Servlet Exception(java.lang.StackOverflowError,用于将表中的值插入另一个]

问题描述 投票:0回答:1

我正在开发一个servlet应用程序,该应用程序在jsp中显示该表,然后将其插入其他表中(将该表的ID插入另一个表中),就像我用调试器检查的那样,从中读取值没有问题第一张表并将其插入到scound表中,但是当我在一段时间后单击插入按钮时,出现此错误(java.lang.StackOverflowError),当我查看mysql表时,该值也比插入它的时间还多就像无限循环一样。关于我正在读取数据的表,我可以说(表名:class,id(type:int(primary),null:no,default:none,Extra:AUTO_INCREMENT),name(type :varchar,null:yes,default:current_timestamp)),而我要插入的表是(表名test,id(type:int(index),null:yes,default:null))

我想再说一遍,除了错误之外,我还有更多插入,我可以在sql表中单击jsp(例如多出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>
java jsp servlets
1个回答
0
投票

如果要在表中插入多个记录,为什么不使用for loop而不是递归方法。

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