[bash] How to get the first line of a file in a bash script?

I have to put in a bash variable the first line of a file. I guess it is with the grep command, but it is any way to restrict the number of lines?

The question didn't ask which is fastest, but to add to the sed answer, -n '1p' is badly performing as the pattern space is still scanned on large files. Out of curiosity I found that 'head' wins over sed narrowly:

# best:
head -n1 $bigfile >/dev/null

# a bit slower than head (I saw about 10% difference):
sed '1q' $bigfile >/dev/null

# VERY slow:
sed -n '1p' $bigfile >/dev/null

to read first line using bash, use read statement. eg

read -r firstline<file

firstline will be your variable (No need to assign to another)

line=$(head -1 file)

Will work fine. (As previous answer). But

line=$(read -r FIRSTLINE < filename)

will be marginally faster as read is a built-in bash command.

This suffices and stores the first line of filename in the variable $line:

read -r line < filename

I also like awk for this:

awk 'NR==1 {print; exit}' file

To store the line itself, use the var=$(command) syntax. In this case, line=$(awk 'NR==1 {print; exit}' file).

Or even sed:

sed -n '1p' file

With the equivalent line=$(sed -n '1p' file).

See a sample when we feed the read with seq 10, that is, a sequence of numbers from 1 to 10:

$ read -r line < <(seq 10) 
$ echo "$line"
1

$ line=$(awk 'NR==1 {print; exit}' <(seq 10))
$ echo "$line"
1

Just echo the first list of your source file into your target file.

echo $(head -n 1 source.txt) > target.txt