ga libog nako mga master, na stuck up nako.... unsay angay nako buhaton if level 1, dapat mo display cya ug salary na 1380.00....ako nani g suwayan ug if statement sulod sa loop (if salaryLvl == "1") sysout("1380.00"); wa may mo gawa....nag libug ko ani kay mo tawag man ko ug string sa txt file, murag sayup to ako pamaagi na g == nako ug "1"
mao ni sulod sa akong text file...kanang naay 1,2,3 sa kinalasan mao na ilang level
--------------------------
Juan_De_La_Cruz,A001,1
Miriam_Bismonte,A002,1
Gloria_Macapagal,A003,3
Manny_Pacquiao,A004,3
Jose_Rizal,A005,2
-------------------------
mao ni akong code
import java.io.IOException;
import java.io.RandomAccessFile;
import java.util.Scanner;
public class Payroll
{
public Payroll()
{
}
static Scanner input = new Scanner(System.in);
static RandomAccessFile instream;
public static void main(String [] args)
{
System.out.println("enter:");
String data;
String id = "", name = "",id2 = "", salaryLvl = "";
boolean found = false;
String idToFind = input.next();
try
{
instream = new RandomAccessFile("employee.txt","r");
int numberOfRecords = 5;
instream.seek(0);
for(int i = 0; i < numberOfRecords; i++)
{
data = instream.readLine();
String [] record;
record = data.split(",");
for(int x = 0; x < record.length;x++)
id = record[1];
name = record[0];
salaryLvl = record[2];
if(id.equals(idToFind))
{
System.out.println(name);
System.out.println(id);
System.out.println(salaryLvl);
if(salaryLvl == "1")
{
System.out.println("weee");
}
found = true;
break;
}
}
if(!found)
System.out.println("not found!");
}//try
catch(IOException e)
{
}
}
}