Integer.parseInt 没有将字符串解析为整数

Question

Scanner sc = new Scanner(System.in);
length = sc.nextLine();
length_to_play = Integer.parseInt(length);

我曾尝试使用 length.trim() 和 length.replaceAll() 来丢弃空格，但没有奏效。我在线程 "main" java.lang.NumberFormatException.

中遇到异常

Exception in thread "main" java.lang.NumberFormatException: For input string: "y"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:492)
    at java.lang.Integer.parseInt(Integer.java:527)
    at speedwords.first_activity(speedwords.java:27)
    at speedwords.main(speedwords.java:338)

Answer 1

我认为您误解了

的功能

Integer.parseInt();

来自 Java 文档

  Parses the string argument as a signed decimal integer. The 
  characters in the string must all be decimal digits, except that 
  the first character may be an ASCII minus sign <code>'-'</code> 
  (<code>'&#92;u002D'</code>) to indicate a negative value. The resulting 
  integer value is returned, exactly as if the argument and the radix 
  10 were given as arguments to the 
  {@link #parseInt(java.lang.String, int)} method.

  @param s       a <code>String</code> containing the <code>int</code>
              representation to be parsed
  @return     the integer value represented by the argument in decimal.
  @exception  NumberFormatException  if the string does not contain a
                parsable integer.

public static int parseInt(String s) throws NumberFormatException {
    return parseInt(s,10);
    }

Answer 2

当用户输入数字时，你应该使用sc.nextInt()。这样你就不需要自己写解析了。

如果您需要字符串并希望将它们转换为 int，您可以对输入字符串的每个字符分别使用 Character.getNumericValue(myChar)。

Answer 3

你应该看看 Integer.parseInt(String s)

上的文档

如您所见，解析后的字符串被转换为整数值，如果该值不是整数，则会抛出 NumberFormatException.

Why are you expecting it should return ASCII integer value?

Answer: For your kind information we are passing String, not a char which has independent ASCII value. If you want the ASCII of each character in your String then you can have a look at this question.

Answer 4

好的，我发布了我发现最适合我想要的案例的答案。不过谢谢大家的帮助。

Scanner sc = new Scanner(System.in);
length = sc.nextLine();
//length_to_play = Integer.valueOf(length); didn't work
char ch = length.charAt(0);
length_to_play=Character.getNumericValue(ch); //worked

Integer.parseInt 没有将字符串解析为整数

Integer.parseInt not parsing a string into integer

java

string

parsing

integer