将 sed 命令行字符串作为变量传递给 bash

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

Answer 1

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")