Show uncommon part of the line - linux

Question

Hi I have two files which contain paths. I want to compare the two files and show only uncommon part of the line.
1.txt:
/home/folder_name/abc
2.txt:
/home/folder_name/abc/pqr/xyz/mnp
Output I want:
/pqr/xyz/mnp
How can I do this?

Answer 1

This bit of awk does the job:
$ awk 'NR==FNR {a[++i]=$0; next}
{
b[++j]=$0;
if(length(a[j])>length(b[j])) {t=a[j]; a[j]=b[j]; b[j]=t}
sub(a[j],"",b[j]);
print b[j]
}' 2.txt 1.txt # or 2.txt 1.txt, it doesn't matter
Write the line from the first file to the array a.
Write the line from the second to b.
Swap a[j] and b[j] if a[j] is longer than b[j] (this might not be necessary if the longer text is always in b).
Remove the part found in a[j] from b[j] and print b[j].
This is a general solution; it makes no assumption that the match is at the start of the line, or that the contents of one file's line should be removed from the other. If you can afford to make those assumptions, the script can be simplified.
If the match may occur more than once on the line, you can use gsub rather than sub to perform a global substitution.

Answer 2

Considering you have strings in 1.txt and in 2.txt following code will do.
paste 1.txt 2.txt |
while read a b;
do
if [[ ${#a} -gt ${#b} ]];
then
echo ${a/$b};
else
echo ${b/$a};
fi;
done;
This is how it works on my system,
shiplu#:~/test/bash$ cat 1.txt
/home/shiplu/test/bash
/home/shiplu/test/bash/hello/world
shiplu#:~/test/bash$ cat 2.txt
/home/shiplu/test/bash/good/world
/home/shiplu/test/bash
shiplu#:~/test/bash$ paste 1.txt 2.txt |
> while read a b;
> do
> if [[ ${#a} -gt ${#b} ]];
> then
> echo ${a/$b};
> else
> echo ${b/$a};
> fi;
> done;
/good/world
/hello/world

Answer 3

This script will compare all lines in the file and only output the change in the line.
First it counts the number of lines in the first file.
Then i start a loop that will iterate for the number of lines.
Declare two variable that are the same line from both files.
Compare the lines and if they are the same output that they are.
If they are not then replace duplicate parts of the string with nothing(effectively removing them)
I used : as the seperator in sed as your variables contain /. So if they contain : then you may want to consider changing them.
Probably not the most efficient solution but it works.
#!/bin/bash
NUMOFLINES=$(wc -l < "1.txt")
echo $NUMOFLINES
for ((i = 1 ; i <= $NUMOFLINES ; i++)); do
f1=$(sed -n $i'p' 1.txt)
f2=$(sed -n $i'p' 2.txt)
if [[ $f1 < $f2 ]]; then
echo -n "Line $i:"
sed 's:'"$f1"'::' <<< "$f2"
elif [[ $f1 > $f2 ]]; then
echo -n "Line $i:"
sed 's:'"$f2"'::' <<< "$f1"
else
echo "Line $i: Both lines are the same"
fi
echo ""
done

Answer 4

If you happen to use bash, you could try this one:
echo $(diff <(grep -o . 1.txt) <(grep -o . 2.txt) \
| sed -n '/^[<>]/ {s/^..//;p}' | tr -d '\n')
It does a character-by-character comparison using diff (where grep -o . gives an intermediate line for each character to be fed to line-wise diff), and just prints the differences (intermediate diff output lines starting with markers < or > omitted, then joining lines with tr).
If you have multiple lines in your input (which you did not mention in your question) then try something like this (where % is a character not contained in your input):
diff <(cat 1.txt | tr '\n' '%' | grep -o .) \
<(cat 2.txt | tr '\n' '%' | sed -e 's/%/%%/g' | grep -o .) \
| sed -n '/^[<>]/ {s/^..//;p}' | tr -d '\n' | tr '%' '\n'
This extends the single-line solution by adding line end markers (e.g. %) which diff is forced to include in its output by adding % on the left and %% on the right.

Answer 5

If both the files have always a single line in each, then below works:
perl -lne '$a=$_ if($.==1);print $1 if(/$a(.*)/ && $.==2)' 1.txt 2.txt
Tested Below:
> cat 1.txt
/home/folder_name/abc
> cat 2.txt
/home/folder_name/abc/pqr/xyz/mnp
> perl -lne '$a=$_ if($.==1);print $1 if(/$a(.*)/ && $.==2)' 1.txt 2.txt
/pqr/xyz/mnp
>