java - 新手指针 rs.next

Question

我只想看到产品用户正在寻找它们，但是当第二个 if 执行时，它将推送（指针或那里的任何东西）到下一个 ID（我拥有唯一的 ID，因此它将推送到无处）并且结果为 null . 我希望你能理解我的问题:)。

    if (stmt.execute(
                        "SELECT * FROM products where ID=" + removeName)) {
                    rs = stmt.getResultSet();
    if (!rs.next()) {
                       m = "ID not found.";
                        return m;
                   }

score 1 · Accepted Answer

在您的情况下，您可以使用PreparedStatement来避免SQL 注入问题。

  PreparedStatement prodsQuery= con.prepareStatement("SELECT * FROM products where ID=?");
  prodsQuery.setInt(1,removeName);
  ResultSet rs = prodsQuery.executeQuery();
  if(!rs.next())
  {
        m = "ID not found.";
        return m;
   }

score 0 · Accepted Answer

问题是您正在阅读第一个结果以了解是否至少有一个结果，然后尝试使用下一个结果并错过第一个结果（改编自您的问题描述）。我在这里解释了它是如何工作的。

此问题的一个可能解决方案是假设查询执行没有问题并且您有结果，然后检索数据（或List数据）并作为最后一步验证数据是否不为空或List数据不为空。

改编自Naveen 的答案的代码以显示建议的解决方案

PreparedStatement prodsQuery =
    con.prepareStatement("SELECT * FROM products where ID=?");
prodsQuery.setInt(1,removeName);
ResultSet rs = prodsQuery.executeQuery();

假设只有一个结果可以得到：

//also assuming you will set the results in a Data class (yes, this can be replaced)
Data data = null;
if (rs.next()) {
    //logic to retrieve data...
    data = new Data();
    data.setSomething(rs.get(1));
    //more and more code to fill the data...

    //because it looks that you need it as String (wonder why you return a String as well)
    return data.toString();
}
//note: I use an else statement to check if indeed there were no results at all
//else statement added using a line separator for code explanation purposes
else {
    m = "ID not found.";
    return m;
}

假设有一个要获得的结果列表：

//also assuming you will set the results in a Data class (yes, this can be replaced)
List<Data> dataList = new ArrayList<Data>();
while (rs.next()) {
    //logic to retrieve data...
    Data data = new Data();
    data.setSomething(rs.get(1));
    //more and more code to fill the data...

    //because it looks that you need it as String (wonder why you return a String as well)
    dataList.add(data);
}
//in this case, there's no validation in order to know if there's any result
//the validation must be in the client of this class and method checking if
//the result list is empty using if(!List#isEmpty) { some logic... }
return dataList;

score 0 · Accepted Answer

首先，您的方法容易受到 SQL 注入的攻击。请前往 PreparedStatement。
看看这个使用 PreparedStatement 的简单示例

你应该这样做：

ResultSet rs = stmt.executeQuery("SELECT * FROM products where ID=" + removeName);
if (!rs.next()) {
      m = "ID not found.";
      return m;
}

java - 新手指针 rs.next

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