Java 如何将整数转换为 base64 (0-9A-Za-z)
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How to convert integers to base64 (0-9A-Za-z)
提问by pondigi
I have been trying to reduce the length of the way. I represent some integer ID's in my program. For Example
我一直在努力缩短路的长度。我在我的程序中表示一些整数 ID。例如
2
3
15
26
63
...
151564852
I would like them to be represented as such (0-9A-Za-z only)
我希望他们这样表示（仅限 0-9A-Za-z）
2
3
F
Q
z
...
vDF25a //For example
The approach I thought of is to have 63 if statements which each of the mappings from 0-63 to 0-z respectively and for anything above 64 do a recursion on the value minus 63.
我想到的方法是有 63 个 if 语句，其中每个从 0-63 到 0-z 的映射分别和高于 64 的任何东西对值减去 63 进行递归。
Needless to say, I think my approach is very flawed and impractical. What would be a more appropriate way of doing it?
不用说，我认为我的方法是非常有缺陷和不切实际的。什么是更合适的方法？
Update:
更新：
Following fge'ssuggestion I've got the encoder to work correctly, however my decode function only works for up-to length 2 strings, in cases where the string is larger the sum becomes erroneous. For example for 3840 to 3845 this is the output
按照fge 的建议，我已经让编码器正常工作，但是我的解码功能仅适用于最长 2 个字符串，在字符串较大的情况下，总和变得错误。例如对于 3840 到 3845，这是输出
// Encoded
zw
x
zy
zz
100
// Decoded
3840
3841
3842
3843
124 //Invalid decoding
Here is my code for the decode function
这是我的解码功能代码
public static int decode(String value)
{
String revStr = new StringBuilder(value).reverse().toString();
int sum = 0;
for (int i=1; i < revStr.length(); i++)
{
for (int j=0; j < ALPHABET.length; j++)
{
if (ALPHABET[j] == revStr.charAt(i))
{
sum += (ALPHABET.length * j) * i;
break;
}
}
}
for (int j=0; j < ALPHABET.length; j++)
{
if (ALPHABET[j] == revStr.charAt(0))
{
sum += j;
break;
}
}
return sum;
}
采纳答案by fge
This is not base64; base64 encodes binary data.
这不是 base64；base64 编码二进制数据。
Anyway, you don't need a s*load of ifstatements; use an array:
无论如何，您不需要 as*loadif语句；使用数组：
public final class AlphabetEncoder
{
private static final char[] ALPHABET = { '0', '1', '2', ...., 'z' };
private static final int ENCODE_LENGTH = ALPHABET.length;
public static String encode(int victim)
{
final List<Character> list = new ArrayList<>();
do {
list.add(ALPHABET[victim % ENCODE_LENGTH]);
victim /= ENCODE_LENGTH;
} while (victim > 0);
Collections.reverse(list);
return new String(list.toArray(new char[list.size()],
StandardCharsets.UTF_8);
}
public int decode(final String encoded)
{
int ret = 0;
char c;
for (int index = 0; index < encoded.length(); index++) {
c = encoded.charAt(index);
ret *= ENCODE_LENGTH;
ret += Arrays.binarySearch(ALPHABET, c);
}
return ret;
}
}
NOTE ABOUT THE DECODE FUNCTION: it is possible to use Arrays.binarySearch()here since the alphabet has the nice property of being naturally sorted (0 < 1 < 2 < ... < z). However, a test should probably be added that its return code not be negative!
关于解码功能的注意事项：可以在Arrays.binarySearch()此处使用，因为字母表具有自然排序的良好特性（0 < 1 < 2 < ... < z）。但是，可能应该添加一个测试，它的返回码不是负数！
回答by ctutte
Depending on the language you use, there should already be a module which converts a string from/to base64. Check this other post: Base64 Encoding in Java
根据您使用的语言，应该已经有一个模块可以将字符串从 base64 转换为 base64。查看其他帖子：Java 中的 Base64 编码
回答by Uday Shankar
You can refer to already existing logic for converting from Decimal [0-9] to Hexadecimal conversion present in Integerclass and extend the logic for your Base 64 converison. Refer
您可以参考类中现有的用于从十进制 [0-9] 转换为十六进制转换Integer的逻辑，并扩展 Base 64 转换的逻辑。参考
Integer.toHexString(int i)
This maybe the efficient implementation for conversion.
这可能是转换的有效实现。
