Unix command to list the portion of the end of a log file from a line containing only hyphens to end of file

Question

I have a long log file where each entry begins with a line containg only hyphens.

Answer 1

You can do it with a shell script thus:

#!/bin/bash
if [[ -z "$1" ]] ; then
    echo Usage: $0 '<inputFile>'
    exit 1
fi
line=$(grep -n '^--*$' "$1" | tail -1 | sed 's/:.*//')
if [[ -z "${line}" ]] ; then
    cat "$1"
else
    sed "1,${line}d" "$1"
fi

Given the input file:

this is line 1
-------
this is line 3
-------
this is line 5
this is line 6

it produces:

this is line 5
this is line 6

By way of explanation, the grep -n produces a series of lines like:

2:-------
4:-------

where the 2 and 4 are the line numbers. The tail -1 then just filters out all but the last and the sed strips out everything from the colon to the end of the line, leaving just the line number

Then, if there was no lines with the desired pattern, it just outputs the entire file. Otherwise it deletes all the lines between 1 and the last hyphen line.

---

As an aside, my original answer included this awk snippet which will process the file only once:

awk '/^--*$/{s=""}{s=s$0"\n";}END{print s}'

However, keep in mind that it works by accumulating lines into a string and clearing the string out whenever it finds a hyphen line. Then, at the end, it simply outputs the string (all the lines after the last hyphen line).

While at first glance, this may appear to be more efficient, it doesn't seem to be in reality. In (admittedly non-exhaustive) tests on my system, it actually ran quite a bit slower, I think to do with the many string appends going on. The fact is that the script solution seems to be faster despite the fact that it makes multiple passes of the data (possibly because each pass is very limited in what it does).

Answer 2

awk -vRS="-+" 'END{print}' ORS="" file

Answer 3

You can also do it with sed:

% cat t.txt
this is line 1
this is line 2
-------
this is line 3
----
this is line 4
-------
this is line 5
this is line 6
% sed -n -e '/^---*/{h;d;}' -e H -e '${g;p;}' t.txt
-------
this is line 5
this is line 6
% 

(with some seds, those semicolons would have to be newlines).

Answer 4

I think this can be easily done using sed. You want a command to find the final (i.e. last) line of only-hyphens, and you want to print from that point to the end of file.

Unfortunately, I'm not very good with sed. Hoping someone else can elaborate.

---

EDIT

OK, sed is not ideal. Here's how to do it with ex, the text-only twin of vi:

ex filename
$
?----------
.,$p
q

Answer 5

tac file | grep -B 10000 -m 1 -- '------' | tac

Answer 6

This is probably not the most efficient solution:

#!/bin/bash

file=$1
pattern='^-+$'
declare -i count=0
declare -i index=0

while read -r line
do
    count+=1
    [[ $line =~ $pattern ]] && index=$count
done < "$file"

tail -n "$((count - index))" "$file"

Answer 7

Use tac and sed:

$ cat log-file 
---
first
------
second
---
last

$ tac log-file | sed -e '/^-\+$/,$d' | tac
last

Answer 8

echo "`sed -n '/^--*$/=' <file> | tail -1`,\$p" <file>  | xargs sed -n

But I like Norman Gray's solution much better. May like it even more if he explained it :-)