【发布时间】:2016-11-10 00:30:54
【问题描述】:
我怀疑这可能是误报,但我不能确定,所以我有点困惑。我正在使用 Eclipse Neon,问题出现在我第三次准备声明时。我在下面做了几乎相同的事情,没有错误。
try{
Connection con = MySQL.connection;
PreparedStatement ps = con.prepareStatement("SELECT * from UsernameData "
+ "WHERE UUID = '" + player.getUniqueId() + "'");
ResultSet rs = ps.executeQuery();
if(rs.next() == true){
ps = con.prepareStatement("update UsernameData set UUID = ?, Username = ? where UUID = ?");
ps.setString(1, uuid);
ps.setString(2, username);
ps.setString(3, uuid);
ps.execute();
ps.close();
rs.close();
return;
}
ps = con.prepareStatement("insert into UsernameData(UUID, Username)"
+ " values (?, ?)");
ps.setString(1, uuid);
ps.setString(2, username);
ps.execute();
ps.close();
rs.close();
return;
}catch(SQLException e){
Bukkit.getServer().getLogger().warning("SQL Error: " + e);
}
【问题讨论】:
标签: java mysql prepared-statement resultset