Variables are not passed from one script calling another with Bash heredoc

Question

I have a shell script kept at my local user called (executeAdM.sh), and when I execute this script I am switching to a sudo user by taking commands from an instruction file. But when I execute this script I am also passing parameters, which is actually some directory path of the sudo user. See the script below.

Script for local user (executeADM.sh):

#!/bin/bash
echo password | sudo -S -l
sudo /usr/bin/su - user  <<\EOF
#ls -lrt
pwd
for entry in $(ls -r)
  do
  if [ "$entry" = "testingADM.sh" ];then
    ./"$entry" "$1"
  fi
done
EOF

I'm executing the above as:

./executeADM.sh "/global/u70/globe/ADM"

When the above script executes, it switches successfully to the other user, and with the sudo user I execute a for loop which searches for another script called testingADM.sh.

Once the script is found I execute that script with the parameter passed from the local user, and testingADM.sh should set the path for me.

This is not working – it is not reading the parameter passed from the local user.

The sudo user script (testingADM.sh):

#!/bin/bash
changepath=$1
cd "$changepath"
pwd

When I hardcode the path variable in the script, all works fine. But I don't want that:

 #!/bin/bash
 changepath=somepath
 cd "$changepath"
 pwd

The issue is instead of going to "/global/u70/globe/ADM" it takes the path as "/global/u70/globe/", Which is the path after I do sudo. It is just not passing the variable.

---

With the below suggestions, this is what I ended up with:

#!/bin/bash
echo password | sudo -S -l
sudo /usr/bin/su - user  <<EOF
#ls -lrt
#pwd
echo "$1"
./testingADM.sh "$1"
EOF

The echo command prints nothing; a blank space is shown. It worked when I changed \EOF to EOF.

Can anyone explain the difference between:

\EOF , "EOF" , 'EOF' , EOF

or do the first three have the same meaning? Then what is difference between them?

Answer 1

Do not quote or escape your heredoc limit string (i.e. use EOF instead of \EOF or "EOF"), as otherwise, special variables like $ will not be interpreted. See here for more info and examples (e.g. Example 19-7).

Here is a minimal script that works:

$ cat test.sh
#!/bin/bash
sudo su - $(whoami) <<EOF
./test2.sh "$1"
EOF

$ cat test2.sh
#!/bin/bash
foo=$1
echo "$foo"

$ ./test1.sh "/path/to/file"
"/path/to/file"

Note that:

• There is no need for the heredoc at all. Just call ./test2.sh directly after sudo: sudo -u $(whoami) test2.sh "$1"
• You should not loop over the output of ls.
• There is no need to do a for loop to search for a single file. Just call the script directly with its name.

You should also quote your variables, that is:

• Execute your script with ./executeADM.sh "/path/with white/space"
• Call ./"$entry" "$1" in your script
• In your other script, use cd "$path"

In general:

• Always double-quote variables. (Don't single-quote them -- they won't expand.)
• Don't use a variable called path. (Much as you wouldn't use variables called user, home, etc.)

Answer 2

Edit
From back of my head. Does your path contain spaces?

If yes you need to execute it as ./executeADM.sh 'somePath'.

It is always better to save the $1 into a variable when you are passing it further it makes it easier to read. I would add path=$1 then in batch file you would have ./$entry "$path"