shell script - copy stdout to file
 

hello,
I have 2 scripts. at the script1 I have this line
at script2 I print a lot and I need all to be at the stdout(because it will go out to $outputfile), and at some point at script2 I need to use all I have printed till now at script2.
cause I have at script2 line like this
I tried something like that (the first line is at the beginnig of the script2, and the 2 last lines are at the end of script2, so it will go to $outputfile)
what's wrong with that(showed me empty lines at $outputfile)?
Is there a way where I can copy the stdout to tmp5(that it will be both on stdout and tmp5) - its the best solution for me.

I think the problem is that you still have tmp5 open for writing when you cat it. Maybe try
Possibly you can use tee if it's okay to run script3 in parallel, although since "...." will go in a subshell it might give you problems depending on what's in "...":Actually, would avoid the subshell problems? There's even more concurrency, though.

Can you plz explain me what does the parameters(and &) mean?
Should I put this line together? exec 1>&- 2>&- 1>&3- 2>&4-<tmp5 script3
or it whould be 2 lines?
and what does it means <tmp5 script3
Thanks !

We could probably just use { } instead of ( ) for that. After all stdout returns back to original when code is no longer around { }.
This one causes bad file descriptor error. I think it's no longer necessary.
@kobigeu Try to read more about Redirection and Process Substitution in the bash manual.

See Redirections, please ask specific questions if anything is unclear.

2 lines.

It means the same as cat tmp5 | script3, it's more usual to write script3 <tmp5, but I wrote it like that to emphasize the similarity to the cat version.


But it would still be in a subshell because it's part of a pipeline so it makes no difference, right?

Hmm, I must have been thinking that the second part of the script doesn't have output, but that makes no sense, oops. :o

Since the context of it doesn't need to be placed on a subshell unlike those placed after pipe (|), it won't. This would still echo "Was in current shell.":
Or

I'm not sure if we are talking about the same "it"? The second example echoes nothing, which I think is what applies.

Now that's odd. With pipe you don't need to fork twice to create two processes on the background and share each other. Perhaps it's necessary to avoid errors or conflicts in other implementations of the shell.

The part of the code or the context of the script that doesn't run in a subshell by default like those in { }.

Since we can't use pipes to maintain the code not running on a subshell we could just use process substitution twice: