print variable in a string of a string

Question

I don't know how to print a variable inside a string of a string.

I first started off without a variable like this and it worked perfectly:

#!/bin/bash

ssh 1.1.1.1 $'sudo -H -u apache bash -c \'cd ~/html; echo development > stuff.text\''

When I login to my server at 1.1.1.1, I can see that the file stuff.text has the word development. Perfect.

Then I made this bash script:

#!/bin/bash

BRANCH=development
ssh 1.1.1.1 $'sudo -H -u apache bash -c \'cd ~/html; echo ${BRANCH} > stuff.text\''

But running this bash script causes an empty stuff.text file. I also tried each of these commands, but they all gave syntax/parse errors:

ssh 1.1.1.1 $`sudo -H -u apache bash -c 'cd ~/html; echo ${BRANCH} > stuff.text'`
ssh 1.1.1.1 ${`sudo -H -u apache bash -c 'cd ~/html; echo ${BRANCH} > stuff.text'`}
ssh 1.1.1.1 ${sudo -H -u apache bash -c 'cd ~/html; echo ${BRANCH} > stuff.text'}
ssh 1.1.1.1 ${"sudo -H -u apache bash -c 'cd ~/html; echo ${BRANCH} > stuff.text'"}

How do I write a variable inside the string of another string?

Does this answer your question? [$VAR vs ${VAR} and to quote or not to quote](https://unix.stackexchange.com/questions/4899/var-vs-var-and-to-quote-or-not-to-quote) — roaima, May 13 '21 at 12:28
The notation you are using doesn't make sense. What exactly are you saying that your first command did? What do you think this is doing: `ssh 1.1.1.1 $'sudo -H -u apache bash -c \'cd ~/html; echo development > stuff.text\''`? You cannot escape single quotes, this is just a very complex notation to do something actually quite simple. Please [edit] your question and explain what you are trying to do. — terdon, May 13 '21 at 12:42
@terdon, you can't escape single quotes within plain `'...'`, but you _can_ within a `$'...'`. Of course the syntax highlighting doesn't know it here. — ilkkachu, May 13 '21 at 12:53
@ilkkachu thanks! I had never seen a shell context that allowed escaping single quotes like that before. Of course, the OP has no reason to use `$'...'` at all, but that's a different issue. — terdon, May 13 '21 at 12:59
Instead of just throwing a bunch of syntax at the wall, I'd encourage you to read about bash quoting: [3.1.2 Quoting](https://www.gnu.org/software/bash/manual/bash.html#Quoting) — glenn jackman, May 13 '21 at 13:17

score 0 · Answer 1 · answered May 13 '21 at 12:42

0

After a bunch of trial an error, I found that this worked:

#!/bin/bash

BRANCH=development
ssh 1.1.1.1 $"sudo -H -u apache bash -c 'cd ~/html; echo ${BRANCH} > stuff.text'"

Now i can see the word development in the file ~/html/stuff.text

answered May 13 '21 at 12:42

John

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I think what you are looking for is `ssh 1.1.1.1 "sudo -H -u apache bash -c 'cd ~/html; echo $BRANCH > stuff.text'"`, The `$` isn't doing anything useful. – terdon May 13 '21 at 12:44
3

`$"..."` is meant for locale translation of strings, you probably don't need it, just use regular `"...`"`. The problem in the code in your Q is that variables aren't expanded within `'...'` or `$'...'` See [What is the difference between the “…”, '…', $'…', and $“…” quotes in the shell?](https://unix.stackexchange.com/questions/503013/what-is-the-difference-between-the-and-quotes-in-th) – ilkkachu May 13 '21 at 12:55

score 0 · Accepted Answer · answered May 13 '21 at 12:56

You are using a needlessly complicated notation. The $ isn't needed at all here, ssh takes a string that is executed as a command on the remote server. You also don't need to cd anywhere. Try this:

#!/bin/bash

## avoid CAPS for shell variable names.
branch=development

ssh 1.1.1.1 "sudo -H -u apache bash -c 'echo $branch > ~/html/stuff.text'"

print variable in a string of a string

2 Answers2