How to remove the last line from a variable in bash or sh?

Question

I have a variable that has few lines. I would like to remove the last line from the contents of the variable. I searched the internet but all the links talk about removing the last line from a file. Here is the content of my variable

$echo $var
$select key from table_test
UNION ALL
select fob from table_test
UNION ALL
select cal from table_test
UNION ALL
select rot from table_test
UNION ALL
$

I would like to get rid of UNION ALL appearing in the last line alone.

`echo $var` would not produce the output you show, although `echo "$var"` would. — chepner, Apr 05 '17 at 12:12

fzd · Accepted Answer · 2017-04-05T12:13:56.147

14

sed can do it the same way it would do it from a file :

> echo "$var" | sed '$d'

EDIT : $ represents the last line of the file, and d deletes it. See here for details

edited Apr 05 '17 at 12:13

answered Apr 05 '17 at 12:10

fzd

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May be: `var=$(sed '$d' <<< "$var")` – sat Apr 05 '17 at 12:13
I suspect your variable to be on only one line. Does `echo "$var" | wc -l` return 1 ? – fzd Apr 05 '17 at 12:31
Yes. It returns 1. – Alex Raj Kaliamoorthy Apr 05 '17 at 12:41
Ok..I have managed to fix it. I have added this command `selectstmt=$(echo -e $var)` and then executed this command `echo "$var"| sed '$d'` – Alex Raj Kaliamoorthy Apr 05 '17 at 12:47

score 2 · Answer 2 · edited Apr 18 '20 at 15:23

2

echo $var | head -n -1

Get all but the last line.

edited Apr 18 '20 at 15:23

Ondrej K.

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answered Apr 18 '20 at 15:00

Леонид Никифоренко

21
2

score 1 · Answer 3 · answered Apr 05 '17 at 12:21

1

You could try to cut off the last line.

Count=$(echo "$Var" | wc -l)
echo "$Var" | head -n $(($Count -1))

head -n $(($Count -1)) describes how many rows you want to show.

answered Apr 05 '17 at 12:21

Mario

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score 1 · Answer 4 · answered Apr 05 '17 at 12:35

1

Try this:

last_line=`echo "${str##*$'\n'}"` # "${str##*$'\n'}" value gives the last line for 'str'
str=${str%$last_line} # subtract last line from 'str'
echo "${str}"

answered Apr 05 '17 at 12:35

Ashish K

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score 1 · Answer 5 · answered Nov 13 '20 at 23:12

1

Bash way:

echo "${var%$'\n'*}"

This prints $var with all characters incl. and after the last LF removed.

answered Nov 13 '20 at 23:12

svobodb

56
5

How to remove the last line from a variable in bash or sh?

5 Answers5