using a Bash variable in place of a file as input for an executable

Question

I have an executable that is used in a way such as the following:

executable -v -i inputFile.txt -o outputFile.eps

In order to be more efficient, I want to use a Bash variable in place of the input file. So, I want to do something like the following:

executable -v -i ["${inputData}"] -o outputFile.eps

Here, the square brackets represent some clever code.

Do you know of some trick that would allow me to pipe information into the described executable in this way?

Many thanks for your assistance

It's a bit unclear just what you want to happen. Like, what is value of inputData? — hyde, Feb 12 '13 at 19:55
"${inputData}" is simple, ASCII data which I want to 'package' as a file without actually using a file (which may need to be stored in non volatile memory). The user mikyra suggests an approach involving process substitution which I think is exactly I need. Thanks for your help. — d3pd, Feb 13 '13 at 21:13
Ah, now I got it... Cool, I didn't know of that. Apparently it is a *bash* feature, not supported by plain *sh* or *dash*, so better remember to use `#!/bin/bash` when using that in a script. +1+1 — hyde, Feb 13 '13 at 21:21

score 63 · Accepted Answer · edited Sep 16 '21 at 01:26

63

You can use the following construct:

<(command)

So, to have bash create a FIFO with the command as the output for you, instead of your attempted -i ["${inputData}"], you would do:

-i <(echo "$inputData")

Therefore, here is your final total command:

executable -v -i <(echo "$inputData") -o outputFile.eps

edited Sep 16 '21 at 01:26

Gabriel Staples

answered Feb 12 '13 at 19:49

mikyra

2

This process substitution is exactly what I need. Thank you very much for your assistance! – d3pd Feb 13 '13 at 21:10
1

I encountered some command that need to seek the input file, where fifo pipe doesn't work. Zsh has a =() implemented as a temp file, which works well. – misssprite Jan 29 '15 at 07:40
4

https://www.gnu.org/software/bash/manual/bash.html#Process-Substitution worth to read this section – Sławosz Nov 30 '15 at 08:51
2

This fails if `inputData` contains literal backslashes, or is `-e` – Eric Mar 27 '18 at 17:53
For a solution that works in the cases that @Eric mentions, see [here](https://stackoverflow.com/a/51022016/5353461). – Tom Hale Jun 25 '18 at 11:08

Eric · Answer 2 · 2018-06-25T17:53:11.667

19

To correctly handle pathological cases like inputdata='\ntest' or inputdata='-e', you need

executable -v -i <(cat <<< "$inputData")

In zsh, the cat is not necessary

Edit: even this adds a trailing newline. To output the exact variable contents byte-by-byte, you need

executable -v -i <(printf "%s" "$inputData")

edited Jun 25 '18 at 17:53

answered Mar 27 '18 at 17:54

Eric

Tom Hale · Answer 3 · 2018-06-28T09:06:12.163

3

Note: zsh only:

To get a filename containing the contents of ${variable}, use:

<(<<<${variable})

Note:

So for the OP's case:

executable -v -i <(<<<${inputData}) -o outputFile.eps

edited Jun 28 '18 at 09:06

answered Jun 25 '18 at 11:06

Tom Hale

2

This also emits a trailing newline after the contents of `variable`, which in most cases doesn't matter, but might be worth drawing attention to – Eric Jun 28 '18 at 17:22

Raz Kheli · Answer 4 · 2022-10-17T20:21:03.800

1

I couldn't find a better solution than making temporary file and reference it into a variable. For example:

key=$(mktemp);    # Creating a random file in the tmp folder
echo "some data" > $key;    # Filling the file with content
ssh -i $key user@localhost

I assume when the bash session is close, the variable and temporary file is cleared

edited Oct 17 '22 at 20:21

answered Oct 17 '22 at 20:15

Raz Kheli

Actually, `mktemp` is an external utility, so it's up to your program to clean up the file when it's done with it – JM0 Jul 10 '23 at 19:09

4 Answers4