I'm trying to, if stringNumber (randomized) is one, make num 1. If string number is two, make num two, etc. using a method. This is an example, and the real program will be on a much larger scale.
import java.util.Random;
public static void main(String[] args) {
int num;
int num2;
int num3;
num = num2string(StringNumber);
num2 = num2string(StringNumber);
num3 = num2string(StringNumber);
System.out.println((num + 10)); //should print out 11, 12, or 13
System.out.println(num2);
System.out.println(num3);
}
//Picks a number.
public static String numPicker() {
Random rgen = new Random();
int rNumber;
rNumber = rgen.nextInt(3) + 1;
if (rNumber == 1)
return "Number 1";
else if (rNumber == 2)
return "Number 2";
else if (rNumber == 3)
return "Number 3";
else
return null; //Should never happen
}
private static int num2string() {
//In this case x = One
if (StringNumber.equals("Number 1"))
return num as 1;
else if (StringNumber.equals("Number 2"))
return num as 2;
else if (StringNumber.equals("Number 3"))
return num as 3;
else
return null; //should never happen
}
The correct version of this would do the following: 11, 12, or 13 should print out. Then 1, 2, or 3 should print out. Then 1, 2, or 3 should print out again.
答案 0 :(得分:0)
Consider using Integer.parseInt() method, which takes a string as an argument and retrieves int from it... Also, use equals() when comparing 2 strings.
答案 1 :(得分:0)
想通了!我必须将num转换为Strings,然后在完成后对其进行解析。
String StringNumber = RandomNumberGenerator();
String num;
num = num2string(StringNumber);
int num2 = Integer.parseInt(num);
使变量可以调用该方法。
private static String num2string(String StringNumber) {
if (StringNumber.equals("Number 1"))
return "1";
else if (StringNumber.equals("Number 2"))
return "2";
else if (StringNumber.equals("Number 3"))
return "3";
else
return null;
}