我有这样的代码在Eclipse:此转换为什么会出现错误?
String A = String.valueOf(a);
String B = String.valueOf(b);
String C = String.valueOf(c);
String D = String.valueOf(d);
String E = String.valueOf(e);
String F = String.valueOf(f);
String G = String.valueOf(g);
String H = String.valueOf(h);
String I = String.valueOf(i);
String J = String.valueOf(j);
String K = String.valueOf(k);
String rawpassword = A+B+C+D+E+F+G+H+I+J+K;
int password = Integer.parseInt(rawpassword);
System.out.println(password);
,这让我这个错误
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:495)
at java.lang.Integer.parseInt(Integer.java:527)
at com.jakibah.codegenerator.Main.Generate(Main.java:65)
at com.jakibah.codegenerator.Main.run(Main.java:24)
at java.lang.Thread.run(Thread.java:745)
但我不明白为什么。 有人可以帮助我吗?
告诉我们a,b,c ... k和rawPassword – SpringLearner
a = r.nextInt(10); \t \t b = r.nextInt(10); \t \t c = r.nextInt(10); \t \t d = r.nextInt(10); \t \t E = r.nextInt(10); \t \t \t F = r.nextInt(10); 克= r.nextInt(10); \t \t h = r.nextInt(10); \t \t i = r.nextInt(10); \t \t j = r.nextInt(10); \t \t k = r.nextInt(10); rawpassword是codestring对不起,这是我班的旧版本。 – Jakibah
你的rawpassword'String显然不包含有效的数字。 –