我是Java的新手,我在这里遇到了这个问题.我会发布链接并记住,这里的其他类似问题对我没有帮助,因为我有不同的代码,因此我在这里创建了这个帐户.
码:
package secret.package.guys;import java.util.Scanner;public class NewClass {public static void main(String[] args) { System.out.println("Number: "); Scanner scanner = new Scanner(System.in); String data=scanner.nextLine(); System.out.println(data); int a = 0; while(a < 6) { System.out.println(a); a ; } if (a > 6){ System.out.println("SAFE SPACE"); } else { System.out.println("Get in the Safe Space"); System.out.println("PERSON has entered the Safe Space. Safe Space closes instantly."); } String s = new String("Old marks: "); String t = new String("5.5"); String u = new String("4"); String v = new String("3"); String w = new String("2.5"); String x = new String("6.0"); String y = new String("5.2"); String z = new String("4"); String t1 = t.replaceAll("5.5", "6"); String u1 = u.replaceAll("4", "4"); String v1 = v.replaceAll("3", "5"); String w1 = w.replaceAll("2.5", "3"); String x1 = x.replaceAll("6.0", "2"); String y1 = y.replaceAll("5.2", "1.8"); String z1 = z.replaceAll("4", "4.4"); System.out.println("New: " s " " t1 " " u1 " " v1 " " w1 " " x1 " " y1 " " z1); System.out.println("Enter new marks: "); int foo = Integer.parseInt("t1"); int foo1 = Integer.parseInt("u1"); int foo2 = Integer.parseInt("v1"); int foo3 = Integer.parseInt("w1"); int foo4 = Integer.parseInt("x1"); int foo5 = Integer.parseInt("y1"); int foo6 = Integer.parseInt("z1"); System.out.println("foo1 foo2 foo3 foo4 foo5 foo6");}
}
解决方法:
您必须在String变量的值上调用parseInt,而不是在变量名称上调用.
例如,
int foo = Integer.parseInt("t1");
应该
int foo = Integer.parseInt(t1);
这只有在t1包含整数的String表示时才会起作用(这意味着parseInt(z1)和parseInt(y1)仍会失败,因为这些字符串不包含整数).
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