我有這樣的代碼在Eclipse:此轉換爲什麼會出現錯誤?
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);
,這讓我這個錯誤
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)
但我不明白爲什麼。 有人可以幫助我嗎?
告訴我們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對不起,這是我班的舊版本。 – Jakibah
你的rawpassword'String顯然不包含有效的數字。 –