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替换文件中的时间戳 Base64 值

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替换文件中的时间戳 Base64 值

我有一个问题，我有一个文件 test.txt，其内容如下：

dn: serv=CSPS,mscId=167e48dc2b7a42d4acce611c8b477262,ou=multiSCs,dc=three
structuralObjectClass: CP1
objectClass: CP1
objectClass: CUDBServiceAuxiliary
objectClass: CP2
objectClass: CP3
objectClass: CP4
objectClass: CP5
objectClass: CP6
UNKNLOCDATECS:: FQsJ
UNKNLOCDATEPS:: FgMe
ISTTIMESTAMP:: FgMIDyI7
CSULTIME:: HgMWCzYo
CSLOCTIME:: AQQWBA0R
PSULTIME:: HgMWDBco
PSLOCTIME:: HgMWDBco
SCHAR:: AgA=
ICS: 1
CAT: 10
DBSG: 1
OFA: 1
SOCB: 1
PWD: 0000
PWDC: 0
SOCFB: 0

每次找到文本 CSULTIME:: 和 CSLOCTIME:: 时，我想用以下函数替换这些文字后面的值，以将该时间戳解码为可识别的格式（如果我可以在一次文件扫描中替换这两个值，效果更好）我们讨论的是 8 GB 文件，两种情况下的功能是相同的）：

base64 -d | hexdump -v -e '1/1 "%02d" ' | awk 'BEGIN {FS = ""} {print "20" $5 $6 "-" $3 $4 "-" $1 $2 " " $7 $8 ":" $9 $10 ":" $11 $12}'

如果我在 unix 中对这两个值进行 echo：

For CSULTIME the result would be 2022-03-30 11:54:40: echo -n "HgMWCzYo" | base64 -d | hexdump -v -e '1/1 "%02d" ' | awk 'BEGIN {FS = ""} {print "20" $5 $6 "-" $3 $4 "-" $1 $2 " " $7 $8 ":" $9 $10 ":" $11 $12}'
For CSLOCTIME the result would be 2022-04-01 04:13:17: echo -n "AQQWBA0R" | base64 -d | hexdump -v -e '1/1 "%02d" ' | awk 'BEGIN {FS = ""} {print "20" $5 $6 "-" $3 $4 "-" $1 $2 " " $7 $8 ":" $9 $10 ":" $11 $12}'

因此，文件最后将具有以下 CSULTIME 和 CSLOCTIME 值：

dn: serv=CSPS,mscId=167e48dc2b7a42d4acce611c8b477262,ou=multiSCs,dc=three
    structuralObjectClass: CP1
    objectClass: CP1
    objectClass: CUDBServiceAuxiliary
    objectClass: CP2
    objectClass: CP3
    objectClass: CP4
    objectClass: CP5
    objectClass: CP6
    UNKNLOCDATECS:: FQsJ
    UNKNLOCDATEPS:: FgMe
    ISTTIMESTAMP:: FgMIDyI7
    CSULTIME:: 2022-03-30 11:54:40
    CSLOCTIME:: 2022-04-01 04:13:17
    PSULTIME:: HgMWDBco
    PSLOCTIME:: HgMWDBco
    SCHAR:: AgA=
    ICS: 1
    CAT: 10
    DBSG: 1
    OFA: 1
    SOCB: 1
    PWD: 0000
    PWDC: 0
    SOCFB: 0

我完全迷失了，因为我尝试的 sed 的所有组合都无法使它们工作。

提前致谢！！！！

答案1

我会做类似的事情：

perl -MMIME::Base64 -pe 's{^[^:]*TIME:\K: (\S+)}{
  my ($d, $m, $y, @t) = unpack "C*", decode_base64 $1;
  sprintf "20%02d-%02d-%02d %02d:%02d:%02d", $y, $m, $d, @t}e'

答案2

您可以使用以下方法，我们首先设置用于 Base64 解码的设备，然后使用 GNU sed 处理编码数据。

#------------------
#  base64 in sed
#------------------
set -u

#> present century
century=$(date '+%C')

#> format string printf
fmt='%s%s\n'

#> bit, sextet, & octet regex
bit='[01]'
sextet="${bit}{6}"
 octet="${bit}{8}"; byte=$octet

#> base64 charset
b64='/[:alnum:]+'
declare -a b64_chars=({A..Z} {a..z} {0..9} + /)

#> user-defined helper function(s)
oneLine() {
  # collapse stdin into one line
  paste -sd'\0' -
}

esc_rhs() {
  # make stdin pkuggable
  # on the rhs of a s///
  sed -e '
    s:[\/&]:\\&:g
    $!s:$:\\:
  ' -
}

dec2bin() {
  # create an array of size 2^$1
  # Usage: dec2bin 6
  # creates array d2b as:
  # underscore for clarity only
  # $d2b[0] => "000_000"
  # $d2b[1] => "000_001"
  #        ...
  # $d2b[63] => "111_111"

  eval "d2b=($(yes '{0,1}' | sed "$1q" | oneLine))"
}

#> build the encoding lookup table
encode_tbl=$(printf '%s\n' "$(
dec2bin 6
i=0
for c in "${b64_chars[@]}"
do
  printf "$fmt" "$c" "${d2b[$i]}"
  (( i++ ))
done
)" | esc_rhs)

#> build the decoding lookup table
decode_tbl=$(printf '%s\n' "$(
dec2bin 8
for dec in {0..255}
do
  hex=$(printf '%02x' "$dec")
  bin=${d2b[$dec]}
  printf "$fmt" "$bin" "$hex"
done
)" | esc_rhs)

#> hex to bin
h2d=$(for i in {0..255};do
  printf 'x%02x:%02d\n' "$i" "$i"
done | oneLine | esc_rhs)

#######> main()
sed -E "
  s/^\s*(CSULTIME|CSLOCTIME)::\s*/&\n/;T
  h;s/\n.*//
  x;s/.*\n//
  s/[^${b64}]//g
  /^\$/d;s/^|\$/\n/g

  # unencode
  s/\$/$encode_tbl/
  :encode
  /\n\n/!{
    s/((\n)[${b64}])(.*\1(${sextet}))/\4\2\3/
    b encode
  }

  s/\n.*//;:pad
  /^(${octet})+\$/!{
    s/\$/0/;b pad
  }
  # chunk it in 8-bit portions
  s/(${octet})/& /g
  G;s/\$/${decode_tbl}/;t decode
  :decode
    s/(${octet}) (.*\n\1([[:xdigit:]]{2}))/\3\2/
  t decode
  s/\n.*//

  #> hex to decimal
  s/^|\$/\n/g
  s/\$/$h2d/;t hex2int
  :hex2int
  s/\n([[:xdigit:]]{2})(.*\n.*x\1:([[:digit:]]+))/\3\n\2/
  t hex2int
  s/\n.*//

  #> rearrange decoded value into
  #> yyy-mm-dd hh:mm:ss format
  s/(..)(..)(..)/\3\2\1/
  s/../&-/g
  s/(.*)./$century\1/
  s/-/:/3g
  s/:/ /

  #> stitch back the prefixes
  #> CSULTIME, CSLOCTIME
  H;z;x;s/\n//
" file

输出：-

-1e-0===
dn: serv=CSPS,mscId=167e48dc2b7a42d4acce611c8b477262,ou=multiSCs,dc=three
structuralObjectClass: CP1
objectClass: CP1
objectClass: CUDBServiceAuxiliary
objectClass: CP2
objectClass: CP3
objectClass: CP4
objectClass: CP5
objectClass: CP6
UNKNLOCDATECS:: FQsJ
UNKNLOCDATEPS:: FgMe
ISTTIMESTAMP:: FgMIDyI7
CSULTIME:: 2022-03-30 11:54:40
CSLOCTIME:: 2022-04-01 04:13:17
PSULTIME:: HgMWDBco
PSLOCTIME:: HgMWDBco
SCHAR:: AgA=
ICS: 1
CAT: 10
DBSG: 1
OFA: 1
SOCB: 1
PWD: 0000
PWDC: 0
SOCFB: 0

答案3

我明白了，我做得对（使用 sed，不知道如何在 perl 中使用 hexdump），问题是我需要在 sed 内转义的字符。

现在它可以完美地工作，尽管我必须执行两个 sed：

sed  's/\(CSULTIME::\)\(.*\)/echo -n \1" ";echo \2\| base64 -d \| hexdump -v -e '\''1\/1 "%02d" '\'' \| awk '\''BEGIN {FS = ""} {print "20" $5 $6 "-" $3 $4 "-" $1 $2 " " $7 $8 ":" $9 $10 ":" $11 $12}'\'' /ge ' 
sed  's/\(CSLOCTIME::\)\(.*\)/echo -n \1" ";echo \2\| base64 -d \| hexdump -v -e '\''1\/1 "%02d" '\'' \| awk '\''BEGIN {FS = ""} {print "20" $5 $6 "-" $3 $4 "-" $1 $2 " " $7 $8 ":" $9 $10 ":" $11 $12}'\'' /ge '

#------------

您可以在 bash 函数中分解代码，将其导出，然后在 GNU sed 的 ss///e 中调用它

fx() {
  printf '%s' "$2" | base64 -d |
  hexdump -v -e '1/1 "%02d" ' |
  awk -v temp="$1" 'BEGIN {FS = ""} {print temp "20" $5 $6 "-" $3 $4 "-" $1 $2 " " $7 $8 ":" $9 $10 ":" $11 $12}'
}

export -f fx

sed -E "s/((CSULTIME|CSLOCTIME)::\s*)(\S.*)/fx '\1' '\3'/e" file

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