Question

以下是我的代码。我正在查询user_group表并将结果存入List。我想迭代List。但是得到如下例外。

    List<Group> list= empDAO.getStudentList();
    for(Group o :list){
        System.out.println("NAME :"+ o.getFirstName());
    }

这是我的DAO方法

public List<Group> getStudentList() {
    System.out.println("INSIDE DAO");
    List<Group> groups = new ArrayList<Group>();

    List<Group> rows = jdbcTemplate.queryForList("select * from user_group");

    return rows;
}

Group.class

public class Group {
    public Group() {
        // TODO Auto-generated constructor stub
    }

    private String firstName;

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }
}

我收到以下异常

Aug 3, 2012 9:22:47 PM org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet mvc-dispatcher threw exception
java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to com.common.form.Group
        at com.common.controller.HelloWorldController.helloWorld(HelloWorldController.java:28)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
        at java.lang.reflect.Method.invoke(Method.java:597)
        at org.springframework.web.bind.annotation.support.HandlerMethodInvoker.doInvokeMethod(HandlerMethodInvoker.java:421)
        at org.springframework.web.bind.annotation.support.HandlerMethodInvoker.invokeHandlerMethod(HandlerMethodInvoker.java:136)
        at org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter.invokeHandlerMethod(AnnotationMethodHandlerAdapter.java:326)
        at org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter.handle(AnnotationMethodHandlerAdapter.java:313)
        at org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:875)
        at org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:807)
        at org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:571)
        at org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:501)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:617)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
        at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
        at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
        at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
        at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
        at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
        at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
        at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
        at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
        at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:857)
        at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
        at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
        at java.lang.Thread.run(Thread.java:619)

有人可以告诉我们如何修复此错误吗？

Answer 1

行jdbcTemplate.queryForList("select * from user_group");返回HashMap列表，而不是Group项列表（您的IDE可能在该行上显示警告）。

您可能希望阅读有关jdbcTemplate的Spring文档，并且您可能希望使用RowMapper将每一行转换为Group对象。

我还认为您希望将jdbcTemplate.query(String sql, RowMapper rm)方法用于您的用例，请查看javadoc。

Answer 2

jdbcTemplate.queryForList未返回List<Group>，它不知道Group是什么。它返回一个List<Map<String,Object>>，一个行列表，每一行都是一个从列名称到其值的映射。

这样做：

for (Map<String,Object> m : jdbcTemplate.queryForList("select * from user_group")) {
    for (Map.Entry<String,Object> e : m.entrySet()) {
        String columnName = e.key;
        Object columnValue = e.value;
        ...build a Group somehow?...
    }
}

泛型 - 迭代ArrayList时的ClassCastException

2 个答案: