Reputation: 79552

How to execute the output of a command within the current shell?

I'm well aware of the source (aka .) utility, which will take the contents from a file and execute them within the current shell.

Now, I'm transforming some text into shell commands, and then running them, as follows:

$ ls | sed ... | sh

ls is just a random example, the original text can be anything. sed too, just an example for transforming text. The interesting bit is sh. I pipe whatever I got to sh and it runs it.

My problem is, that means starting a new sub shell. I'd rather have the commands run within my current shell. Like I would be able to do with source some-file, if I had the commands in a text file.

I don't want to create a temp file because feels dirty.

Alternatively, I'd like to start my sub shell with the exact same characteristics as my current shell.

update

Ok, the solutions using backtick certainly work, but I often need to do this while I'm checking and changing the output, so I'd much prefer if there was a way to pipe the result into something in the end.

sad update

Ah, the /dev/stdin thing looked so pretty, but, in a more complex case, it didn't work.

So, I have this:

find . -type f -iname '*.doc' | ack -v '\.doc$' | perl -pe 's/^((.*)\.doc)$/git mv -f $1 $2.doc/i' | source /dev/stdin

Which ensures all .doc files have their extension lowercased.

And which incidentally, can be handled with xargs, but that's besides the point.

find . -type f -iname '*.doc' | ack -v '\.doc$' | perl -pe 's/^((.*)\.doc)$/$1 $2.doc/i' | xargs -L1 git mv

So, when I run the former, it'll exit right away, nothing happens.

Upvotes: 120

Answers (9)

ish-west

Reputation: 111

To use the mark4o's solution on bash 3.2 (macos) a here string can be used instead of pipelines like in this example:

. /dev/stdin <<< "$(grep '^alias' ~/.profile)"

Upvotes: 2

Jacek Krysztofik

Reputation: 1366

Try using process substitution, which replaces output of a command with a temporary file which can then be sourced:

source <(echo id)

Upvotes: 56

Juliano

Reputation: 41397

The eval command exists for this very purpose.

eval "$( ls | sed... )"

More from the bash manual:

eval
          eval [arguments]
The arguments are concatenated together into a single command, which is then read and executed, and its exit status returned as the exit status of eval. If there are no arguments or only empty arguments, the return status is zero.

Upvotes: 163

Geoff Nixon

Reputation: 4973

I believe this is "the right answer" to the question:

ls | sed ... | while read line; do $line; done

That is, one can pipe into a while loop; the read command command takes one line from its stdin and assigns it to the variable $line. $line then becomes the command executed within the loop; and it continues until there are no further lines in its input.

This still won't work with some control structures (like another loop), but it fits the bill in this case.

Upvotes: 9

rabensky

Reputation: 2934

Wow, I know this is an old question, but I've found myself with the same exact problem recently (that's how I got here).

Anyway - I don't like the source /dev/stdin answer, but I think I found a better one. It's deceptively simple actually:

echo ls -la | xargs xargs

Nice, right? Actually, this still doesn't do what you want, because if you have multiple lines it will concat them into a single command instead of running each command separately. So the solution I found is:

ls | ... | xargs -L 1 xargs

the -L 1 option means you use (at most) 1 line per command execution. Note: if your line ends with a trailing space, it will be concatenated with the next line! So make sure each line ends with a non-space.

Finally, you can do

ls | ... | xargs -L 1 xargs -t

to see what commands are executed (-t is verbose).

Hope someone reads this!

Upvotes: 44

mark4o

Reputation: 60873

$ ls | sed ... | source /dev/stdin

UPDATE: This works in bash 4.0, as well as tcsh, and dash (if you change source to .). Apparently this was buggy in bash 3.2. From the bash 4.0 release notes: