Question

我正在使用这个方便的Javascript函数来解码base64字符串并获得一个数组作为回报。

这是字符串：

base64_decode_array('6gAAAOsAAADsAAAACAEAAAkBAAAKAQAAJgEAACcBAAAoAQAA')

这是返回的内容：

234,0,0,0,235,0,0,0,236,0,0,0,8,1,0,0,9,1,0,0,10,1,0,0,38,1,0,0,39,1,0,0,40,1,0,0

问题是我真的不懂javascript函数：

var base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'.split("");
var base64inv = {}; 
for (var i = 0; i < base64chars.length; i++) { 
  base64inv[base64chars[i]] = i; 
}
function base64_decode_array (s)
{
  // remove/ignore any characters not in the base64 characters list
  //  or the pad character -- particularly newlines
  s = s.replace(new RegExp('[^'+base64chars.join("")+'=]', 'g'), "");

  // replace any incoming padding with a zero pad (the 'A' character is zero)
  var p = (s.charAt(s.length-1) == '=' ? 
          (s.charAt(s.length-2) == '=' ? 'AA' : 'A') : ""); 

  var r = [];

  s = s.substr(0, s.length - p.length) + p;

  // increment over the length of this encrypted string, four characters at a time
  for (var c = 0; c < s.length; c += 4) {

    // each of these four characters represents a 6-bit index in the base64 characters list
    //  which, when concatenated, will give the 24-bit number for the original 3 characters
    var n = (base64inv[s.charAt(c)] << 18) + (base64inv[s.charAt(c+1)] << 12) +
            (base64inv[s.charAt(c+2)] << 6) + base64inv[s.charAt(c+3)];


    // split the 24-bit number into the original three 8-bit (ASCII) characters
    r.push((n >>> 16) & 255);
    r.push((n >>> 8) & 255);
    r.push(n & 255);


  }
   // remove any zero pad that was added to make this a multiple of 24 bits
  return r;
}

那些“＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;和“＆gt;＆gt;＆gt;”字符。或者Python有这样的函数吗？

Answer 1

谁在乎呢。 Python有更简单的方法来做同样的事情。

[ord(c) for c in '6gAAAOsAAADsAAAACAEAAAkBAAAKAQAAJgEAACcBAAAoAQAA'.decode('base64')]

Answer 2

在Python中我希望你只使用base64 module ...

...但是回答您关于<<和>>>的问题：

<<是左移算子;结果是第一个操作数向左移动了第二个操作数的位数;例如5 << 2是20，因为5是二进制的101，而20是10100。
>>>是非符号扩展的右移运营商;结果是第一个操作数被移位右边第二个操作数的位数......最左边的位总是用0填充。

Answer 3

为什么不呢：

from binascii import a2b_base64, b2a_base64
encoded_data = b2a_base64(some_string)
decoded_string = a2b_base64(encoded_data)

def base64_decode_array(string):
    return [ord(c) for c in a2b_base64(string)]

Answer 4

只是为了好玩/完整，我会更简单地翻译javascript ::）

# No particular reason to make a list of chars here instead of a string.
base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
lookup = dict((c, i) for (i, c) in enumerate(base64chars))

def base64_decode_array(s):
  # Filter out meaningless chars, especially newlines. No need for a regex.
  s = ''.join(c for c in s if c in base64chars + '=')

  # replace any incoming padding with a zero pad (the 'A' character is zero)
  # Their way:
  # p = ('AA' if s[-2] == '=' else 'A') if s[-1] == '=' else ''
  # s = s[:len(s) - len(p)] + p
  # My way (allows for more padding than that; 
  # '=' will only appear at the end anyway
  s = s.replace('=', 'A')

  r = []

  # Iterate over the string in blocks of 4 chars - an ugly hack
  # though we are preserving the original code's assumption that the text length
  # is a multiple of 4 (that's what the '=' padding is for) ;)
  for a, b, c, d in zip(*([iter(s)] * 4)):
    # Translate each letter in the quad into a 6-bit value and bit-shift them
    # together into a 24-bit value
    n = (lookup[a] << 18) + (lookup[b] << 12) + (lookup[c] << 6) + lookup[d]

    # split the 24-bit number into the original three 8-bit (ASCII) characters
    r += [(n >> 16) & 0xFF), (n >> 8) & 0xFF), (n & 0xFF)]

  return r

在Python中将base64数据解码为数组

4 个答案: