如何使用Buffered Reader使用Java读取文件中的整数？

Question

Here is the file im trying to read:

P 1.0 0.0 0.0
80 10
80 30
230 37
280 30
280 10
T
t 100 -75
r 30 0 0
s 0.5 1.5 0 0

Heres part of my code:

File file = new File("coordinatess.txt");
BufferedReader br = new BufferedReader(new FileReader(file));
String st;
glPushMatrix();
while ((st = br.readLine()) != null){
                        String str[] = st.split(" ");
                        if(str[0].equalsIgnoreCase("P")){
                            glColor3f(Float.parseFloat(str[1]), Float.parseFloat(str[2]), Float.parseFloat(str[3]));

                        }
                        else if(str[0].equalsIgnoreCase("T")){
                            br.readLine();
                        }
                        else if(str[0].equalsIgnoreCase("r")){
                            glRotatef(Float.parseFloat(str[1]), Float.parseFloat(str[2]), Float.parseFloat(str[3]), 0);
                        }
                        else if(str[0].equalsIgnoreCase("s")){
                            glScalef(Float.parseFloat(str[1]), Float.parseFloat(str[2]), Float.parseFloat(str[3]));
                        }
                        else if(str[0].equalsIgnoreCase("t")){
                            glTranslatef(Float.parseFloat(str[1]), Float.parseFloat(str[2]), 0);
                        }

                    }

I'm kind of having trouble reading the line without any letters. Its just integers. How would I read those lines specifically, after "P"?

Answer 1

use nested level to parse values Integer.parseInt(value) with exception handling

String str = "T 10 5.7";
String[] strings = str.split(" ");
for (String string : strings) {
    try {
        System.out.println(Integer.parseInt(string));
    } catch (Exception e) {
        try {
            System.out.println(Float.parseFloat(string));
        } catch (Exception e2) {
            System.out.println(string);
        }
    }
}

Answer 2

Try Santaji's way, and in try-catch block include finally as well outside try-catch block. Your final try-catch block should look like this. Any input stream must be closed.

String str = "T 10 5.7";
        String[] strings = str.split(" ");
        for (String string : strings) {
            try {
                System.out.println(Integer.parseInt(string));
            } catch (Exception e) {
                try {
                    System.out.println(Float.parseFloat(string));
                } catch (Exception e2) {
                    System.out.println(string);
                }
            } finally {
                  try {
                    if (file !=null) {
                          file.close();
                 }
            }
           }
        }

But, to convert integer to string, you could possibly use other ways like :
String.valueOf(number) (most preferred)
OR
Integer.toString(number)