Shell Programming and Scripting

I have a text file like the following:

Major=1
Minor=2
Sub=3
Path=4

It is a text file with a full version number of some software. Im trying to get the 1234 to show up just like that on one line.

Ive tried the following:

cat file.txt | awk -F"=" '{print $2}' | sed -e s/\\r\$/\./g (and various different escape options)

&

cat file.txt | awk -F"=" '{print $2}' | tr -d "\r" and all various forms as well...

Any one care to point out how to get this to print out the way I want? end result id like to see is 1.2.3.4, but I can live with just 1234.

Try this:

Code:

awk -F= '{printf("%s%s",NR==1?"":".", $2)}END{print}' file

BTW, the use of cat with awk or sed is redundant.

Regards

Code:

$ cat info.txt
Major=1
Minor=2
Sub=3
Path=4
$ . ./info.txt
$ echo $Major.$Minor.$Sub.$Path
1.2.3.4
$

thanks for the help, I realize cat/sed (even awk) is redundant.. but its just habit when I start trying to break something down like this.

what is the {printf("%s%s",NR==1?"":".", $2)} actually dong here? Im a bit confused? btw this works great.. just trying to learn what I am doing right here

Hi,

This will do ..

Code:

$ cat out.lst
Major=1
Minor=2
Sub=3
Path=4

Code:

$ awk -F"=" '{print $2}' out.lst|tr '\n' '.'|sed 's/.$/\n/g'
1.2.3.4

Thanks
Sha

The printf statement prints two strings, a dot and field 2 but the dot shouldn't print with the first line.
The conditional statement checks for the linenumber NR, if it is the 1st line it prints nothing ("") before the 2nd field otherwise it prints a dot (".") before the the 2nd field.

Regards

Code:

awk -F"=" '{print $2}' ORS="" file