Trying to get file with max lines to print with lines number - linux

Question

So i've been goofing with this since last night and I can get a lot of things to happen just not what I want.
I need a code to find the file with the most lines in a directory and then print the name of the file and the number of lines that file has.
I can get the entire directory's lines to print but can't seem to narrow the field so to speak.
Any help for a fool of a learner?
wc -l $1/* 2>/dev/null
| grep -v ' total$'
| sort -n -k1
| tail -1l
After some pro help in another question, this is where I got to, but it returns them all, and doesn't print their line counts.

Answer 1

Following awk command should do the job for you and you can avoid all redundant piped commands:
wc -l $1/* | awk '$2 != "total"{if($1>max){max=$1;fn=$2}} END{print max, fn}'
UPDATE: To avoid last line of wc's output this might be better awk command:
wc -l $1/* | awk '{arr[cnt++]=$0} END {for (i=0; i<length(arr)-1; i++)
{split(arr[i], a, " "); if(a[1]>max) {max=a[1]; fn=a[2]}} print max, fn}'

Answer 2

you can try:
wc -l $1/* | grep -v total | sort -g | tail -1
actually to avoid the grep that would also remove files containing "total":
for f in $1/*; do wc -l $f; done | sort -g | tail -1
or even better, as suggested in comments:
wc -l $1/* | sort -rg | sed -n '2p'
you can even make it a function:
function get_biggest_file() {
wc -l $* | sort -rg | sed -n '2p'
}
% ls -l
... 0 Jun 12 17:33 a
... 0 Jun 12 17:33 b
... 0 Jun 12 17:33 c
... 0 Jun 12 17:33 d
... 25 Jun 12 17:33 total
% get_biggest_file ./*
5 total
EDIT2: using the function I gave, you can simply output what you need as follows:
get_biggest $1/* | awk '{print "The file \"" $2 "\" has the maximum number of lines: " $1}'
EDIT: if you tried to write the function as you've written it in the question, you should add line continuation character at the end, as follows, or your shell will think you're trying to issue 4 commands:
wc -l $1/* 2>/dev/null \
| grep -v ' total$' \
| sort -n -k1 \
| tail -1l