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，它将打印下表以获取匹配的输出。

现在，它应该打印出每个输出。但是，它只打印出第一个。

我相信下面的代码是原因：

 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 确定结果集是否为空，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

所以我在这里看到了两个问题。最大的是这个：

System.out.println(toString());

这会调用当前类的.toString() 方法，该方法不会输出来自ResultSet 的任何数据。至少，不是基于您显示的任何代码。您将从ResultSet 返回的所有值存储在变量中，但这些变量似乎没有在任何地方使用。您需要以某种方式将这些变量传递给您的.println()。

第二个问题是rs.next() 将光标向前移动了一行。所以当你这样做时：

if(rs.next()) {

这会导致您跳过第一行。这实际上有点难以解决，因为如果不调用.next()，就无法判断ResultSet 是否为空。我可能处理这个问题的方法是将所有结果拉入列表中的对象，然后根据列表而不是 ResultSet 本身进行所有打印。

Answer 3

在使用带有 Prepared 语句的 select 查询时，我也遇到了同样的问题。 例如-

String sqlquerySELECTAgentsWithParam = " select * from AGENTS WHERE AGENT_CODE = ( ? ) ";

它返回了一行，因为我使用了 IF 条件，然后使用了 while(rs.next()) 我无法打印单行，因此，我使用了如下所示的 do-while 循环来打印第一行结果。下面的代码sn-p-希望对您有所帮助！

Class.forName("oracle.jdbc.driver.OracleDriver");
Connection conn = DriverManager.getConnection(oracelDBUrl, oracleDBUser,oracleDBPwd );

if (conn != null) { System.out.println("Connected to the Oracle DB ");  }
else { System.out.println("Failed to Connect to Oracle DB "); }
PreparedStatement pstmt = conn.prepareStatement(sqlquerySELECTAgentsWithParam);
pstmt.setString(1,"<agent_id>");
ResultSet rs = pstmt.executeQuery();
boolean returnResultrs = rs.next();
System.out.println("Fetched Result is = " +returnResultrs);
if(returnResultrs)
{
    do {
        System.out.println("Inside Do - While Loop");
        System.out.println("AGENT_CODE = " + rs.getString(1)+ "has AGENT_NAME = "  +rs.getString(2));
        }
while(rs.next());

}