Passing sed command line string to bash as variable

Question

I have a bash script which uses one multi-line sed command to change a file. From the command line, this line works:

sed -e '1a\' -e '/DELIMITER="|" \' -e '/RESTRICT_ADD_CHANGE=1 \' -e '/FIELDS=UPDATE_MODE|PRIMARYKEYVALUE|PATRONFLAGS' -e '1 d' general_update_01 > general_update_01.mp

I use the same bash script for a variety of files. So I need to pass all of the sed commands from the sending application to the bash script as a single parameter. However, when it passes in from the application, I get only -e.

In the bash script, I have tried a variety of ways to receive the variable as a complete string. None of these store the variable.

sed_instructions=$(echo $6)
sed_instructions=$6
sed_instructions=$(eval "$6")

and a few other configurations.

My command line would use the variable like this:

sed $sed_instructions $filename > $filename.mp

"I need to pass all of the sed commands [...] as a single parameter" -- why do you think that? — glenn jackman
– glenn jackman, Commented Jan 13, 2022 at 19:18
BashFAQ #50 is relevant: I'm trying to put a command in a variable, but the complex cases always fail! — Charles Duffy
– Charles Duffy, Commented Jan 13, 2022 at 19:30

glenn jackman · Accepted Answer · 2022-01-13 19:24:14Z

3

I assume you invoke your shell script like this

script.sh -e '1a' -e '/DELIMITER="|" ' -e '/RESTRICT_ADD_CHANGE=1 ' -e '/FIELDS=UPDATE_MODE|PRIMARYKEYVALUE|PATRONFLAGS' -e '1 d' general_update_01

What you want to do here is store the nth parameter as the filename, and the first n-1 parameters in an array

#!/usr/bin/env

n=$#
filename=${!n}
sed_commands=("${@:1:n-1}")

# Now, call sed
sed "${sed_commands[@]}" "$filename" > "${filename}.mp"

To demonstrate that code in action:

$ set -- one two three four
$ n=$#
$ filename=${!n}
$ args=("${@:1:n-1}")
$ declare -p filename args
declare -- filename="four"
declare -a args=([0]="one" [1]="two" [2]="three")

answered Jan 13, 2022 at 19:24

glenn jackman

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