从结果集 Java 中获取值
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Get value from ResultSet Java
提问by Bui Quang Huy
I want to get value from ResultSet after query to the database so how can I do it. This is some line of my code
我想在查询数据库后从 ResultSet 获取值，所以我该怎么做。这是我的一些代码行
conn = DBConnection.connect();
String SQL = "SELECT * from usertable";
// ResultSet
ResultSet rs = conn.createStatement().executeQuery(SQL);
if (rs.next()) {
System.out.println(rs.getString(1));
}
It printed: "1", not the value I want to get. Could anyone help me about this. This is example data of usertable:
它打印：“1”，不是我想要的值。任何人都可以帮我解决这个问题。这是用户表的示例数据：
采纳答案by java acm
you dont put more details of your problem but this is an example :
you have a person class with this fields you could to make Setter Getter by yourself
你没有提供更多关于你的问题的细节，但这是一个例子：
你有一个带有这些字段的人员类，你可以自己制作 Setter Getter
class Person{
String name;
int id;
}
then in your ResultSet :
think your table have two column ("userid" and "firstname") and first column is "userid"
然后在你的 ResultSet 中：
认为你的表有两列（“userid”和“firstname”），第一列是“userid”
PreparedStatement ps = null;
Connection con = null;
// LIMIT 1 because you have one Person Object to fill otherwise you must be have an Array of Person
String SQL = "SELECT * from usertable LIMIT 1";
try {
con = DBConnection.getConnection();
ps = con.prepareStatement(SQL);
ResultSet rs = ps.executeQuery();
Person p = null;
while (rs.next()) {
p = new Person();
p.id = rs.getInt(1);
// or p.id=rs.getInt("userid"); by name of column
p.name = rs.getString(2);
// or p.name=rs.getString("firstname"); by name of column
}
return p;
} catch (
SQLException ex) {
Logger.getLogger(YourClassName.class.getName()).log(Level.SEVERE, null, ex);
return null;
} finally {
if (con != null) {
try {
con.close();
} catch (SQLException ex) {
Logger.getLogger(YourClassName.class.getName()).log(Level.SEVERE, null, ex);
}
}
if (ps != null) {
try {
ps.close();
} catch (SQLException ex) {
Logger.getLogger(YourClassName.class.getName()).log(Level.SEVERE, null, ex);
}
}
}
if your Result is more than one you must be change Person to Person[] or "ArrayList" Object
如果您的结果不止一个，您必须将 Person 更改为 Person[] 或“ArrayList”对象
