Friday, May 27, 2022

[SOLVED] Keep new line where grep is output

Issue

When I run ls I see the following. I want to delete all instances of Screen Shot..

10-15
BashStorage
blah.csv
LockScreen.app
NetsuiteBFN
Screen Shot 2017-10-26 at 11.09.08 AM.png
Screen Shot 2017-11-02 at 12.24.13 PM.png
Screen Shot 2017-11-10 at 9.20.33 AM.png
Screen Shot 2017-11-10 at 9.21.29 AM.png
Screen Shot 2017-11-20 at 10.26.24 AM.png
Screen Shot 2017-11-20 at 10.29.18 AM.png
Screen Shot 2017-11-20 at 10.30.40 AM.png
Screen Shot 2017-11-20 at 10.31.55 AM.png
Screen Shot 2017-11-20 at 10.34.11 AM.png
Screen Shot 2017-11-20 at 10.55.34 AM.png
Screen Shot 2017-11-20 at 10.56.44 AM.png
Screen Shot 2017-11-20 at 10.56.54 AM.png
Screen Shot 2017-11-20 at 10.59.20 AM.png
finalResourceUrls.txt
good
lockScreen
lockScreen.zip
ls.txt
x

ls | grep -F 'Screen Shot' returns only the screenshot entries each on a new line

The following code tries to find the file screen then shot for $i for i in $(ls | grep -F "Screen Shot"); do rm $i; done

and fails because it uses a space as delimiter

I also observed that grep will print new lines when stdout is the terminal but when I redirect to a file, the new lines are maintained (ls | grep -F "Screen Shot") > shots.txt

However, if I set the output to a variable there are no new lines

shots="$(ls | grep -F "Screen Shot")" 

Screen Shot 2017-10-26 at 11.09.08 AM.png Screen Shot 2017-11-02 at 12.24.13 PM.png Screen Shot 2017-11-10 at 9.20.33 AM.png Screen Shot 2017-11-10 at 9.21.29 AM.png Screen Shot 2017-11-20 at 10.26.24 AM.png Screen Shot 2017-11-20 at 10.30.40 AM.png Screen Shot 2017-11-20 at 10.31.55 AM.png Screen Shot 2017-11-20 at 7.01.32 PM.png Screen Shot 2017-11-21 at 11.49.15 AM.png
  1. How do I iterate over the screenshot entries and delete them?
  2. How do I keep a newline on grep?

EDIT: Cyrus (see comment) has working answers to question 1. Still looking for question 2 answers for grep behavior


Solution

You can still do this, you just need to specify the IFS (internal field separator).

shots="$(ls | grep -F "Screen Shot")"
echo $shots
>some_long_one_line_blob

IFS='$\n'
echo $shots
>line
>by
>line
>output

Or can you the -E argument to echo which will evaluate the newline characters.

echo -E "$shots"
>line
>by
>line
>output


Answered By - Matt Clark
Answer Checked By - Pedro (WPSolving Volunteer)
at