resultset.next（）仅显示第一行的输出

Question

所以，我有这样的事情：

System.out.println("Enter owner's IC no. or plate no. : ");
String update = in.nextLine();

String sql = String.format("SELECT * FROM `vehicle` WHERE ic='%s' OR plate ='%s'",update,update);

ResultSet rs = stmt.executeQuery(sql);

if(rs.next()) {
    System.out.println("RegNo." +"\t\t"+ "Name" + "\t\t" + "IC" +"\t\t" + "Plate No." + "\t" + "Color" + "\t\t" + "Year" + "\t\t" + "Make" + "\t\t" + "Model" +"\t\t"+ "Capacity" + "\t" + "Type" +"\t\t" + "Max Load");
}
else {
    System.out.println("IC and PLate No. not found....");}

while (rs.next()) {
    regno = rs.getInt("regno");
    name = rs.getString("name");
    ic = rs.getString("ic");
    plate = rs.getString("plate");
    color = rs.getString("color");
    year = rs.getInt("year");
    make = rs.getString("make");
    model = rs.getString("model");
    capacity = rs.getDouble("capacity");
    type = rs.getString("type");
    maxload = rs.getDouble("maxload");

    System.out.println(toString());
}   

我要做的是，如果在数据库中找到数据，它将打印下表以查找匹配的输出。

现在，它应该打印出每个输出。但是，它只打印出第一个。

我相信以下代码是原因：

 if(rs.next()) {
    System.out.println("RegNo." +"\t\t"+ "Name" + "\t\t" + "IC" +"\t\t" + "Plate No." + "\t" + "Color" + "\t\t" + "Year" + "\t\t" + "Make" + "\t\t" + "Model" +"\t\t"+ "Capacity" + "\t" + "Type" +"\t\t" + "Max Load");
}
else {
    System.out.println("IC and PLate No. not found....");}

Answer 1

使用MessageFormat格式化输出，使用counter确定是否为空结果集，如下所示：

String strFormat = "RegNo. {0}\tName {1}\tIC {2}\tPlate No. {3}\tColor {4}\tYear {5}\tMake {6}\tModel {7}\tCapacity {8}\tType {9}\tMax Load {10}");

int counter = 0;

while (rs.next()) {
    counter++;

    regno = rs.getInt("regno");
    name = rs.getString("name");
    ic = rs.getString("ic");
    plate = rs.getString("plate");
    color = rs.getString("color");
    year = rs.getInt("year");
    make = rs.getString("make");
    model = rs.getString("model");
    capacity = rs.getDouble("capacity");
    type = rs.getString("type");
    maxload = rs.getDouble("maxload");

    System.out.println(MessageFormat.format(strFormat, regno, name, ic, plate, color, year, make, model, capacity, type, maxload));
}

if (counter == 0) {
    System.out.println("IC and PLate No. not found....");
}

Answer 2

所以我在这里看到两个问题。最大的一个是：

System.out.println(toString());

这会调用当前类的.toString()方法，该方法不会输出ResultSet中的任何数据。至少，不是基于您展示的任何代码。您将从ResultSet返回的所有值存储在变量中，但这些变量似乎并未在任何地方使用。你需要以某种方式将这些变量带到你的.println()。

第二个问题是rs.next()将光标向前移动一行。所以当你这样做时：

if(rs.next()) {

这会导致您跳过第一行。这实际上很难解决，因为没有调用ResultSet就没有好的方法来判断.next()是否为空。我可能处理的方法是将所有结果拉入列表中的对象，然后根据列表进行所有打印，而不是ResultSet本身。