js实现基于Base64的编码及解码

转载于： js实现基于Base64的编码及解码_xiaoxu的博客-CSDN博客

代码如下：

const Base64 = {
  _keyStr: 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=',
  encode: function (e) {
    let t = ''
    let n, r, i, s, o, u, a
    let 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) {
    let t = ''
    let n, r, i
    let s, o, u, a
    let 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')
    let t = ''
    for (let n = 0; n < e.length; n++) {
      const 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) {
    let t = ''
    let n = 0
    let r = 0
    let c2 = 0
    let c3 = 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
  }
}