Kyl*_*yle 4 java eclipse jsp hibernate servlets
当我使用我的搜索功能时,我收到http状态500错误.它声称
错误报告
HTTP状态500 - 内部服务器错误
输入例外报告
消息SearchServlet已由更新版本的Java Runtime(类文件版本53.0)编译,此版本的Java Runtime仅识别最高52.0的类文件版本(无法加载类[SearchServlet])
说明服务器遇到意外情况,无法满足请求.
异常java.lang.UnsupportedClassVersionError:SearchServlet已由更新版本的Java Runtime(类文件版本53.0)编译,此版本的Java Runtime仅识别最高为52.0的类文件版本(无法加载类[SearchServlet]) org.apache.catalina.loader.WebappClassLoaderBase.findClassInternal(WebappClassLoaderBase.java:2286)org.apache.catalina.loader.WebappClassLoaderBase.findClass(WebappClassLoaderBase.java:811)org.apache.catalina.loader.WebappClassLoaderBase.loadClass(WebappClassLoaderBase. java:1260)org.apache.catalina.loader.WebappClassLoaderBase.loadClass(WebappClassLoaderBase.java:1119)org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:488)org.apache.catalina.valves.ErrorReportValve. invoke(ErrorReportValve.java:81)org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:650)org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:342)org.apache.coyote. HTTP 11.Http11Processor.service(Http11Processor.java:803)org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)org.apache.coyote.AbstractProtocol $ ConnectionHandler.process(AbstractProtocol.java:790)org.apache. tomcat.util.net.NioEndpoint $ SocketProcessor.doRun(NioEndpoint.java:1459)org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source )java.util.concurrent.ThreadPoolExecutor $ Worker.run(Unknown Source)org.apache.tomcat.util.threads.TaskThread $ WrappingRunnable.run(TaskThread.java:61)java.lang.Thread.run(Unknown Source)
当我转到我的编译器时,我使用的是编译器合规级别9.我有jre 1.8,我认为这是最新的.当我尝试使用eclipse市场安装java 9时,它告诉我在http://download.eclipse.org/eclipse/updates/none上找不到存储库.
我相信鉴于所有这些我完全是最新的,但我怎么能够在数据库中搜索记录？
<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1" <title>Add contact to Phone Book</title>
<h1>Add contact to Phone Book</h1>
</head>
<body>
<form name="submitInfo" method="get" action="AddEntryServlet" First Name: <input type="text" name="firstName" />
<br/>
<br/>
Last Name: <input type="text" name="lastName" />
<br/>
<br/>
Area Code: <input type="number" name="areaCode" />
<br/>
<br/>
Phone Number: <input type="number" name="phoneNumber" />
<br/>
<br/>
<input type="submit" value="Submit" />
</form>
</body>
<h1>Search for Contact in Phone Book</h1>
<body>
<form name="searchInfo" method="get" action="SearchServlet">
Search First Name: <input type="text" name="searchFirstName" />
<br/>
<br/>
<input type="submit" name="action" value="search" />
</form>
</body>
</html>
搜索servlet代码
import java.io.IOException;
import java.io.PrintWriter;
import java.util.List;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.hibernate.Query; import org.hibernate.Session;
import entities.Tbphonebook; import util.HibernateUtil;
/** * Servlet implementation class SearchServlet */
@SuppressWarnings("deprecation")
@WebServlet("/SearchServlet")
public class SearchServlet extends HttpServlet { private static final long
serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public SearchServlet() {
super();
// TODO Auto-generated constructor stub
}
/** * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response) */
@SuppressWarnings({ "deprecation","unchecked" })
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
String firstName = request.getParameter("searchFirstName");
@SuppressWarnings("rawtypes")
Query query = Session.createQuery("from Tbphonebook where firstname = ?");
List<Tbphonebook> persons = (List<Tbphonebook>) query.getResultList();
session.getTransaction().commit();
session.close();
request.setAttribute("persons", persons);
request.getRequestDispatcher("searchResults.jsp").forward(request,response);
response.sendRedirect("searchResults.jsp");
}
结果jsp
%@ page language="java" contentType="text/html; charset=ISO-8859-1"
pageEncoding="ISO-8859-1"%> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html;
charset=ISO-8859-1"> <title>Insert title here</title> </head>
<h1>Search Results</h1>
<h1>List Books:</h1>
<table class="table table-boardered table-striped" id="searchresults">
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>Area Code</th>
<th>PhoneNumber</th>
</tr>
</table>
<script>
$(document).ready(function(){
$.getJSON("persons", function(data){
var persons_data = '';
$.each(data, function(key, value){
person_date += '<tr>';
person_data += '<td>'+value.Firstname+'</td>';
person_data += '<td>'+value.Lastname+'</td>';
person_data += '<td>'+value.Areacode+'</td>';
person_data += '<td>'+value.Phonenumber+'</td>';
person_date += '</tr>';
}
);
$('#searchresults').append(persons_data);
});
});
</script>
<%
String message = (String) request.getAttribute("message");
if(!(message == null)){
out.println(message);
}else{
message= "";
out.println(message);
}
%>
</html>
"当我使用我的编译器时,我使用的是编译器合规级别9." 版本52.0是Java 8; 我假设53.0是Java 9.您正在尝试在Java 8上运行为Java 9编译的代码,这会导致您获得错误.将合规性版本更改为Java 8.
要暂时更正问题，请执行以下操作：
Project menu> Properties> Java Compiler> Compiler compliance level>1.8
永久性修复可能涉及安装 JDK 9。
仅供参考 1.8 就是 Java 8 的名称。
我最近在尝试 C# 后回到 Java，并将 Eclipse Oxygen 安装到一个干净的系统上，该系统以前从未安装过 Java。这种默认的一切都与一个全新的安装Eclipse氧但不知何故或其他Eclipse不能让自己的参数匹配的已安装了JDK。这是我创建的第二个项目，也是我第二次遇到这种头痛。
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