I have the following simple bash script which takes input from stdin and prints the third line given as input.
#!/bin/bash
var=$(cat)
echo $var | head -n 3 | tail -n 1
The problem with this script is that it prints all the lines but here is the funny part, when I type the commands individually on the command line I am getting the expected result i.e. the third line. Why this anomaly? Am I doing something wrong here?
The aim of head -n 3 | tail -n 1 is to keep the third line into variable
It will be more efficient to use read builtin
read
read
read var
echo "${var}"
Or to keep heading white-spaces
IFS= read
and not join lines ending with \ or not give special meaning to \
read -r
You don't need $(cat) in your script. If script is reading data from stdin then just have this single line in your script:
head -n 3 | tail -n 1
And run it as:
bash myscript.sh < file.txt
This will print 3rd line from file.txt
PS: You can replace head + tail with this faster sed to print 3rd line from input:
sed '3q;d'
The shell is splitting the var variable so echo get multiple parameters. You need to quote your variable to prevent this to happen:
#!/bin/bash
var=$(cat)
echo "$var" | head -n 3 | tail -n 1
This should do the trick, as far as I understand your question:
#!/bin/bash
var=$(cat)
echo "$var" | head -n 3 | tail -n 1
var=$(cat) will not allow you to escape out of stdin mode. you need to specify the EOF for the script to understand to stop reading from stdin.
read -d '' var << EOF
echo "$var" | head -n 3 | tail -n 1
echo "$var" line and delimited by end of file.
echo | head -n 3 | tail -n 1 into var, which is hardly what you want.$var.
catto input something and store into var ?sed -n '3{p;q}'or similar?