微信小游戏中 window.atob 和 window.btoa 方法未定义。

这两个方法是用来做 base64 加密和解密的，可以自己实现一下 atob 和 btoa 方法。

// decoder
function atob(input) {
    var chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';
    // ExtendScript bad parse of /=
    var str = (String (input)).replace (/[=]+$/, '');

    for (
        // initialize result and counters
        var bc = 0, bs, buffer, idx = 0, output = '';
        // get next character
        buffer = str.charAt (idx++); // eslint-disable-line no-cond-assign
        // character found in table? initialize bit storage and add its ascii value;
        ~buffer && (bs = bc % 4 ? bs * 64 + buffer : buffer,
        // and if not first of each 4 characters,
        // convert the first 8 bits to one ascii character
        bc++ % 4) ? output += String.fromCharCode (255 & bs >> (-2 * bc & 6)) : 0
    ) {
        // try to find character in table (0-63, not found => -1)
        buffer = chars.indexOf (buffer);
    }
    return output;
}

// encoder
function btoa(input) {
    var chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';
    var str = String (input);
    for (
        // initialize result and counter
        var block, charCode, idx = 0, map = chars, output = '';
        // if the next str index does not exist:
        //   change the mapping table to "="
        //   check if d has no fractional digits
        str.charAt (idx | 0) || (map = '=', idx % 1);
        // "8 - idx % 1 * 8" generates the sequence 2, 4, 6, 8
        output += map.charAt (63 & block >> 8 - idx % 1 * 8)
    ) {
        charCode = str.charCodeAt (idx += 3 / 4);
        if (charCode > 0xFF) {
            throw new InvalidCharacterError ("'btoa' failed: The string to be encoded contains characters outside of the Latin1 range.");
        }
        block = block << 8 | charCode;
    }
    return output;
}