Bash: How to pass input file content as command argument

Question

I have a command command that takes an input file as argument. Is there a way to call command without actually creating a file?

I would like to achieve the following behavior

$ echo "content" > tempfile
$ command tempfile
$ rm tempfile

if possible:

as a one-liner,
without creating a file,
using either a bash (or sh) feature or a "well-known" command (as standard as xargs)

It feels like there must be an easy way to do it but I can't find it.

Relevant: unix.stackexchange.com/questions/505828/…

Flux
– Flux

2019-03-12 10:31:10 +00:00
Commented Mar 12, 2019 at 10:31 — Flux
– Flux, Commented Mar 12, 2019 at 10:31

kojiro · Accepted Answer · 2015-07-04 14:18:18Z

6

Just use a process substitution.

command <(echo "content")

Bash will create a FIFO or other type of temporary file in /dev for the standard output of whatever happens in the process. For example:

$ echo <(echo hi)
/dev/fd/63

answered Jul 4, 2015 at 14:18

kojiro

77.7k20 gold badges151 silver badges217 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Eugeniu Rosca Over a year ago

Here string would be another option: command <<< "content". Then, a pipe: echo "content" | command

kojiro Over a year ago

@EugeniuRosca I think that misrepresents the original question. Presumably OP has a command that expects a file, not simply standard input. Although, many such commands will look in standard input if the file name is -.

Eugeniu Rosca Over a year ago

You are right. If the OP's command expects a filename as argument and the command cannot be altered by OP to accept stdin, then my suggestions are wrong. Thanks for feedback.

Collectives™ on Stack Overflow

Bash: How to pass input file content as command argument

1 Answer 1

3 Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Related