我怀疑这可能是假阳性,但我不能确定,所以我有点困惑。我正在使用Eclipse Neon,问题出现在我准备声明的第三次。我在下面做了一些几乎相同的事情,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);
}
当您在插入时踩踏ps
时,您不会关闭第一组资源。
你还应该考虑使用try-with-resources:
try (Connection con = MySQL.connection;
PreparedStatement ps = con.prepareStatement("SELECT * from UsernameData "
+ "WHERE UUID = '" + player.getUniqueId() + "'");
PreparedStatement ps2 = con.prepareStatement("update UsernameData set UUID = ?, Username = ? where UUID = ?");
PreparedStatement ps3 = con.prepareStatement("insert into UsernameData(UUID, Username)"
+ " values (?, ?)");
ResultSet rs = ps.executeQuery()) {
if (rs.next() == true) {
ps2.setString(1, uuid);
ps2.setString(2, username);
ps2.setString(3, uuid);
ps2.execute();
return;
}
ps3.setString(1, uuid);
ps3.setString(2, username);
ps3.execute();
return;
} catch (SQLException e) {
Bukkit.getServer().getLogger().warning("SQL Error: " + e);
}
是的,第二个和第三个PreparedStatement可能被浪费了。如果你愿意,你可以把它们包装在自己的try-with-resources中。
但是问题的关键是你践踏了ps
变量