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('');
    };
})();

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('');
    };
})();

到目前为止，对我来说最简单的方法是:

//1. Create or fetch the Uint8Array to use in the example
const bufferArray = new Uint8Array([10, 10, 10])

//2. Turn the Uint8Array into a regular array
const array = Array.from(bufferArray);

//3. Stringify it (option A)
JSON.stringify(array);


//3. Stringify it (option B: uses @serdarsenay code snippet to decode each item in array)
let binArrayToString = function(binArray) {
    let str = "";
    for (let i = 0; i < binArray.length; i++) {        
        str += String.fromCharCode(parseInt(binArray[i]));
    }
    return str;
}

binArrayToString(array);

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!

我正在使用这个函数，它对我有用:

function uint8ArrayToBase64(data) {
    return btoa(Array.from(data).map((c) => String.fromCharCode(c)).join(''));
}

对于ES6和UTF8字符串

decodeURIComponent(escape(String.fromCharCode(...uint8arrData)))