1

我有一个看起来像这样的 ArrayList：

[
1 2011-05-10  1  22.0, 
2 2011-05-10  2  5555.0, 
3 2011-05-11  3  123.0, 
4 2011-05-11  2  212.0, 
5 2011-05-30  1  3000.0, 
6 2011-05-30  1  30.0, 
7 2011-06-06  1  307.0, 
8 2011-06-06  1  307.0, 
9 2011-06-06  1  307.0, 
10 2011-06-08  2  3070.0, 
11 2011-06-03  2  356.0, 
12 2011-05-10  2  100.0, 
13 2011-05-30  1  3500.0, 
14 2011-05-10  3  1000.0, 
15 2011-05-10  3  1000.0, 
16 2011-05-07  1  5000.0, 
17 2011-05-07  4  500.0, 
18 2011-08-07  3  1500.0, 
19 2011-08-08  6  11500.0, 
20 2011-08-08  4  11500.0, 
21 2011-08-08  7  11500.0, 
22 2011-06-07  8  3000.0]

这是我如何得到这个数组列表的代码：

@Override
    public ArrayList<Expenses> getExpenses() {
        ArrayList<Expenses> expenses = new ArrayList<Expenses>();
        try {
            Statement stmt = myConnection.createStatement();
            ResultSet result = stmt.executeQuery("SELECT * FROM expenses");
            while(result.next()){

                Expenses expense = new Expenses();
                expense.setNum(result.getInt(1));
                expense.setPayment(result.getString(2));
                expense.setReceiver(result.getInt(3));
                expense.setValue(result.getDouble(4));

                expenses.add(expense);

                }
        }
            catch (SQLException e){
                 System.out.println(e.getMessage());
             }
        return expenses;
    }

但我想得到一个数组列表，这样数组的每个元素都不是表格的行（我现在拥有的），但表格的每个单独元素都应该是数组的元素（[1, 2011-05 -10, 1, 22.0, 2, 2011-05-10, 2, 5555.0, 3, 2011-05-11, 3, 123.0,]. 有人能帮我吗？

4

3 回答 3

6

您可以添加到不同类型的 ArrayList 元素的唯一方法是将它们视为一般对象。但是，您已经拥有的代码要好得多。

@Override
public ArrayList<Object> getExpenses() {
    ArrayList<Object> expenses = new ArrayList<Object>();
    try {
        Statement stmt = myConnection.createStatement();
        ResultSet result = stmt.executeQuery("SELECT * FROM expenses");
        while(result.next()) {

            expenses.add(new Integer(result.getInt(1)));

            expenses.add(result.getString(2));
            expenses.add(new Integer(result.getInt(3)));
            expenses.add(result.getDouble(4));
        }
    }
    catch (SQLException e) {
        System.out.println(e.getMessage());
    }
    return expenses;
}
于 2012-05-15T06:13:18.230 回答
2

由于您在评论中说您希望获得ArrayListof Strings，因此以下内容应该有效：

@Override
public ArrayList<String> getExpenses() {
    ArrayList<String> expenses = new ArrayList<String>();
    try {
        Statement stmt = myConnection.createStatement();
        ResultSet result = stmt.executeQuery("SELECT * FROM expenses");
        while (result.next()) {
            expenses.add(result.getString(1));
            expenses.add(result.getString(2));
            expenses.add(result.getString(3));
            expenses.add(result.getString(4));
        }
    }
    catch (SQLException e) {
        System.out.println(e.getMessage());
    }
    return expenses;
}

JDBC 规范要求（在附录 B 中）大多数 SQL 类型（包括此处使用的所有类型）都可以使用Statement.getString().

PS 完成后立即关闭每个 JDBC 语句被认为是一种很好的做法。这通常在一个finally块中完成（或在 Java 7 中，使用try-with-resources 构造）。见https://stackoverflow.com/a/10514079/367273

于 2012-05-15T06:15:48.457 回答
1 
@Override
public ArrayList<String> getExpenses() {
    ArrayList<String> expenses = new ArrayList<String>();
    try {
        Statement stmt = myConnection.createStatement();
        ResultSet result = stmt.executeQuery("SELECT * FROM expenses");
        while(result.next()){

            String expense = null;
            setExpense(result.getInt(1).toString());
expenses.add(getExpense());                
setExpense(result.getString(2));
expenses.add(getExpense());                
setExpense(result.getInt(3).toString());
expenses.add(getExpense());                
setExpense(result.getDouble(4).toString());
expenses.add(getExpense());                


            }
    }
        catch (SQLException e){
             System.out.println(e.getMessage());
         }
    return expenses;
}
// make getter setters of expense accordingly
于 2012-05-15T06:16:14.860 回答