#!/bin/bash
mainDir="$1"
fileName="$2"
preCommand="$3"
echo ${mainDir}
echo ${fileName}
echo ${preCommand}
I am trying to insert text and arguments from earlier calls into arguments for another script.
./script.sh "dir1" "filename" "text with ealier agrument ${mainDir}"
The echo from my script always returns: "text with ealier agrument"
Is it possible to do this in a script?
If you want the first argument to be part of the third, you need to to so explicitly. Either
./script.sh "dir1" "filename" "text with ealier agrument dir1"
or
first=dir1
./script.sh "$first" "filename" "text with earlier argument $first"
You can use the "source" command
Example:
source ./script.sh "dir1" "filename" "text with ealier agrument ${mainDir}"
When you run a script, you are spawning a new shell process as a child for your current shell (changes in child do not affect parent). When you use "source", you are running in the current shell.
Every time you run the above command, "text with earlier argument ${mainDir}", will then echo back to you what was supplied to the command previously.
source change anything relevant? The text in the argument was already substituted.
${mainDir}to be replaced withdir1in this example.${mainDir}is expanded by the shell before your script is called, and it won't be expanded inside the shell.