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Java Map小结以及前后端加密传输

Java Map小结以及前后端加密传输

作者: 长弘羲雨 | 来源:发表于2020-05-19 16:01 被阅读0次

    1、通过采用 LinkedHashMap 将 list 中按照指定的 key 进行分组排序(HashMap 会乱序)

    public LinkedHashMap<String,List<XqPersonTransfer>> qryHistoryEventList() {
            LinkedHashMap<String, List<XqPersonTransfer>> resultMap = new LinkedHashMap<>(100);
    
            List<XqPersonTransfer> resultList = xqPersonTransferMapper.qryHistoryEventList();
            if(resultList != null && resultList.size()>0){
                try{
                   for(XqPersonTransfer xqPersonTransfer : resultList){
                        //map中eventId已存在,将该数据存放到同一个key(key指的就是eventId)的map中
                        if(resultMap.containsKey(xqPersonTransfer.getEventId().toString())){
                            resultMap.get(xqPersonTransfer.getEventId().toString()).add(xqPersonTransfer);
                        }else{//map中不存在,新建key,用来存放数据
                            List<XqPersonTransfer> tmpList = new ArrayList<XqPersonTransfer>();
                            tmpList.add(xqPersonTransfer);
                            resultMap.put(xqPersonTransfer.getEventId().toString(), tmpList);
    
                        }
                    }
    
    
                }catch(Exception e){
                    try {
                        throw new Exception("按照eventID对数据进行分组时出现异常", e);
                    } catch (Exception e1) {
                        logger.error(e.getMessage(),e);
                    }
                }
            }
            return resultMap;
        }
    

    2、关于避免抓包前后端加密传输一些敏感信息

    • 前端 JS base64 加密算法
    var keyStr =
      "ABCDEFGHIJKLMNOP" +
      "QRSTUVWXYZabcdef" +
      "ghijklmnopqrstuv" +
      "wxyz0123456789+/" +
      "=";
    function encode64(input) {
      var output = "";
      var chr1,
        chr2,
        chr3 = "";
      var enc1,
        enc2,
        enc3,
        enc4 = "";
      var i = 0;
      do {
        chr1 = input.charCodeAt(i++);
        chr2 = input.charCodeAt(i++);
        chr3 = input.charCodeAt(i++);
        enc1 = chr1 >> 2;
        enc2 = ((chr1 & 3) << 4) | (chr2 >> 4);
        enc3 = ((chr2 & 15) << 2) | (chr3 >> 6);
        enc4 = chr3 & 63;
        if (isNaN(chr2)) {
          enc3 = enc4 = 64;
        } else if (isNaN(chr3)) {
          enc4 = 64;
        }
        output =
          output +
          keyStr.charAt(enc1) +
          keyStr.charAt(enc2) +
          keyStr.charAt(enc3) +
          keyStr.charAt(enc4);
        chr1 = chr2 = chr3 = "";
        enc1 = enc2 = enc3 = enc4 = "";
      } while (i < input.length);
    
      return output;
    }
    
    • 后端 Java 解密算法
    public static String decode(String str) {
            byte[] base64DecodeChars = new byte[] { -1, -1, -1, -1, -1,
                    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
                    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
                    -1, -1, -1, -1, 62, -1, -1, -1, 63, 52, 53, 54, 55, 56, 57, 58, 59,
                    60, 61, -1, -1, -1, -1, -1, -1, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
                    10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, -1,
                    -1, -1, -1, -1, -1, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37,
                    38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, -1, -1, -1,
                    -1, -1 };
    
            byte[] data = str.getBytes();
            int len = data.length;
            ByteArrayOutputStream buf = new ByteArrayOutputStream(len);
            int i = 0;
            int b1, b2, b3, b4;
    
            while (i < len) {
                do {
                    b1 = base64DecodeChars[data[i++]];
                } while (i < len && b1 == -1);
                if (b1 == -1) {
                    break;
                }
    
                do {
                    b2 = base64DecodeChars[data[i++]];
                } while (i < len && b2 == -1);
                if (b2 == -1) {
                    break;
                }
                buf.write((int) ((b1 << 2) | ((b2 & 0x30) >>> 4)));
    
                do {
                    b3 = data[i++];
                    if (b3 == 61) {
                        return new String(buf.toByteArray());
                    }
                    b3 = base64DecodeChars[b3];
                } while (i < len && b3 == -1);
                if (b3 == -1) {
                    break;
                }
                buf.write((int) (((b2 & 0x0f) << 4) | ((b3 & 0x3c) >>> 2)));
    
                do {
                    b4 = data[i++];
                    if (b4 == 61) {
                        return new String(buf.toByteArray());
                    }
                    b4 = base64DecodeChars[b4];
                } while (i < len && b4 == -1);
                if (b4 == -1) {
                    break;
                }
                buf.write((int) (((b3 & 0x03) << 6) | b4));
            }
            return new String(buf.toByteArray());
        }
    

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