try
{
PreparedStatement s = (PreparedStatement) conn.prepareStatement("SELECT voters.Check,count(*) FROM voting.voters where FirstName="+first+"and LastName="+last+" and SSN="+voter_ID);
//java.sql.Statement k = conn.createStatement();
rs=s.executeQuery();
//s.executeQuery("SELECT voters.Check,count(*) FROM voting.voters where FirstName="+first+"and LastName="+last+" and SSN="+voter_ID);
System.out.println(rs.first());
c=rs.getInt(1);
d=rs.getInt(2);
System.out.println(c);
System.out.println(d);
if(c==1 && d==1)
{
s.executeUpdate("update cand set total=total+1 where ssn="+can_ID);
System.out.println("Succeful vote");
System.out.println("after vote");
s.executeUpdate("update voters set voters.Check=1 where ssn="+voter_ID);
toclient=1;
PreparedStatement qw = (PreparedStatement) conn.prepareStatement("select FirstName from cand where ssn="+can_ID);
// rs=k.executeQuery("select FirstName from cand where ssn="+can_ID);
rs1 = qw.executeQuery();//Error Here Plz help me
String name1= (String) rs1.getString(1);
System.out.println(name1);
s.executeUpdate("update voters set VTO="+name1+"where ssn="+voter_ID);
System.out.println(rs.getString(1));
}
else
{
if(c != -1)
toclient =2;
if(d ==0)
toclient =3;
if(d>1)
toclient =4;
}
System.out.println("out-----------");
rs.close();
s.close();
}
catch (SQLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
錯誤是:使用ResultSet
前的MySQL + Java異常:在結果的開始時設置
java.sql.SQLException: Before start of result set
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1072)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:986)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:981)
恕我直言,張貼代碼塊和一些例外,沒有任何解釋嘗試是不是很好:/ – dangerstat 2010-02-14 09:21:48
我更喜歡這個「我有一個問題,請幫助」,沒有任何代碼;)。而這個很容易,所以不是那麼麻煩。 – Bozho 2010-02-14 09:24:14
我明白你的意思了,雖然我看不出*有一點解釋的危害。我想有一個妥協。 – dangerstat 2010-02-14 13:11:30