shell script to compare files and print formatted output

Question

I'm trying to write a shell script which will compare two files, and if there are no differences between then, it will indicate that there was a success, and if there are differences, it will indicate that there was a failure, and print the results. Here's what I have so far:

result = $(diff -u file1 file2)

if [ $result = "" ]; then
    echo It works!
else
    echo It does not work
    echo $result
fi

Anybody know what I'm doing wrong???

Answer 1

result=$(diff -u file1 file2)

if [ $? -eq 0 ]; then
    echo "It works!"
else
    echo "It does not work"
    echo "$result"
fi

Suggestions:

• No spaces around "=" in the variable assignment for results
• Use $? status variable after running diff instead of the string length of $result.
• I'm in the habit of using backticks for command substitution instead of $(), but @Dennis Williamson cites some good reasons to use the latter after all. Thanks Dennis!
• Applied quotes per suggestions in comments.
• Changed "=" to "-eq" for numeric test.

Answer 2

First, you should wrap strings being compared with quotes.

Second, "!" cannot be use it has another meaning. You can wrap it with single quotes.

So your program will be.

result=$(diff -u file1 file2)

if [ "$result" == "" ]; then
    echo 'It works!'
else
    echo It does not work
    echo "$result"
fi

Enjoy.

Answer 3

Since you need results when you fail, why not simply use 'diff -u file1 file2' in your script? You may not even need a script then. If diff succeeds, nothing will happen, else the diff will be printed.

Answer 4

bash string equivalence is "==". -n is non-zero string, -z is zero length string, wrapping in quotes because the command will complain if the output of diff is longer than a single string with "too many arguments". so

if [ -n "$(diff $1 $2)" ]; then
    echo "Different"
fi

or

if [ -z "$(diff $1 $2)" ]; then
    echo "Same"
fi