grep -r (recursive), remove/hide all directories

I think with the default capabilities of grep there in no way of doing this.

You could go with something like this, which is just a "small" regex:

grep -r asdf | sed '#^.*/##' | sort --unique

Note: This approach will not work if the search-pattern contains a /

Try this,

grep -r abcd | awk -F ':' '{gsub(/.*\//,"",$1) }1' | awk '!seen[$1]++'

• gsub will remove directory structure. (/.*\//,"",$1) will remove("") all(.*) before the last match of '/' in first field($1).
• !seen[$1]++ will uniq the filename.

Note: directory name must not have : in it.

This groups by basename and grep output:

   ]# grep -ro '#include' include/ |sed -E 's|.*/(.*:)|\1|' |uniq -c |sort|tail -n7
         28 kvm_host.h:#include
         28 mm.h:#include
         29 ib_verbs.h:#include
         31 net_namespace.h:#include
         32 sock.h:#include
         44 fs.h:#include
         48 drmP.h:#include

I used grep -o to get some duplicates. Same time it leaves out the slashes...

If the names contain : the sed will not work correctly. The regex first throws away everything until the last /, then stores everything until a : as \1.

I used -E because of the (subexpression) and | because of the slash.

The subexpression (.*:) is a bit simple (will fail if a grepped line contains a colon). If you leave out the colon, it will fail when the line contains a slash.

Looking at this output I say this is impossible in theory (to parse grep's output in that way):

]# grep -r "" d*
d:/ir:/afile...in file "d"
d:/ir:/afile...in file "ir"

This is identical. I needed a dir with a colon at the end and a file with overlapping name and contents.

]# ls d* 
d

'd:':
ir

grep --color makes the difference!

The include directory is the one from the linux kernel source. One full line in one include-file looks like this.

]# grep -rH '#incl' include/linux/aio.h 
include/linux/aio.h:#include <linux/aio_abi.h>

With perl -pe 's|.*?([^/]*:)|\1|' it seems to work with colons and slashes in the files (content, not name!). Two lines below have both.

]# grep -r ':' include |head -n3 | perl -pe 's|.*?([^/]*:)|\1|'
cistpl.h:/* SPDX-License-Identifier: GPL-2.0-only */
ss.h:/* SPDX-License-Identifier: GPL-2.0-only */
ss.h: * are three options:

Is this mind-bending? The idea seems simple:

Keep all non-slash chars in front of the first colon.

The regex |.*?([^/]*:)| has these parts:

• .*? match anything, but as little as possible, then...
• ([^/]*:) this subexpression which is...
• [^/]*: as much non-slash as possible, and then colon.

At least it bends in the right direction.

With output from grep --null, the following GNU awk program should work with any filename:

BEGIN {
    # OFS will be printed between
    # each filename and matched line
        OFS = ":"

    # Use null-byte as a record separator
    # to use with output from grep --null
        RS = "\0"

    # Apart from the first record,
    # everything up to the first newline
    # of a record is the matched line from grep
    # Everything after first newline is a filename
        FPAT = "^[^\n]*|\n.*$"

}
NR == 1 {
    # Entire first record
    # is the first filename
    # set `file` to the basename
        file = gensub(".*/","",1)
        next
}
! seen[file, $1]++ {
    # If filename+match
    # not seen, print it
        print file, $1
}
{
    # Get basename of file
    # from next match
        file = gensub(".*/","",1,$2)
}

grep --null -rF asdf . | awk -f see_above.gawk