Shell Programming and Scripting

Hello!!


I'm trying to get each line of a file, using cut and new line "\n" inside a script, but everytime I do the command it returns everything, not limited by the new line.
Am I using a wrong format to new line? Is there another way to do it?


cut -d"\n" -f1 < myFile


Thanks in advance!!

From the man page -d is a field delimiter... and you are giving it nothing. which means the default is a tab or space.

If your lines are uniform in length, then you can use -c #-# to get a range of characters say, 1-80...

Also, you might have to put in a printf statement to get the \n to print a newline. Or you can revert to my favorite echo ""

Yes..

-d is the delimiter, but why it isn't recognizing the "\n" ?
Unfortunately the lines are not uniform in length..

I suspect it isnt recognized by -d. I doubt you can use a construct like a newline... maybe you can use the Ascii equiv if it is even supported.

Otherwise you will have to get creative like use a quoted double space \" \"

You can use awk where you can chose $0 which will select the whole line if you dont select a delimiter..

awk '{ print $0 \n }' <Myfile


I think the syntax is wrong but close .


EDIT: I forgot the quotes!!! this works.

awk '{ print $0 "\n" }' < vglist.sh

Hi Kelam_Magnus


I tried the

awk '{ print $0 "\n" }' < myFile

and it still returns all the lines...
Don't know if makes any difference, but myFile is not a sh file...

please post a snippet of your input file and another snippet of what your output should look like ...

I think , you are trying to get/cut the first line of the file.
use the following alternatives.

Code:

head -1 file1

Code:

sed -n '1p' file1