编码和解码跳过字符

问题描述 投票：0回答：3

我正在尝试使用一些字符串的编码格式将它们存储到数据库中。但是当检索回来时，字符串格式错误。

这是我的代码示例，您可以轻松地看到传递给编码的字符串与解码后的字符串不同。为什么会出现这种情况？

还有其他库可以帮助我编码和解码吗？任何有关相同的建议都会有帮助。

var Base64 = {
    _keyStr : "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=",
    encode : function(e) {
        var t = "";
        var n, r, i, s, o, u, a;
        var f = 0;
        e = Base64._utf8_encode(e);
        while (f < e.length) {
            n = e.charCodeAt(f++);
            r = e.charCodeAt(f++);
            i = e.charCodeAt(f++);
            s = n >> 2;
            o = (n & 3) << 4 | r >> 4;
            u = (r & 15) << 2 | i >> 6;
            a = i & 63;
            if (isNaN(r)) {
                u = a = 64
            } else if (isNaN(i)) {
                a = 64
            }
            t = t + this._keyStr.charAt(s) + this._keyStr.charAt(o)
                    + this._keyStr.charAt(u) + this._keyStr.charAt(a)
        }
        return t
    },
    decode : function(e) {
        var t = "";
        var n, r, i;
        var s, o, u, a;
        var f = 0;
        e = e.replace(/[^A-Za-z0-9+/=]/g, "");
        while (f < e.length) {
            s = this._keyStr.indexOf(e.charAt(f++));
            o = this._keyStr.indexOf(e.charAt(f++));
            u = this._keyStr.indexOf(e.charAt(f++));
            a = this._keyStr.indexOf(e.charAt(f++));
            n = s << 2 | o >> 4;
            r = (o & 15) << 4 | u >> 2;
            i = (u & 3) << 6 | a;
            t = t + String.fromCharCode(n);
            if (u != 64) {
                t = t + String.fromCharCode(r)
            }
            if (a != 64) {
                t = t + String.fromCharCode(i)
            }
        }
        t = Base64._utf8_decode(t);
        return t
    },
    _utf8_encode : function(e) {
        e = e.replace(/rn/g, "n");
        var t = "";
        for (var n = 0; n < e.length; n++) {
            var r = e.charCodeAt(n);
            if (r < 128) {
                t += String.fromCharCode(r)
            } else if (r > 127 && r < 2048) {
                t += String.fromCharCode(r >> 6 | 192);
                t += String.fromCharCode(r & 63 | 128)
            } else {
                t += String.fromCharCode(r >> 12 | 224);
                t += String.fromCharCode(r >> 6 & 63 | 128);
                t += String.fromCharCode(r & 63 | 128)
            }
        }
        return t
    },
    _utf8_decode : function(e) {
        var t = "";
        var n = 0;
        var r = c1 = c2 = 0;
        while (n < e.length) {
            r = e.charCodeAt(n);
            if (r < 128) {
                t += String.fromCharCode(r);
                n++
            } else if (r > 191 && r < 224) {
                c2 = e.charCodeAt(n + 1);
                t += String.fromCharCode((r & 31) << 6 | c2 & 63);
                n += 2
            } else {
                c2 = e.charCodeAt(n + 1);
                c3 = e.charCodeAt(n + 2);
                t += String.fromCharCode((r & 15) << 12 | (c2 & 63) << 6 | c3
                        & 63);
                n += 3
            }
        }
        return t
    }
}


var str = "background:url(/XYX/player_assets/skin_0/XYXA_default_skinRightCorner.png) ;"

var encoded = Base64.encode(str);
//console.log(encoded);

var decoded = Base64.decode(encoded);
console.log(str,"......Input");
console.log(decoded,".....Output");

javascript html base64 decode encode
3个回答
2
投票

常见浏览器中的Base64编码

在JavaScript中有两个函数分别用于解码和编码base64字符串：

atob()

btoa()

atob()
函数对使用base-64编码的数据字符串进行解码。相反，
btoa()
函数从二进制数据“字符串”创建 Base-64 编码的 ASCII 字符串。

1
投票

使用 

atob
和 
btoa
。

const foo = "bar"

const encodedFoo = btoa(foo)
const decodedFoo = atob(encodedFoo)

console.log(encodedFoo)
console.log(decodedFoo)

您可以在这里阅读更多相关信息。

1
投票
这里错了（_utf8_encode）：


e = e.replace(/rn/g, "n");


正确如下：


e = e.replace(/\r\n/g, "\n");






var Base64 = {
    _keyStr : "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=",
    encode : function(e) {
        var t = "";
        var n, r, i, s, o, u, a;
        var f = 0;
        e = Base64._utf8_encode(e);
        while (f < e.length) {
            n = e.charCodeAt(f++);
            r = e.charCodeAt(f++);
            i = e.charCodeAt(f++);
            s = n >> 2;
            o = (n & 3) << 4 | r >> 4;
            u = (r & 15) << 2 | i >> 6;
            a = i & 63;
            if (isNaN(r)) {
                u = a = 64
            } else if (isNaN(i)) {
                a = 64
            }
            t = t + this._keyStr.charAt(s) + this._keyStr.charAt(o)
                    + this._keyStr.charAt(u) + this._keyStr.charAt(a)
        }
        return t
    },
    decode : function(e) {
        var t = "";
        var n, r, i;
        var s, o, u, a;
        var f = 0;
        e = e.replace(/[^A-Za-z0-9+/=]/g, "");
        while (f < e.length) {
            s = this._keyStr.indexOf(e.charAt(f++));
            o = this._keyStr.indexOf(e.charAt(f++));
            u = this._keyStr.indexOf(e.charAt(f++));
            a = this._keyStr.indexOf(e.charAt(f++));
            n = s << 2 | o >> 4;
            r = (o & 15) << 4 | u >> 2;
            i = (u & 3) << 6 | a;
            t = t + String.fromCharCode(n);
            if (u != 64) {
                t = t + String.fromCharCode(r)
            }
            if (a != 64) {
                t = t + String.fromCharCode(i)
            }
        }
        t = Base64._utf8_decode(t);
        return t
    },
    _utf8_encode : function(e) {
        e = e.replace(/\r\n/g, "\n");
        var t = "";
        for (var n = 0; n < e.length; n++) {
            var r = e.charCodeAt(n);
            if (r < 128) {
                t += String.fromCharCode(r)
            } else if (r > 127 && r < 2048) {
                t += String.fromCharCode(r >> 6 | 192);
                t += String.fromCharCode(r & 63 | 128)
            } else {
                t += String.fromCharCode(r >> 12 | 224);
                t += String.fromCharCode(r >> 6 & 63 | 128);
                t += String.fromCharCode(r & 63 | 128)
            }
        }
        return t
    },
    _utf8_decode : function(e) {
        var t = "";
        var n = 0;
        var r = c1 = c2 = 0;
        while (n < e.length) {
            r = e.charCodeAt(n);
            if (r < 128) {
                t += String.fromCharCode(r);
                n++
            } else if (r > 191 && r < 224) {
                c2 = e.charCodeAt(n + 1);
                t += String.fromCharCode((r & 31) << 6 | c2 & 63);
                n += 2
            } else {
                c2 = e.charCodeAt(n + 1);
                c3 = e.charCodeAt(n + 2);
                t += String.fromCharCode((r & 15) << 12 | (c2 & 63) << 6 | c3
                        & 63);
                n += 3
            }
        }
        return t
    }
}


var str = "background:url(/xyx-courses/player_assets/skin_0/XYX_default_skinRightCorner.png) ;"

var encoded = Base64.encode(str);
//console.log(encoded);

var decoded = Base64.decode(encoded);
console.log(str,"......Input");
console.log(decoded,".....Output");






另外，我认为这里根本不需要这行
e = e.replace(/\r\n/g, "\n");
，因为任务是将字符串作为utf8编码为base64，而不是将额外的
CRLF
转换为
LF
（换行符）

    



      

      
    

  

  

    
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