用于将值从 table 插入到另一个值的 Servlet 异常 ( java.lang.StackOverflowError )

Question

我正在开发一个在 jsp 中显示 table 的 servlet 应用程序，而不是在其他 table 中插入 table 的值（插入来自table 到另一个 table) 当我用调试器检查时，从第一个 table 读取一个值并将其插入到 scound table 没有问题但是当我点击插入按钮时过了一会儿，我也有这个错误（java.lang.WhosebugError），当我检查 mysql table 我明白了，我的价值比我插入它的时间要多，就像无限 loop.about 我正在读取数据的 table 我可以说（table 的名称 :class,id( type:int(primary),null:no,default:none,额外:AUTO_INCREMENT),名称(类型: varchar,null:yes,default:current_timestamp)) 和 table 我插入的值是(table name test,id(type:int(index),null:是的，默认值：null))

我想再说一遍，除了错误之外，我在 sql jsp 中单击了更多插入 table（大约 100 倍）

// servlet class
public class add_course extends HttpServlet {
 private dbutil dbutil;
 @Resource(name="jdbc/web_student_tracker")
 private DataSource dataSource;
 @Override
 public void init() throws ServletException {
  //dbutil= new dbutil(dataSource);
  super.init();
  try {
   dbutil=new dbutil(dataSource);
  }
  catch(Exception exc) {
  throw new ServletException(exc);
  }
 }
 protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
//  List<student> student;
//  try {
//   student = dbutil.getcourse();
//   request.setAttribute("select",student);
//   RequestDispatcher dispatcher = request.getRequestDispatcher("/course.jsp");
//  dispatcher.forward(request,response);
//  } catch (Exception e) {   // TODO Auto-generated catch block
//   e.printStackTrace();
//  }
  try {
   String thecommand=request.getParameter("command");
   if(thecommand==null) {
    thecommand="LIST";
   }
   switch(thecommand) {
   case"LIST":
    listcourse(request,response);
    break;
   case"insert":
    insertcourse(request,response);
    break;
   }
  }
  catch(Exception exc) {
   throw new ServletException(exc);
  }
  
 
 }
 private void insertcourse(HttpServletRequest request, HttpServletResponse response) throws Exception {
  int courseid = Integer.parseInt(request.getParameter("courseid"));
  student thestudent=new student(courseid);
  dbutil.insetcourse(thestudent);
  insertcourse(request,response);
   

 }
 private void listcourse(HttpServletRequest request, HttpServletResponse response) throws Exception {
  List<student> student=dbutil.getcourse();
 request.setAttribute("select",student);
 RequestDispatcher dispatcher = request.getRequestDispatcher("/course.jsp");
 dispatcher.forward(request,response);
 }

}

// db class
public List <student> getcourse() throws Exception{
 List<student> course=new ArrayList<>();
 Connection myConn = null;
 Statement myStmt = null;
 ResultSet myRs = null;
 try {
  
  myConn=dataSource.getConnection();
  String sql="select id from class";
  myStmt=myConn.createStatement();
  myRs=myStmt.executeQuery(sql);
  while (myRs.next()) {
 int id = myRs.getInt("id");
 student tempstudent = new student(id);

    course.add(tempstudent);
 }
 return course;
}
 finally {
  // close JDBC objects
  close(myConn, myStmt, myRs);
 }  
}

public void insetcourse(student thestudent)throws SQLException  {
 Connection myConn = null;
 PreparedStatement myStmt=null;
 /////////
 try {
  myConn = dataSource.getConnection();
  String sql="insert into test"+"(id)"+"value(?)";
  myStmt=myConn.prepareStatement(sql);

  myStmt.setInt(1,thestudent.getId());
  
  myStmt.execute();
  
 }
 finally {
  close(myConn,myStmt,null);
 }
}
}

//jsp page 
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
        <%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
    
<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<table>
<c:forEach var="tempstudent" items="${select}">
<c:url var="insert" value="add_course">
  <c:param name="command" value="insert"/>
    <c:param name="courseid" value="${tempstudent.id}"/>
   </c:url> 
<tr>
<td>${tempstudent.id}</td>
<td> 
  <a href="${insert}" 
       onclick="if (!(confirm('Are you sure you want to insert this student?'))) return false"> 
      
      insert</a>  
       </td> 
</tr>

</c:forEach>




</table>

</body>
</html>

Answer 1

如果要在 table 中插入多条记录，为什么不使用 for loop 而不是递归方法。

private void insertcourse(HttpServletRequest request, HttpServletResponse response) throws Exception {

    // Assuming you are getting multiple courseid in your request, 
    Object[] courseid = request.getParameter("courseid");
    for(Object obj : courseid)
    {
          student thestudent=new student(courseid);
           dbutil.insetcourse(thestudent);
     }
    request.setAttribute("message", "Records loaded successfully");
 RequestDispatcher dispatcher = request.getRequestDispatcher("/course.jsp");
dispatcher.forward(request,response);


}

插入完成后，将结果发送给 forntend，以便通知用户，

如果您只想插入一条记录，只需调用 insert 一次。

但是查看您的 jsp 代码，您正在为每条记录生成 insert hyper link，因此您一次只能插入一条记录，在这种情况下，您可以按照下面的代码。

  int courseid = Integer.parseInt(request.getParameter("courseid"));
  student thestudent=new student(courseid);
  dbutil.insetcourse(thestudent);
  request.setAttribute("message", "Records loaded successfully");
  RequestDispatcher dispatcher = request.getRequestDispatcher("/course.jsp");
  dispatcher.forward(request,response);

用于将值从 table 插入到另一个值的 Servlet 异常 ( java.lang.StackOverflowError )

Servlet Exception( java.lang.StackOverflowError ) for inserting a value from table to other one

java

jsp

servlets