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Accepted
Sergio Jimenez
Asked: 2022-04-29 06:17:36 +0800 CST

替换文件中的时间戳 Base64 值

  • 772

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之后的值，以将该时间戳解码为识别格式（如果我可以在文件的单次扫描中替换这两个值更好作为我们谈论的是 8 GB 文件，两种情况下的功能相同）：

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 中对这两个值进行回显：

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，最后文件将具有 CSULTIME 和 CSLOCTIME 的这些值：

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 sed 的所有组合都无法使它们起作用。

提前致谢！！！！

sed base64

3 个回答

  1. Best Answer

    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'
    • 1

  2. 我明白了，我做对了（使用 sed，不知道如何在 perl 中使用 hexdump）问题是我需要在 sed 中转义的字符。

    现在它可以完美地工作，尽管我必须做两个 seds：

    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 '

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

    您可以将代码分解为 bash 函数，将其导出，然后在 GNU sed 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
    • 0

  3. 您可以使用以下方法，我们首先设置用于 base64 解码的工具，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
    • 0

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