Question

我正在创建一个查询数据库的Web服务，并返回数据库中对象的列表。我收到错误：NullPointerAccess：变量“varname”在此位置只能为null。无论我把变量放在哪里，我都会得到相同的警告。无论我放在变量中，它都返回null。以下是它出现的方法：

    public List<Contacts> getUsers()
  {     String test = null;
        String username = "root";
        String password = "ticket";
        String tablename = "users";
        String fieldname = "*";
        String query = "SELECT " + fieldname + " FROM " + "android." + tablename + ";";

        Contacts cont = new Contacts();
        List<Contacts> lstc = null;

        /* this chnk of code can appear more or less verbatim */
        /* in your database apps (including JSPs) */
        String url = "jdbc:mysql://"my IP address":3306/android";
        try{
        Class.forName("com.mysql.jdbc.Driver").newInstance();
        Connection con = DriverManager.getConnection(url, username, password);
        Statement stmt = con.createStatement();
        ResultSet rs = stmt.executeQuery(query);


        while (rs.next()){
            if (!(rs.getString("username") == null)){
             cont.setUsername(rs.getString("username"));
            }
            if (!(rs.getString("location") == null)){
             cont.setLocation(rs.getString("location"));
            }
            if (!(rs.getString("updated_at") == null)){
             cont.setUpdated_at(rs.getDate("updated_at"));
            }
            if (!(rs.getString("idUsers") == null)){
             cont.setId(rs.getInt("idUsers"));
            }
        }

        rs.close();
        stmt.close();
        con.close();
        }catch(Exception e){
            test = e.toString();
        }
         lstc.add(cont); //THIS IS THE VARIABLE THAT IS GIVING THE WARNINGS
        test = "blahbadslf";
        return lstc;
        }

获取警告的变量是我的List lstc变量。当我尝试将一个Object添加到List时，我得到了错误。任何帮助将非常感激。以下是Class：以防万一：

public class Contacts{
private int id;
private String username;
private String location;
private Date updated_at;

public void setId(int id) {
    this.id = id;
}
public int getId() {
    return id;
}
public void setUsername(String username) {
    this.username = username;
}
public String getUsername() {
    return username;
}
public void setLocation(String location) {
    this.location = location;
}
public String getLocation() {
    return location;
}
public void setUpdated_at(Date updated_at) {
    this.updated_at = updated_at;
}
public Date getUpdated_at() {
    return updated_at;
}

}

谢谢！

Answer 1

你收到的消息是因为你有：

List<Contacts> lstc = null;

你的意思是：

List<Contacts> lstc = new ArrayList<Contacts>();

顺便说一句：

你有第二个错误：

    Contacts cont = new Contacts();  // you only create one of these

更好：

    while (rs.next()){
        Contacts cont = new Contacts();  // move to within the loop so one is created for every row in the result set

Answer 2

有这个：

List<Contacts> lstc = null;

然后没有任何对象分配给lstc。失败：（

考虑new ArrayList<Contacts>()某处意味着什么（例如在声明中）。

Answer 3

您没有实例化列表。当您想要访问它时，该列表仍为空。

所以你应该使用它：

List<Contacts> lstc = new List<Contacts>();

错误：NullPointerAccess：变量“”在此位置只能为null

3 个答案: