Compare text in Unix

Question

I am trying to compare two text in Unix. I tried the below. It didn't work. In need to compare the first and last lines of a file.

firstline=`head -1 test.txt`
echo $firstline
lastline=`tail -1 test.txt`
echo $lastline
if [ $firstline == $lastline ]
then
   echo "Found"
fi

Surely, am missing something. Please help.

Answer 1

Assuming you are using "some sort" of bourne shell, you should (a) quote the variables and (b) need to use a single = :

if [ "$firstline" = "$lastline" ]
then
   echo "Found"
fi

Update In response to some comments, this will also work if $firstline is -z . Even in this case the if statement is not interpreted as if [ -z ... ] , at least in the ksh (Korn Shell) or in Bash (I don't have a system with a plain bourne shell sh available).

Answer 2

Perhaps simpler...

bash-3.2$ if [ "$(sed -n '1p' file)" = "$(sed -n '$p' file)" ]; then
    echo 'First and last lines are the same'
else
    echo 'First and last lines differ'
fi

---

Update to answer Jan's questions.

bash-3.2$ cat file
-z
-G
bash-3.2$ if [ "$(sed -n '1p' file)" = "$(sed -n '$p' file)" ]; then
>     echo 'First and last lines are the same'
> else
>     echo 'First and last lines differ'
> fi
First and last lines differ

I prefer sed for grabbing the first and last lines of a file because the same command-line works on Linux, Mac OS and Solaris. The head and tail command-lines are different between Linux and Solaris.

Answer 3

Should be if [ "$firstline" = "$lastline" ]

If you omit double quotes it will not work if the line(s) contain white characters.

Answer 4

At the very least, you have to quote the variable expansions. Plus you should add prefix to avoid problems if the strings start with - . And the correct operator is = . So it should be

if [ "x$firstline" = "x$lastline" ]