如何从ArrayBuffer获取二进制字符串?

时间:2013-05-03 16:18:43

标签: javascript arraybuffer

在JavaScript中从ArrayBuffer获取二进制字符串的方法是什么?

我不想对字节进行编码,只需将二进制表示形式化为String。

提前致谢!

4 个答案:

答案 0 :(得分:16)

以下代码会始终将ArrayBuffer转换为String,然后再返回,而不会丢失或添加任何其他字节。

function ArrayBufferToString(buffer) {
    return BinaryToString(String.fromCharCode.apply(null, Array.prototype.slice.apply(new Uint8Array(buffer))));
}

function StringToArrayBuffer(string) {
    return StringToUint8Array(string).buffer;
}

function BinaryToString(binary) {
    var error;

    try {
        return decodeURIComponent(escape(binary));
    } catch (_error) {
        error = _error;
        if (error instanceof URIError) {
            return binary;
        } else {
            throw error;
        }
    }
}

function StringToBinary(string) {
    var chars, code, i, isUCS2, len, _i;

    len = string.length;
    chars = [];
    isUCS2 = false;
    for (i = _i = 0; 0 <= len ? _i < len : _i > len; i = 0 <= len ? ++_i : --_i) {
        code = String.prototype.charCodeAt.call(string, i);
        if (code > 255) {
            isUCS2 = true;
            chars = null;
            break;
        } else {
            chars.push(code);
        }
    }
    if (isUCS2 === true) {
        return unescape(encodeURIComponent(string));
    } else {
        return String.fromCharCode.apply(null, Array.prototype.slice.apply(chars));
    }
}

function StringToUint8Array(string) {
    var binary, binLen, buffer, chars, i, _i;
    binary = StringToBinary(string);
    binLen = binary.length;
    buffer = new ArrayBuffer(binLen);
    chars  = new Uint8Array(buffer);
    for (i = _i = 0; 0 <= binLen ? _i < binLen : _i > binLen; i = 0 <= binLen ? ++_i : --_i) {
        chars[i] = String.prototype.charCodeAt.call(binary, i);
    }
    return chars;
}

我通过在这个jsfiddle中绕过以下值来测试它:http://jsfiddle.net/potatosalad/jrdLV/

(String) "abc" -> (ArrayBuffer) -> (String) "abc"
(String) "aΩc" -> (ArrayBuffer) -> (String) "aΩc"
(Uint8Array) [0,1,255] -> (ArrayBuffer) -> (String) -> (Uint8Array) [0,1,255]
(Uint16Array) [0,1,256,65535] -> (ArrayBuffer) -> (String) -> (Uint16Array) [0,1,256,65535]
(Uint32Array) [0,1,256,65536,4294967295] -> (ArrayBuffer) -> (String) -> (Uint32Array) [0,1,256,65536,4294967295]

答案 1 :(得分:2)

这将为您提供来自类型化数组的二进制字符串

var bitsPerByte = 8;
var array = new Uint8Array([0, 50, 100, 170, 200, 255]);
var string = "";

function repeat(str, num) {
    if (str.length === 0 || num <= 1) {
        if (num === 1) {
            return str;
        }

        return '';
    }

    var result = '',
        pattern = str;

    while (num > 0) {
        if (num & 1) {
            result += pattern;
        }

        num >>= 1;
        pattern += pattern;
    }

    return result;
}

function lpad(obj, str, num) {
    return repeat(str, num - obj.length) + obj;
}

Array.prototype.forEach.call(array, function (element) {
    string += lpad(element.toString(2), "0", bitsPerByte);
});

console.log(string);

输出

000000000011001001100100101010101100100011111111

jsfiddle

或许你在问这个问题?

function ab2str(buf) {
    return String.fromCharCode.apply(null, new Uint16Array(buf));
}

注意:以这种方式使用apply意味着您可以达到参数限制(大约16000个元素),然后您将不得不循环遍历数组元素。

html5rocks

答案 2 :(得分:2)

近年来,通过添加JavaScript,使这一过程变得更加简单-这是一种将Uint8Array转换为二进制编码字符串的单行方法:

const toBinString = (bytes) =>
  bytes.reduce((str, byte) => str + byte.toString(2).padStart(8, '0'), '');

示例:

console.log(toBinString(Uint8Array.from([42, 100, 255, 0])))
// => '00101010011001001111111100000000'

如果您从ArrayBuffer开始,请创建缓冲区的Uint8Array“视图”以传递给此方法:

const view = new Uint8Array(myArrayBuffer);
console.log(toBinString(view));

来源:Libauth库(binToBinString method

答案 3 :(得分:-1)

function string2Bin(s) {
  var b = new Array();
  var last = s.length;
  for (var i = 0; i < last; i++) {
    var d = s.charCodeAt(i);
    if (d < 128)
      b[i] = dec2Bin(d);
    else {
      var c = s.charAt(i);
      alert(c + ' is NOT an ASCII character');
      b[i] = -1;
    }
  }
  return b;
}

function dec2Bin(d) {
  var b = '';
  for (var i = 0; i < 8; i++) {
    b = (d%2) + b;
    d = Math.floor(d/2);
  }
  return b;
}