11.Http11AprProtocol$Http11ConnectionHandler.process(Http11AprProtocol.java:594)
at org.apache.tomcat.util.net.AprEndpoint$Worker.run(AprEndpoint.java:1675)
at java.lang.Thread.run(Thread.java:662)

Answer 1

希望您使用的是MySQL DB ..

确保将正确的MySQL Java Connector添加为外部库，并将Tomcat添加为外部服务器（在Eclipse的项目构建路径中）。

（您还可以将$ TOMCAT_HOME / $ CATALINA_HOME的lib文件夹中的所有库复制到构建路径。

这应解决此问题。

感谢。

Answer 2

您可以查看日志：Apache Tomcat/6.0.35 logs

但错误消息本身非常清楚：数据库连接为空，可能是您的连接字符串/参数不正确。

Answer 3

我认为conn已打开。

为什么不打印错误消息，而不是打印堆栈跟踪？

Answer 4

行：

PreparedStatement stmt = conn.prepareStatement(strQuery);

“conn”未在任何地方创建

尝试这样的方法来创建连接：

public class MyConnection {
Connection conn;
String username;
String password;
public Connection getConnection(){
    try {
        String url = "jdbc:mysql://localhost/yourDBName?";
        Class.forName("com.mysql.jdbc.Driver");
        try {
            conn = (Connection) DriverManager.getConnection(url,username,password);
        } catch (Exception e) {
        System.out.println(e);
        }

    } catch (SQLException e) {
        System.out.println(e);
    }
    return conn;
}

}

我的网络应用程序中的Null.Pointer.Exception

4 个答案: