Question

为什么我总是得到numberformatexception？我尝试将数据库中的值检索到我的jsp。我想在我的jsp中查看我的连接表。

我有2个实体

项目类

public class Item {

@Id
@Column(name="p_id")
private String pid;

@Column(name="p_name")
private String pname;


@Column(name="catId")
private String catid;

public String getPid() {
    return pid;
}

public void setPid(String pid) {
    this.pid = pid;
}

public String getPname() {
    return pname;
}

public void setPname(String pname) {
    this.pname = pname;
}

public String getCatid() {
    return catid;
}

public void setCatid(String catid) {
    this.catid = catid;
}}

类别类

public class Category {

@Id
@Column(name="c_id")
private int cid;

@Column(name="catId")
private String catid;

@Column(name="c_name")
private String cname;

@Column(name="c_brand")
private String cbrand;

public int getCid() {
    return cid;
}

public void setCid(int cid) {
    this.cid = cid;
}

public String getCname() {
    return cname;
}

public void setCname(String cname) {
    this.cname = cname;
}

public String getCbrand() {
    return cbrand;
}

public void setCbrand(String cbrand) {
    this.cbrand = cbrand;
}

public String getCatid() {
    return catid;
}

public void setCatid(String catid) {
    this.catid = catid;
}}

我的Item类中的btw catId是Category的外键。

我尝试在我的DaoImpl中创建hql并加入两个表，看起来像这样

public List<Item> getItem() {
    EntityManager entityManager = transactionManager.getEntityManagerFactory().createEntityManager();
    Query sqlQuery = entityManager.createQuery("SELECT i.pname,c.cname,c.cbrand FROM Item i,Category c where i.catid= c.catid");

    List<Item>iList = sqlQuery.getResultList();
    if (entityManager.isOpen())
        entityManager.close();

    return iList;
}

@Controller

    @RequestMapping(value = "/list")
public String result(HttpServletRequest request, ModelMap map) {

    List<Item> iList = catService.getItem();
    map.addAttribute("iList", iList);

    return "new/list";
}

@my view jsp看起来像这样

<c:if test="${not empty iList}">
        <table border="1">
        <tr>

            <th>Product name</th>
            <th>Category</th>
            <th>Brand</th>

            <c:forEach var="p" items="${iList}">

                    <td>${p.pname}</td>
                    <td>${p.cname}</td><td>${p.cbrand}</td>

            </c:forEach>
            </tr>
        </table>
</c:if>

我总是得到这个错误：

Stacktrace:] with root cause
java.lang.NumberFormatException: For input string: "pname"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at javax.el.ArrayELResolver.coerce(ArrayELResolver.java:158)
    at javax.el.ArrayELResolver.getValue(ArrayELResolver.java:45)
    at org.apache.jasper.el.JasperELResolver.getValue(JasperELResolver.java:104)
    at org.apache.el.parser.AstValue.getValue(AstValue.java:183)
    at org.apache.el.ValueExpressionImpl.getValue(ValueExpressionImpl.java:184)
    at org.apache.jasper.runtime.PageContextImpl.proprietaryEvaluate(PageContextImpl.java:967)
    at org.apache.jsp.pages.new_.list_jsp._jspx_meth_c_005fforEach_005f0(list_jsp.java:243)
    at org.apache.jsp.pages.new_.list_jsp._jspx_meth_c_005fif_005f0(list_jsp.java:204)
    at org.apache.jsp.pages.new_.list_jsp._jspService(list_jsp.java:98)
    at org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:727)
    at org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432)
    at org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:395)
    at org.apache.jasper.servlet.JspServlet.service(JspServlet.java:339)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:727)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:303)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
    at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:241)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
    at org.apache.catalina.core.ApplicationDispatcher.invoke(ApplicationDispatcher.java:748)
    at org.apache.catalina.core.ApplicationDispatcher.processRequest(ApplicationDispatcher.java:486)
    at org.apache.catalina.core.ApplicationDispatcher.doForward(ApplicationDispatcher.java:411)
    at org.apache.catalina.core.ApplicationDispatcher.forward(ApplicationDispatcher.java:338)
    at org.springframework.web.servlet.view.InternalResourceView.renderMergedOutputModel(InternalResourceView.java:238)
    at org.springframework.web.servlet.view.AbstractView.render(AbstractView.java:263)
    at org.springframework.web.servlet.DispatcherServlet.render(DispatcherServlet.java:1208)
    at org.springframework.web.servlet.DispatcherServlet.processDispatchResult(DispatcherServlet.java:992)
    at org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:939)
    at org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:856)
    at org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:936)
    at org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:827)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:620)
    at org.springframework.web.servlet.FrameworkServlet.service(FrameworkServlet.java:812)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:727)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:303)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
    at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:241)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:220)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:122)
    at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:503)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:170)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:103)
    at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:950)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:116)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:421)
    at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1070)
    at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:611)
    at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:314)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
    at java.lang.Thread.run(Unknown Source)

有人可以帮我解决这个问题吗？提前谢谢！

Answer 1

我不知道这是一个好的做法还是对，但这段代码有效......

public List<Customer> getView( int startPage, int maxResults) {

    EntityManager entityManager = transactionManager.getEntityManagerFactory().createEntityManager();
    Session session = entityManager.unwrap(Session.class);
    String sql = "SELECT customer.firstname, customer.lastname, customer.address, account.ano, account.bank FROM customer, account WHERE customer.cid = account.cid";
            SQLQuery query = session.createSQLQuery(sql);

            query.setResultTransformer(Criteria.ALIAS_TO_ENTITY_MAP);

            List<Customer> vList = query.list();
            if (entityManager.isOpen())
                entityManager.close();
                // session.close();

            return vList;


}

java.lang.NumberFormatException：对于输入字符串：“pname”

1 个答案: