Java android应用程序的arraylist中的返回值不正确
此函数返回两个对象Java android应用程序的arraylist中的返回值不正确,java,android,arraylist,return,Java,Android,Arraylist,Return,此函数返回两个对象TMBase。customer@40fb1588,TMBase。customer@40fb1ad8,但我需要返回这些值emy,samera。我需要知道问题出在哪里 public ArrayList<TMBase.Customer> GetWorkerNameOfCustomerByID( int LineID) { ArrayList<TMBase.Customer> CustomerArr = new ArrayList<TMBa
TMBase。customer@40fb1588
,TMBase。customer@40fb1ad8
,但我需要返回这些值emy
,samera
。我需要知道问题出在哪里
public ArrayList<TMBase.Customer> GetWorkerNameOfCustomerByID( int LineID)
{
ArrayList<TMBase.Customer> CustomerArr = new ArrayList<TMBase.Customer>();
connecttodatabase qq = new connecttodatabase();
qq.dbconnect();
if (qq.con != null)
{
String result = "";
ResultSet rs = null;
try {
CallableStatement cstmt = null;
Statement st = qq.con.createStatement();
cstmt = qq.con.prepareCall("{CALL GetWorkerNameOfCustomerByID(?)}");
cstmt.setInt(1, LineID);
rs = cstmt.executeQuery();
ResultSetMetaData rsmd = rs.getMetaData();
while(rs.next())
{
TMBase.Customer CustomerObj=new TMBase.Customer();
CustomerObj.setWorkerName(rs.getString("WorkerName"));
CustomerArr.add(CustomerObj);
}
} catch (SQLException e) {
e.printStackTrace();
}
qq.closeConnection();
}
return CustomerArr;
}
您需要重写
Customer
对象的toString()
方法
我假设您正在使用ArrayAdapter
在ListView
中显示这些内容,该视图对数据集中的每个项目调用toString()
,以显示在屏幕上
通过重写toString()
,您可以返回希望在列表视图中显示的任何数据,例如:
public String toString(){
return getWorkerName();
}
public String toString(){
return getWorkerName();
}