Java 从随机整数到实际字符串

这是一个接收ASCII表中所有打印字符的数组的代码。我试图使任何整数形式的字符串消息（例如，转换为97098097的字符串“aba”可以放回其原始字符串形式。100101101可以被提取并制作回“dee”。我确实尝试过这个方法，但它似乎不起作用，特别是当涉及数字等问题时，请帮助我。顺便说一下，它是Java的，我正在使用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;

公共静态字符串IntToString（）{
int n=0；
字符串消息=”；
字符串message2=null；
，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，，C、D、E、F、G、H、I、J、K、L、M、N、O、P、Q、R、S、t、u、v、w、x、y、z、“{”、“|”、“}”、“~”}；
字符串IntMessage=result.toString（）；
String firstChar=IntMessage.substring（0,2）；
if（IntMessage.substring（0,1）=“1”&&IntMessage.length（）%3==0）
{
对于（intx=（IntMessage.length（）%3-3）%3；x

看来你的编码方案是，呃，疯狂的

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

所以

“abc”=>{97,98,99}=>“979899”。

但是，由于您使用的是ASCII，其值可以为100或更多，因此如果整数小于100，则使用0填充整数：

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

但你只是偶尔决定这么做，因为

"aba" => "97098097"

也就是说，第一个“a”变成“97”，但最后一个“a”变成“097”

我想你应该先修改你的编码方案

另外，希望这些不是“随机整数”，因为您正试图将它们转换为合理的字符串。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，它们可能没有多大意义

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。

publicstaticvoidmain（String[]srgs）{
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;
}

字符串aaa=“100101101”；
字符串[]a=拆分（aaa，3）；
字符串s=“”；
对于（int i=0；嗯…为什么？你想用它实现什么？必须有一种更明智的方法来解决你的根本问题。我知道我这样做是为了一个项目，但我无法让它工作我想要的是：97098097=>“aba“仅此而已。您可以修复此代码中的格式设置吗？您粘贴的标签导致了各种奇怪的缩进问题。目前，您的代码示例无法阅读。而且，初始的
for
循环“太聪明了”，可能会导致此处的代码审查失败。”。