Question

这是一个代码，它接收ASCII表的所有打印字符的数组。我试图使它以整数形式的任何String消息（例如转换为97098097的字符串“aba”可以放回其原始的字符串形式.100101101可以被取回并重新变成“dee”。我'我真的很努力地使用这种方法，但它似乎没有用，特别是当涉及到数字时，请帮助我。顺便说一句是Java，我使用的是Eclipse。

public static String IntToString (){


int n = 0;
String message = "";
String message2 = null;
String [] ASCII = {" ","!","\"","#","$","%","&","\'","(",")","*","+",",","-",".","/","0","1","2","3","4","5","6","7","8","9",":",";","<","=",">","?","@","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","[","\\","]","^","_","`","a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","{","|","}","~"};
String IntMessage = result.toString();
String firstChar = IntMessage.substring(0,2);
if (IntMessage.substring(0,1)=="1" && IntMessage.length()%3==0)
{
    for (int x = (IntMessage.length() % 3 - 3) % 3; x < IntMessage.length()-2; x += 3)
        n = Integer.parseInt(IntMessage.substring(Math.max(x, 0), x + 3));
        message=message.concat(ASCII[n-31]);
return message;
}
else if (IntMessage.length()%3==2)
message2=ASCII[(Integer.parseInt(firstChar))-31];
        for (int x = 2; x < IntMessage.length()-2; x += 3)
            n = Integer.parseInt(IntMessage.substring(x, x + 3));
            message=message2+=ASCII [n - 31];
return message;

Answer 1

您的编码方案似乎是，呃，疯了。

首先，您获取字符串的ASCII值，然后获取该ASCII值的字符表示，然后将其存储为字符串。

所以"abc" => {97, 98, 99} => "979899".

但是因为你使用的是ASCII值，它的值可以是100或者更高，所以如果你的整数低于100，你就用0填充你的整数：

"abc" => {97, 98, 99} => {"097", "098", "099"} => "097098099"

但你有时候只是决定这样做，因为不知何故

"aba" => "97098097"

即，第一个“a”变成“97”，但最后一个“a”变成“097”。

我会说你应该先修改你的编码方案。

此外，这些希望不是“随机整数”，因为你试图把它们变成合理的字符串。否则，像base64这样的简单映射很容易将任何整数映射到字符串，它们可能没什么用处。

事实上，它们甚至不是真正的整数。您将编码的字符串存储为字符串。

Answer 2

public static void main(String[] srgs){
    String aaa = "100101101";
    String[] a = split(aaa, 3);

    String s = "";

    for(int i=0;i<a.length;i++){
        char c = (char)Integer.parseInt(a[i]);
        s += Character.toString(c);
    }
    System.out.println(s);
}

public static String[] split(String str, int groupIndex){
    int strLength = str.length();
    int arrayLength = strLength/groupIndex;
    String[] splitedArray = new String[strLength/groupIndex];

    for(int i=0;i<arrayLength;i++){
        String splitedStr = str.substring(0, groupIndex);
        str = str.substring(groupIndex, str.length());
        arrayLength = str.length();
        splitedArray[i] = splitedStr;
    }
    return splitedArray;
}

最重要的是ASCII字符串转换为Char值，而不是将其转换为字符串中的真实字符值。在这种情况下，ASCII码长度需要固定为3可能会有所帮助。

从随机整数到实际的String

2 个答案: