java - String 中的文本在提交并返回后重新排列

Question

我正在创建一个程序，它采用 32 - 126 的 ASCII 代码之间的单个字符并将结果作为二进制返回。这是一堆带有 .contains("")的if语句。该代码在返回单个字符时有效，但在输入单词时，字符串以二进制重新字母顺序返回。因此，如果我输入 Lauren 的名称，它会像这样以二进制形式出现：010011000110000101100101011011100111001001110101-Laenru（大写字母仍然是第一位的）。

这是我的所有代码：

主.java

    import java.util.Scanner;

    public class Main {

         public static Scanner input = new Scanner(System.in);

         public static void main (String[] args) {

              Letters l = new Letters();

              l.Letters();
         }
    }

字母.java

公共类字母扩展主要{

@SuppressWarnings("null")
public static void Letters () {
        System.out.println("Enter your letter(s)");

        String givenLetters = input.next();

            if (givenLetters.contains("A"))
                System.out.println("01000001");

            if (givenLetters.contains("B"))
                System.out.println("01000010");

            if (givenLetters.contains("C"))
                System.out.println("01000011");

            if (givenLetters.contains("D"))
                System.out.println("01000100");

            if (givenLetters.contains("E"))
                System.out.println("01000101");

            if (givenLetters.contains("F"))
                System.out.println("01000110");

            if (givenLetters.contains("G"))
                System.out.println("01000111");

            if (givenLetters.contains("H"))
                System.out.println("01001000");

            if (givenLetters.contains("I"))
                System.out.println("01001001");

            if (givenLetters.contains("J"))
                System.out.println("01001010");

            if (givenLetters.contains("K"))
                System.out.println("01001011");

            if (givenLetters.contains("L"))
                System.out.println("01001100");

            if (givenLetters.contains("M"))
                System.out.println("01001101");

            if (givenLetters.contains("N"))
                System.out.println("01001110");

            if (givenLetters.contains("O"))
                System.out.println("01001111");

            if (givenLetters.contains("P"))
                System.out.println("01010000");

            if (givenLetters.contains("Q"))
                System.out.println("01010001");

            if (givenLetters.contains("R"))
                System.out.println("01010010");

            if (givenLetters.contains("S"))
                System.out.println("01010011");

            if (givenLetters.contains("T"))
                System.out.println("01010100");

            if (givenLetters.contains("U"))
                System.out.println("01010101");

            if (givenLetters.contains("V"))
                System.out.println("01010110");

            if (givenLetters.contains("W"))
                System.out.println("01010111");

            if (givenLetters.contains("X"))
                System.out.println("01011000");

            if (givenLetters.contains("Y"))
                System.out.println("01011001");

            if (givenLetters.contains("Z"))
                System.out.println("01011010");

            if (givenLetters.contains(" "))
                System.out.println("00100000");

            if (givenLetters.contains("a"))
                System.out.println("01100001");

            if (givenLetters.contains("b"))
                System.out.println("01100010");

            if (givenLetters.contains("c"))
                System.out.println("01100011");

            if (givenLetters.contains("d"))
                System.out.println("01100100");

            if (givenLetters.contains("e"))
                System.out.println("01100101");

            if (givenLetters.contains("f"))
                System.out.println("01100110");

            if (givenLetters.contains("g"))
                System.out.println("01100111");

            if (givenLetters.contains("h"))
                System.out.println("01101000");

            if (givenLetters.contains("i"))
                System.out.println("01101001");

            if (givenLetters.contains("j"))
                System.out.println("01101010");

            if (givenLetters.contains("k"))
                System.out.println("01101011");

            if (givenLetters.contains("l"))
                System.out.println("01101100");

            if (givenLetters.contains("m"))
                System.out.println("01101101");

            if (givenLetters.contains("n"))
                System.out.println("01101110");

            if (givenLetters.contains("o"))
                System.out.println("01101111");

            if (givenLetters.contains("p"))
                System.out.println("01110000");

            if (givenLetters.contains("q"))
                System.out.println("01110001");

            if (givenLetters.contains("r"))
                System.out.println("01110010");

            if (givenLetters.contains("s"))
                System.out.println("01110011");

            if (givenLetters.contains("t"))
                System.out.println("01110100");

            if (givenLetters.contains("u"))
                System.out.println("01110101");

            if (givenLetters.contains("v"))
                System.out.println("01110110");

            if (givenLetters.contains("w"))
                System.out.println("01110111");

            if (givenLetters.contains("x"))
                System.out.println("01111000");

            if (givenLetters.contains("y"))
                System.out.println("01111001");

            if (givenLetters.contains("z"))
                System.out.println("01111010");
        }

 }

如果这意味着我必须添加代码以重新排列字母或者我需要修复错误代码中的某些内容，我希望字符串以用户输入的方式出现。先谢谢了！:)

score 2 · Accepted Answer

尝试

  String givenLetters = input.next();
  for(char c : givenLetters.toCharArray()) {
       if (c == 'A')
           System.out.println("01000001");
....

或更好地创建地图

    static Map<Character, String> map = new HashMap<>();

    static {
        map.put('A', "01000001");
        ...
    }

那么代码将是

 for(char c : givenLetters.toCharArray()) {
       String s = map.get(c);
       if (s != null) {
           System.out.print(s);
       }
 ...

score 1 · Accepted Answer

您的问题是，如果在字符串中的任何位置找到子字符串，则givenLetters.contains("X")返回 true 。正如 Evgeniy Dorofeev 建议的那样，您应该做的是一个接一个地遍历字符串中的字符。"X"givenLetters

当然，使用庞大的if语句列表（甚至是庞大的数组）是一种将字符转换为二进制的愚蠢方法。相反，您可以一点一点地完成（ideone.com 上的演示）：

char[] bits = new char[8];
for (char c : givenLetters.toCharArray()) {
    for (int i = 0; i < 8; i++) {
        int bitmask = 1 << (7-i);
        bits[i] = ((c & bitmask) == 0 ? '0' : '1');
    }
    System.out.println(bits);
}

（请注意，Java 字符实际上是 16 位长；此代码将仅打印每个字符的最低 8 位。）

score 0 · Accepted Answer

两条评论：

1）遍历字符串中的字符...

for(int i=0; i<givenLetters.length(); i++) {
  char c = givenLetters.charAt(i);
  System.out.println(code[c]);
}

2) 用你的代码创建一个数组以避免那个讨厌的 if 语句

private static String code[255];

static {
    code['A'] = "01000001";
    code['B'] = "01000010";

（等等）

这使您可以System.out.println(code[c])代替那个巨大的 if 链。如果您想从字面上将二进制文件放在同一行，请替换println为print

java - String 中的文本在提交并返回后重新排列

3 回答 3

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