sed | 脚本中的grep怪异行为

Question

THE ISSUE

I have two files,

File1

INT1;INT2;INT3
INT4;INT5;INT6
INT7;INT7;INT9

File2

INT1;INT2;INT3

Next I'll grep the difference between the files and only take the integers of third column.

  DIFFERENCE=`grep -vxFf File1 File2 | awk 'BEGIN { FS = ";" } ; { print $3 }'`

resulting in

INT6 INT9

Next I want to substitute the spaces with line breaks

echo $DIFFERENCE | sed 's/ /;\n/g'

which results in

INT6;
INT9

Just as it should.

Instead, when I do it in the script, it returns

INT6
INT9

Why does it do this in script, and is there solution to this / how can I modify my result easily?

ORIGINAL CODE - FOR CLARIFICATION

Original code and output here

     CODE=`grep -vxFf $FOUND $COMPARETO | awk 'BEGIN { FS = ";" } ; { print $3 }'` 
     echo "$CODE;" | sed 's/ /;\n/g' > "testfile"   

8000070118157
8002820000804
3394700015011;

Answer 1

Your intermediate output is not INT6 INT9 on single line but already two lines, therefore sed doesn't replace anything.

You can do all of this in awk itself, for example

$ awk -F';' 'NR==FNR{a[$0];next} !($0 in a){print $3 FS}' file2 file1
INT6;
INT9;

if you don't want the last ; , perhaps easier to pipe to sed '$ s/;$//'