Print second last line from variable in bash - linux

VAR="1\n2\n3"
I'm trying to print out the second last line. One liner in bash!
I've gotten so far: printf -- "$VAR" | head -2
It however prints out too much.
I can do this with a file no problem: tail -2 ~/file | head -1


You almost done this task by yourself. Try
VAR="1\n2\n3"; printf -- "$VAR"|tail -2|head -1


Here is one pure bash way of doing this:
readarray -t arr < <(printf -- "$VAR") && echo "${arr[-2]}"
2
You may also use this awk as a single command:
VAR="1\n2\n3"
awk -F '\\\\n' '{print $(NF-1)}' <<< "$VAR"
2


maybe more efficient using a temporary variable and using expansions
var=$'1\n2\n3' ; tmpvar=${var%$'\n'*} ; echo "${tmpvar##*$'\n'}"


Use echo -e for backslash interpretation and to translate \n to newlines and print the interested line number using NR.
$ echo -e "${VAR}" | awk 'NR==2'
2
With multiple lines and do, tail and head can be used to print any particular line number.
$ echo -e "$VAR" | tail -2 | head -1
2
or do a fancy sed, where you keep the previous line in the buffer-space (x) to print and keep deleting until the last line,
$ echo -e "$VAR" | sed 'x;$!d'
2

Related

Linux: Extract string from a line including delimiter character using sed command [duplicate]

For example
echo "abc-1234a :" | grep <do-something>
to print only abc-1234a

I think these are closer to what you're getting at, but without knowing what you're really trying to achieve, it's hard to say.
echo "abc-1234a :" | egrep -o '^[^:]+'
... though this will also match lines that have no colon. If you only want lines with colons, and you must use only grep, this might work:
echo "abc-1234a :" | grep : | egrep -o '^[^:]+'
Of course, this only makes sense if your echo "abc-1234a :" is an example that would be replace with possibly multiple lines of input.
The smallest tool you could use is probably cut:
echo "abc-1234a :" | cut -d: -f1
And sed is always available...
echo "abc-1234a :" | sed 's/ *:.*//'
For this last one, if you only want to print lines that include a colon, change it to:
echo "abc-1234a :" | sed -ne 's/ *:.*//p'
Heck, you could even do this in pure bash:
while read line; do
field="${line%%:*}"
# do stuff with $field
done <<<"abc-1234a :"
For information on the %% bit, you can man bash and search for "Parameter Expansion".
UPDATE:
You said:
It's the characters in the first line of input before the colon. The
input could have multiple line though.
The solutions with grep probably aren't your best choice, then, since they'll also print data from subsequent lines that might include colons. Of course, there are many ways to solve this requirement as well. We'll start with sample input:
$ function sample { printf "abc-1234a:foo\nbar baz:\nNarf\n"; }
$ sample
abc-1234a:foo
bar baz:
Narf
You could use multiple pipes, for example:
$ sample | head -1 | grep -Eo '^[^:]*'
abc-1234a
$ sample | head -1 | cut -d: -f1
abc-1234a
Or you could use sed to process only the first line:
$ sample | sed -ne '1s/:.*//p'
abc-1234a
Or tell sed to exit after printing the first line (which is faster than reading the whole file):
$ sample | sed 's/:.*//;q'
abc-1234a
Or do the same thing but only show output if a colon was found (for safety):
$ sample | sed -ne 's/:.*//p;q'
abc-1234a
Or have awk do the same thing (as the last 3 examples, respectively):
$ sample | awk '{sub(/:.*/,"")} NR==1'
abc-1234a
$ sample | awk 'NR>1{nextfile} {sub(/:.*/,"")} 1'
abc-1234a
$ sample | awk 'NR>1{nextfile} sub(/:.*/,"")'
abc-1234a
Or in bash, with no pipes at all:
$ read line < <(sample)
$ printf '%s\n' "${line%%:*}"
abc-1234a

It is possible to do what you want with only sed.
Here is an example:
#!/bin/sh
filename=$1
pattern=yourpattern
# flag -n disables print everyline (default behavior)
sed -n "
1,/$pattern/ {
/$pattern/n # skip line containing pattern
p # print lines ranging from line 1 untill pattern
}
" $filename
exit 0
This works at least for GNU's sed. It should work for other sed too, except
regarding the comments (some implementations of sed don't support comments).
Source: https://www.grymoire.com/Unix/Sed.html

concatenate the result of echo and a command output

I have the following code:
names=$(ls *$1*.txt)
head -q -n 1 $names | cut -d "_" -f 2
where the first line finds and stores all names matching the command line input into a variable called names, and the second grabs the first line in each file (element of the variable names) and outputs the second part of the line based on the "_" delim.
This is all good, however I would like to prepend the filename (stored as lines in the variable names) to the output of cut. I have tried:
names=$(ls *$1*.txt)
head -q -n 1 $names | echo -n "$names" cut -d "_" -f 2
however this only prints out the filenames
I have tried
names=$(ls *$1*.txt
head -q -n 1 $names | echo -n "$names"; cut -d "_" -f 2
and again I only print out the filenames.
The desired output is:
$
filename1.txt <second character>
where there is a single whitespace between the filename and the result of cut.
Thank you.

Best approach, using awk
You can do this all in one invocation of awk:
awk -F_ 'NR==1{print FILENAME, $2; exit}' *"$1"*.txt
On the first line of the first file, this prints the filename and the value of the second column, then exits.
Pure bash solution
I would always recommend against parsing ls - instead I would use a loop:
You can avoid the use of awk to read the first line of the file by using bash built-in functionality:
for i in *"$1"*.txt; do
IFS=_ read -ra arr <"$i"
echo "$i ${arr[1]}"
break
done
Here we read the first line of the file into an array, splitting it into pieces on the _.

Maybe something like that will satisfy your need BUT THIS IS BAD CODING (see comments):
#!/bin/bash
names=$(ls *$1*.txt)
for f in $names
do
pattern=`head -q -n 1 $f | cut -d "_" -f 2`
echo "$f $pattern"
done

If I didn't misunderstand your goal, this also works.
I've always done it this way, I just found out that this is a deprecated way to do it.
#!/bin/bash
names=$(ls *"$1"*.txt)
for e in $names;
do echo $e `echo "$e" | cut -c2-2`;
done

Getting a specific line from a string where the line number I must get is stored in a variable?

I'm trying to get a specific line of a variable. The line I must get is stored in i. My code looks like this right now.
$(echo "$data" | sed '$iq;d')
It looks like I'm putting i in there wrong, Putting a number in for i works fine but $i gets me the entire string.
I haven't found a solution that works with a variable yet and I'm not too familiar with bash and would appreciate help,
Edit: a bit of context
i=5
data=$(netstat -a | grep ESTAB)
line=$(echo "$data" | sed "${i}p")
echo $line

Use sed -n "${i}p" instead.
Example:
i=4; seq 1 10 | sed -n "${i}p"
Output:
4
Bonus:
i=5
readarray -O 1 -t data < <(exec netstat -a | grep ESTAB) ## Stores data as an array of lines starting at index 1
line=${data[i]}
echo "$line"
# printf '%s\n' "${data[#]}" ## Prints whole data.

Here is way you can do this in BASH itself:
IFS=$'\n' arr=($data)
echo "${arr[$i]}"

print a line which has a digit repeated n times in the third field

I have a file with contents:
20120619112139,3,22222288100597,01,503352786544597,,W,ROAMER,,,,0,mme2
20120703112557,3,00000000000000,,503352786544021,,B,,8,2505,,U,
20120611171517,3,22222288100620,,503352786544620,11917676228846,B,ROAMER,8,2505,,U,
20120703112557,3,00000000000000,,503352786544021,,B,,8,2505,,U,
20120703112557,3,00000000000000,,503352786544021,,B,,8,2505,,U,
20120611171003,3,22222288100618,02,503352786544618,,W,ROAMER,8,2505,,0,
20120611171046,3,00000000000000,02,503352786544618,11917676228846,W,ROAMER,8,2505,,0,
20120611171101,3,22222288100618,02,503352786544618,11917676228846,W,ROAMER,8,2505,,0,
20120611171101,3,22222222222222,02,503352786544618,11917676228846,W,ROAMER,8,2505,,0,
I need to check if the third field of any line has one digit repeated all through 14 times, like:00000000000000 and print such lines to another file
I tried this code:
awk '$3 ~ /[0-9]{14}/' myfile > output.txt
But this prints lines having "22222288100618" such values as well.
Also i tried:
for i in `cat myfile`
do
if [ `echo $i | cut -d"," -f 3 | egrep "^[0-9]{14}$"` ];
then echo $i >> output.txt;
fi
done
This doesn't help as well.This also prints all the lines.
But I only need these lines in the output file.
20120703112557,3,00000000000000,,503352786544021,,B,,8,2505,,U,
20120703112557,3,00000000000000,,503352786544021,,B,,8,2505,,U,
20120703112557,3,00000000000000,,503352786544021,,B,,8,2505,,U,
20120611171046,3,00000000000000,02,503352786544618,11917676228846,W,ROAMER,8,2505,,0,
20120611171101,3,22222222222222,02,503352786544618,11917676228846,W,ROAMER,8,2505,,0,
Thanks in advance for any immediate help

Don't know if this can be done with awk but this should work:
perl -aF, -nle '$F[2]=~/(\d)\1{13}/&& print'

You can use an expression like 0{14}|1{14}.... Try this:
$ for i in 0 1 2 3 4 5 6 7 8 9; do re=$re${re:+|}$i{14}; done
$ awk -F, --posix \$3~/$re/ myfile
(gawk requires --posix to recognize the interval expression {14}. This may not be necessary with all awk.)

Using grep:
grep -E "[0-9]+,[0-9]+,([0-9])\1{13}" myfile

sed -n '/^[^,]+,[^,]+,([0-9])\1{13}/p' input_file

How to read the second-to-last line in a file using Bash?

I have a file that has the following as the last three lines. I want to retrieve the penultimate line, i.e. 100.000;8438; 06:46:12.
.
.
.
99.900; 8423; 06:44:41
100.000;8438; 06:46:12
Number of patterns: 8438
I don't know the line number. How can I retrieve it using a shell script? Thanks in advance for your help.

Try this:
tail -2 yourfile | head -1

A short sed one-liner inspired by https://stackoverflow.com/a/7671772/5287901
sed -n 'x;$p'
Explanation:
-n quiet mode: dont automatically print the pattern space
x: exchange the pattern space and the hold space (hold space now store the current line, and pattern space the previous line, if any)
$: on the last line, p: print the pattern space (the previous line, which in this case is the penultimate line).

Use this
tail -2 <filename> | head -1

ed and sed can do it as well.
str='
99.900; 8423; 06:44:41
100.000;8438; 06:46:12
Number of patterns: 8438
'
printf '%s' "$str" | sed -n -e '${x;1!p;};h' # print last line but one
printf '%s\n' H '$-1p' q | ed -s <(printf '%s' "$str") # same
printf '%s\n' H '$-2,$-1p' q | ed -s <(printf '%s' "$str") # print last line but two

From: Useful sed one-liners by Eric Pement
# print the next-to-the-last line of a file
sed -e '$!{h;d;}' -e x # for 1-line files, print blank line
sed -e '1{$q;}' -e '$!{h;d;}' -e x # for 1-line files, print the line
sed -e '1{$d;}' -e '$!{h;d;}' -e x # for 1-line files, print nothing
You don't need all of them, just pick one.

tail +2 <filename>
This prints from second line to last line.

To clarify what has already been said:
ec2thisandthat | sort -k 5 | grep 2012- | awk '{print $2}' | tail -2 | head -1
snap-e8317883
snap-9c7227f7
snap-5402553f
snap-3e7b2c55
snap-246b3c4f
snap-546a3d3f
snap-2ad48241
snap-d00150bb
returns
snap-2ad48241

tac <file> | sed -n '2p'

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