【问题标题】:one to many relationship in java google ap engine caused error ?java谷歌应用引擎中的一对多关系导致错误?
【发布时间】:2012-07-14 16:57:12
【问题描述】:

我已经实现了一个系统来保存有关用户的信息。 所以我在我的课堂上有聚合,所以用户类有来自联系人类的列表......等等 在第一页“测试它只是为了通过电话号码注册用户”必须将用户保存在数据库中但是当我在 Google 应用程序引擎中部署我的项目时这会导致错误 500 服务器错误 >

另一个页面用于更新之前已经注册的现有用户,所以这里将添加到用户对象拥有的列表中。

用户类

@PersistenceCapable
public class User implements Serializable{
@PrimaryKey
@Persistent(valueStrategy = IdGeneratorStrategy.IDENTITY)
private Key key;

 @Persistent
public String userNumber;

 @Persistent (mappedBy = "userC")
 public List<Contact> UserContacts =new  ArrayList<Contact>();

 public void AddContact(String name,String number) {
     Contact C=new Contact();
     C.ContactName=name;
     C.PhoneNumber=number;
     this.UserContacts.add(C);   }
     }

联系班级

    @PersistenceCapable
public class Contact implements Serializable{
    @PrimaryKey
    @Persistent(valueStrategy = IdGeneratorStrategy.IDENTITY)
    private Key key;

     @Persistent
        public String ContactName;

     @Persistent
    public String PhoneNumber;  
             @Persistent
    public User userC; }

此页面导致用户测试注册将获得用户电话号码,注册应使用此号码创建新用户 测试页面

@SuppressWarnings("serial")

公共类测试扩展 HttpServlet {

public void doGet(HttpServletRequest req, HttpServletResponse resp)
        throws IOException {
        resp.setContentType("text/html");
        resp.getWriter().println("<html><body><form  method = \"POST\"     action=\"/signup\">" + "please enter ur number :"+"<h4><label>name :  <input name =  \"userphoneNUMBER\" type =  \"text \" size =  \"25 \" />  </label>"+"<p> <input type = \"submit\" value = \"Submit\" />"+ "</form></body></html>");

    }
}

本页取号创建用户

@SuppressWarnings("serial")
public class SignUP extends HttpServlet {
    public void doPost(HttpServletRequest req, HttpServletResponse resp)
            throws IOException {
        resp.setContentType("text/plain");
        String user_PhoneNumber=req.getParameter("userphoneNUMBER");
        User obj = new User();
    obj.userNumber=user_PhoneNumber;
    resp.getWriter().println(obj.userNumber );
        PersistenceManager pm = PMF.get().getPersistenceManager();

        try {
            pm.makePersistent(obj);
        } finally {
            pm.close();

        } }  }

此页面继续更新已经存在的用户对象的值

    @SuppressWarnings("serial")
public class Testinfo extends HttpServlet {
    public void doGet(HttpServletRequest req, HttpServletResponse resp)
            throws IOException {
        resp.setContentType("text/html");
        resp.getWriter().println("<html><body><form  method = \"POST\" action=\"/saveinfo\">" +
                "<center> <h2>please fill this form :</h2> </br> "+
"<h4><label> ContactName :  <input name =  \"ContactName\" type =  \"text \" size =  \"25 \" />  </label>"
+
"<p> <input type = \"submit\" value = \"Submit\" />"+
                "</form></body></html>");
    }

}

此页面用于保存导致错误的信息,并且不会在应用引擎中保存任何值

    @SuppressWarnings("serial")
public class SaveInfo extends HttpServlet {
    public void doPost(HttpServletRequest req, HttpServletResponse resp)
            throws IOException {
        resp.setContentType("text/plain");

        String ContactName = req.getParameter("ContactName");
        String ContactNumber = req.getParameter("ContactNumber");
                   PersistenceManager pm = PMF.get().getPersistenceManager();
        Query query = pm.newQuery("select from " + User.class.getName());
        List<User> list = (List<User>) query.execute();
        for (User obj : list) {


            if (obj.userNumber.equals("111")) {
                 pm.currentTransaction().begin();
                                     obj.AddContact(ContactName, ContactNumber);

                pm.makePersistent(obj);

                 pm.currentTransaction().commit(); }
            }
                          pm.close(); }  }

这个 111 用于测试,我之前输入的用户电话号码。

那么我该如何处理列表和聚合问题?

更新用户信息时出现此错误

Uncaught exception from servlet
javax.jdo.JDOUserException: Identifier expected at character 1 in "*" at         org.datanucleus.jdo.NucleusJDOHelper.getJDOExceptionForNucleusException(NucleusJDOHelper.java:375)
at org.datanucleus.jdo.JDOQuery.execute(JDOQuery.java:230)
at sstooree.SaveInfo.doPost(SaveInfo.java:44)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:637)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.mortbay.jetty.servlet.ServletHolder.handle(ServletHolder.java:511)
at org.mortbay.jetty.servlet.ServletHandler$CachedChain.doFilter(ServletHandler.java:1166)
at com.google.apphosting.utils.servlet.ParseBlobUploadFilter.doFilter(ParseBlobUploadFilter.java:102)
at org.mortbay.jetty.servlet.ServletHandler$CachedChain.doFilter(ServletHandler.java:1157)
at com.google.apphosting.runtime.jetty.SaveSessionFilter.doFilter(SaveSessionFilter.java:35)
at org.mortbay.jetty.servlet.ServletHandler$CachedChain.doFilter(ServletHandler.java:1157)
at com.google.apphosting.utils.servlet.TransactionCleanupFilter.doFilter(TransactionCleanupFilter.java:43)

【问题讨论】:

  • “服务器错误”显然会伴随着原因、堆栈跟踪等。将它们保密不会帮助人们向您建议问题出在哪里
  • 这个错误发生在 2012-07-15 14:16:11.786 /saveinfo 500 57ms 0kb Mozilla/5.0 (Windows NT 6.1) AppleWebKit/536.11 (KHTML, like Gecko) Chrome/20.0.1132.57 Safari /536.11 92.253.89.127 - - [15/Jul/2012:14:16:11 -0700] "POST /saveinfo HTTP/1.1" 500 0 "url" "Mozilla/5.0 (Windows NT 6.1) AppleWebKit/536.11 (KHTML , 像壁虎) Chrome/20.0.1132.57 Safari/536.11" "myurl" ms=58 cpu_ms=19 api_cpu_ms=0 cpm_usd=0.000613 instance=00c61b117c530c9a537b0089084cba4cb66e15 W 2012-07-15 /14:16:14:16
  • javax.jdo.JDOUserException:在 org.datanucleus.jdo.NucleusJDOHelper.getJDOExceptionForNucleusException(NucleusJDOHelper.java:375) 在 org.datanucleus.jdo.JDOQuery.execute 中的字符 1 处需要标识符(JDOQuery.java:230) 在 sstooree.SaveInfo.doPost(SaveInfo.java:44) 在 javax.servlet.http.HttpServlet.service(HttpServlet.java:637) 在 javax.servlet.http.HttpServlet.service(HttpServlet. java:717)
  • 所以您使用的查询是错误的......这不是完整的堆栈跟踪。真正的例外是嵌套的。所以把它放在你的帖子中而不是在 cmets 中,并格式化以便它的可读性。
  • 好的,但我认为问题是由于@Persistent(mappedBy = "otherclass")

标签: google-app-engine servlets persistence jdo datanucleus


【解决方案1】:

查询中的问题

改变这个

Query query = pm.newQuery("select from " + User.class.getName());

Query query = pm.newQuery("select * from " + User.class.getName());

您没有选择任何列。这会导致 sql 语法错误

【讨论】:

  • 这样做。这是 JDOQL,您的建议无效。
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