我正在使用这个Typescript代码片段:

function UInt8ArrayToString(uInt8Array: Uint8Array): string
{
    var s: string = "[";
    for(var i: number = 0; i < uInt8Array.byteLength; i++)
    {
        if( i > 0 )
            s += ", ";
        s += uInt8Array[i];
    }
    s += "]";
    return s;
}

如果需要JavaScript版本，则删除类型注释。 希望这能有所帮助!

做什么@Sudhir说，然后得到一个字符串，逗号分隔的数字列表使用:

for (var i=0; i<unitArr.byteLength; i++) {
            myString += String.fromCharCode(unitArr[i])
        }

这会给你想要的字符串， 如果还相关的话

Uint8Array to String

let str = Buffer.from(key.secretKey).toString('base64');

String到Uint8Array

let uint8arr = new Uint8Array(Buffer.from(data,'base64')); 

如果你不能使用TextDecoder API，因为它不支持IE:

你可以使用Mozilla开发者网络网站推荐的faststsmallsttextencoderdecoder polyfill; 你也可以在MDN网站上使用此功能:

function utf8ArrayToString(aBytes) { var sView = ""; for (var nPart, nLen = aBytes.length, nIdx = 0; nIdx < nLen; nIdx++) { nPart = aBytes[nIdx]; sView += String.fromCharCode( nPart > 251 && nPart < 254 && nIdx + 5 < nLen ? /* six bytes */ /* (nPart - 252 << 30) may be not so safe in ECMAScript! So...: */ (nPart - 252) * 1073741824 + (aBytes[++nIdx] - 128 << 24) + (aBytes[++nIdx] - 128 << 18) + (aBytes[++nIdx] - 128 << 12) + (aBytes[++nIdx] - 128 << 6) + aBytes[++nIdx] - 128 : nPart > 247 && nPart < 252 && nIdx + 4 < nLen ? /* five bytes */ (nPart - 248 << 24) + (aBytes[++nIdx] - 128 << 18) + (aBytes[++nIdx] - 128 << 12) + (aBytes[++nIdx] - 128 << 6) + aBytes[++nIdx] - 128 : nPart > 239 && nPart < 248 && nIdx + 3 < nLen ? /* four bytes */ (nPart - 240 << 18) + (aBytes[++nIdx] - 128 << 12) + (aBytes[++nIdx] - 128 << 6) + aBytes[++nIdx] - 128 : nPart > 223 && nPart < 240 && nIdx + 2 < nLen ? /* three bytes */ (nPart - 224 << 12) + (aBytes[++nIdx] - 128 << 6) + aBytes[++nIdx] - 128 : nPart > 191 && nPart < 224 && nIdx + 1 < nLen ? /* two bytes */ (nPart - 192 << 6) + aBytes[++nIdx] - 128 : /* nPart < 127 ? */ /* one byte */ nPart ); } return sView; } let str = utf8ArrayToString([50,72,226,130,130,32,43,32,79,226,130,130,32,226,135,140,32,50,72,226,130,130,79]); // Must show 2H₂ + O₂ ⇌ 2H₂O console.log(str);

Albert给出的解决方案，只要不经常调用所提供的函数，并且只用于大小适中的数组，就能很好地工作，否则效率非常低。下面是一个增强的普通JavaScript解决方案，它适用于Node和浏览器，具有以下优点:

•有效工作于所有八字节数组大小

•不生成中间丢弃字符串

•在现代JS引擎上支持4字节字符(否则“?”将被替换)

var utf8ArrayToStr = (function () {
    var charCache = new Array(128);  // Preallocate the cache for the common single byte chars
    var charFromCodePt = String.fromCodePoint || String.fromCharCode;
    var result = [];

    return function (array) {
        var codePt, byte1;
        var buffLen = array.length;

        result.length = 0;

        for (var i = 0; i < buffLen;) {
            byte1 = array[i++];

            if (byte1 <= 0x7F) {
                codePt = byte1;
            } else if (byte1 <= 0xDF) {
                codePt = ((byte1 & 0x1F) << 6) | (array[i++] & 0x3F);
            } else if (byte1 <= 0xEF) {
                codePt = ((byte1 & 0x0F) << 12) | ((array[i++] & 0x3F) << 6) | (array[i++] & 0x3F);
            } else if (String.fromCodePoint) {
                codePt = ((byte1 & 0x07) << 18) | ((array[i++] & 0x3F) << 12) | ((array[i++] & 0x3F) << 6) | (array[i++] & 0x3F);
            } else {
                codePt = 63;    // Cannot convert four byte code points, so use "?" instead
                i += 3;
            }

            result.push(charCache[codePt] || (charCache[codePt] = charFromCodePt(codePt)));
        }

        return result.join('');
    };
})();

var decodedString = decodeURIComponent（escape（String.fromCharCode（...new Uint8Array（err））））; var obj = JSON.parse（decodedString）;