BASH - Best practice to use output from one script in another - bash

I'm developing a BASH script which invokes another BASH script which prints a line to stdout. That output is captured by the first BASH script and used later. It works, but it has the downside that any other output which is printed by the second script will cause this part to behave unexpectedly, because there will be extra content.
main.sh
#!/bin/bash
# Invoke worker.sh and capture its standard output to stats
stats=$(worker.sh --generate-stats)
echo "stats=$stats"
worker.sh
#!/bin/bash
[[ $1 == "--generate-stats" ]] && echo "cpu=90 mem=50 disk=15"
In this over-simplified example, it's not a problem to use this construct, but as worker.sh grows in size and complexity, it's hard to remember that no other command can print to stdout without confounding the behavior, and if someone else works on worker.sh without realizing they can't print to stdout, it can easily get fouled. So what is considered good practice to generate output in one script and use it in the other?
I'm wondering if a fifo would be appropriate, or another file descriptor, or just a plain file. Or if exec should be used in this case, something like what is shown here https://www.tldp.org/LDP/abs/html/x17974.html:
#!/bin/bash
exec 6>&1 # Link file descriptor #6 with stdout.
# Saves stdout.
exec >&2 # stdout now goes to stderr
echo "Didn't know I shouldn't print to stdout"
exec 1>&6 6>&- # Restore stdout and close file descriptor #6.
[[ $1 == "--generate-stats" ]] && echo "cpu=90 mem=50 disk=15"
But I wouldn't want to use that if it's not considered good practice.

Many command-line utilities have quiet and verbose modes; it's generally considered good practice to have the most verbose output (debugging, tracing, etc.) be separated to standard error anyway, but it's common to have normal output be formatted for human legibility (e.g. include table headings and column separators) and quiet mode output be just the bare data for programmatic use. (For one example, see docker images vs docker images -q). So that would be my recommendation - have worker.sh take a flag indicating whether its output is being consumed programmatically, and write it such that its output is all sent via a function that checks that flag and filters appropriately.

Maybe a different approach would be for the second script to test to see if it's stdout is being used programatically:
gash.sh:
#!/bin/bash
data=$(./another.sh)
echo "Received $data"
another.sh:
#!/bin/bash
# for -t see man isatty(3). 1 is file descriptor 1 - stdout
if [ -t 1 ]; then
echo "stdout is a terminal"
else
echo "stdout is not a terminal"
fi
Gives (where $ is a generic keyboard prompt):
$ bash gash.sh
Received stdout is not a terminal
$ bash another.sh
stdout is a terminal
You could then set a flag to change script behaviour (ls(1) does a similar thing). However, you should be prepared for this:
$ bash another.sh|more
stdout is not a terminal
$ bash another.sh > out.txt
$ cat out.txt
stdout is not a terminal

Related

Get version number into a bash variable - Graphviz dot cli [duplicate]

Is there a filename that is assignable to a variable (i.e. not a magic builtin shell token like &1) that will let me redirect to stdout?
What I finally want to do is run something like this in a cron script:
LOG=/tmp/some_file
...
some_command 2>&1 >> $LOG
echo "blah" >> $LOG
...
Conveniently, this lets me turn off log noise by redirecting to /dev/null later when I'm sure there is nothing that can fail (or, at least, nothing that I care about!) without rewriting the whole script. Yes, turning off logging isn't precisely best practice -- but once this script works, there is not much that can conceivably go wrong, and trashing the disk with megabytes of log info that nobody wants to read isn't desired.
In case something unexpectedly fails 5 years later, it is still possible to turn on logging again by flipping a switch.
On the other hand, while writing and debugging the script, which involves calling it manually from the shell, it would be extremely nice if it could just dump the output to the console. That way I wouldn't need to tail the logfile manually.
In other words, what I'm looking for is something like /proc/self/fd/0 in bash-talk that I can assign to LOG. As it happens, /proc/self/fd/0 works just fine on my Linux box, but I wonder if there isn't such a thing built into bash already (which would generally be preferrable).
Basic solution:
#!/bin/bash
LOG=/dev/null
# uncomment next line for debugging (logging)
# LOG=/tmp/some_file
{
some_command
echo "blah"
} | tee 1>$LOG 2>&1
More evolved:
#!/bin/bash
ENABLE_LOG=0 # 1 to log standard & error outputs
LOG=/tmp/some_file
{
some_command
echo "blah"
} | if (( $ENABLE_LOG ))
then
tee 1>$LOG 2>&1
fi
More elegant solution from DevSolar's idea:
#!/bin/bash
# uncomment next line for debugging (logging)
# exec 1> >(tee /tmp/some_file) 2>&1
some_command
echo "blah"
Thanks to the awesome hints by olibre and suvayu, I came up with this (for the record, version that I'm using now):
# log to file
# exec 1>> /tmp/logfile 2>&1
# be quiet
# exec 1> /dev/null 2>&1
# dump to console
exec 2>&1
Just need to uncomment one of the three, depending on what is desired, and don't worry about anything else, ever again. This either logs all subsequent output to a file or to console, or not at all.
No output duplicated, works universally the same for every command (without explicit redirects), no weird stuff, and as easy as it gets.
If I have understood your requirement clearly, the following should do what you want
exec 2>&1
exec >> $LOG
Stdout and stderr of all subsequent commands will be appended to the file $LOG.
Use /dev/stdout
Here's another SO answer that mentions this solution: Difference between stdout and /dev/stdout

Explain the bash command "exec > >(tee $LOG_FILE) 2>&1"

My intent was to have all the output of my bash script displayed on the console and logged to a file.
Here is my script that works as expected.
#!/bin/bash
LOG_FILE="test_log.log"
touch $LOG_FILE
# output to console and to logfile
exec > >(tee $LOG_FILE) 2>&1
echo "Starting command ls"
ls -al
echo "End of script"
However I do not understand why it works that way.
I expected to have exec >>(tee $LOG_FILE) 2>&1 work but it fails although exec >>$LOG_FILE 2>&1 indeed works.
I could not find the reason for the construction exec > >(command ) in the bash manual nor in advanced bash scripting. Can you explain the logic behind it ?
The >(tee $LOG_FILE) is an example of Process substitution, you might wish to search for that. Advanced Shell Scriptng and Bash manual
Using the syntax, <(program) for capturing output and >(program) for feeding input, we can pass data just one record at a time. It is more powerful than command substitution (backticks, or $( )) because it substitutes for a filename, not text. Therefore anywhere a file is normally specified we can substitute a program's standard output or input (although process substitution on input is not all that common).
This is particularly useful where a program does not use standard streams for what you want.
Note that in your example you are missing a space, exec >>(tee $LOG_FILE) 2>&1 is wrong (you will get a syntax error). Rather,
exec > >(tee $LOG_FILE) 2>&1
is correct, that space is critical.
So, the exec > part changes file descriptor 1 (the default), also known as stdout or standard output, to refer to "whatever comes next", in this case it is the process substitution, although normally it would be a filename.
2>&1 redirects file descriptor 2 (stderr or standard error) to refer to the same place as file descriptor 1 (stdout or standard out). Important: if you omit the & you end-up with a file called 1 rather than successful redirection.
Once you have called the exec line above, then you have changed the current process's standard output, so output from the commands which follow go to that tee process instead of to regular stdout.

Trick an application into thinking it's a pipe, not interactive

I'm looking for an opposite of this:
Trick an application into thinking its stdin is interactive, not a pipe
I'd like to get the output of a command on stdout, but make it think it's writing into a pipe.
The usual solution is to | cat but I have the additional requirement that this is cross platform (ie sh, not bash) and returns a valid exit code if the command fails. Normally I would use pipefail but this isn't available everywhere.
I've tried various incantations of stty but haven't been successful.
You could always use a named pipe:
mkfifo tmp.pipe
# Reader runs in background
cat tmp.pipe &
# Producer in foreground
your_command > tmp.pipe
command_rtn=$?
rm tmp.pipe
Alternately, if you don't need the output in realtime and the output is reasonably small:
output=$(your_command)
command_rtn=$?
echo "${output}"
Or you can write the exit status to a file doing something terrible like:
{ your_command; echo $? > rtn_file; } | cat
command_rtn=$(cat rtn_file)

Open a shell in the second process of a pipe

I'm having problems understanding what's going on in the following situation. I'm not familiar with UNIX pipes and UNIX at all but have read documentation and still can't understand this behaviour.
./shellcode is an executable that successfully opens a shell:
seclab$ ./shellcode
$ exit
seclab$
Now imagine that I need to pass data to ./shellcode via stdin, because this reads some string from the console and then prints "hello " plus that string. I do it in the following way (using a pipe) and the read and write works:
seclab$ printf "world" | ./shellcode
seclab$ hello world
seclab$
However, a new shell is not opened (or at least I can't see it and iteract with it), and if I run exit I'm out of the system, so I'm not in a new shell.
Can someone give some advice on how to solve this? I need to use printf because I need to input binary data to the second process and I can do it like this: printf "\x01\x02..."
When you use a pipe, you are telling Unix that the output of the command before the pipe should be used as the input to the command after the pipe. This replaces the default output (screen) and default input (keyboard). Your shellcode command doesn't really know or care where its input is coming from. It just reads the input until it reaches the EOF (end of file).
Try running shellcode and pressing Control-D. That will also exit the shell, because Control-D sends an EOF (your shell might be configured to say "type exit to quit", but it's still responding to the EOF).
There are two solutions you can use:
Solution 1:
Have shellcode accept command-line arguments:
#!/bin/sh
echo "Arguments: $*"
exec sh
Running:
outer$ ./shellcode foo
Arguments: foo
$ echo "inner shell"
inner shell
$ exit
outer$
To feed the argument in from another program, instead of using a pipe, you could:
$ ./shellcode `echo "something"`
This is probably the best approach, unless you need to pass in multi-line data. In that case, you may want to pass in a filename on the command line and read it that way.
Solution 2:
Have shellcode explicitly redirect its input from the terminal after it's processed your piped input:
#!/bin/sh
while read input; do
echo "Input: $input"
done
exec sh </dev/tty
Running:
outer$ echo "something" | ./shellcode
Input: something
$ echo "inner shell"
inner shell
$ exit
outer$
If you see an error like this after exiting the inner shell:
sh: 1: Cannot set tty process group (No such process)
Then try changing the last line to:
exec bash -i </dev/tty

How to save the command you are about to execute in bash?

Is there a better way to save a command line before it it executed?
A number of my /bin/bash scripts construct a very long command line. I generally save the command line to a text file for easier debugging and (sometimes) execution.
My code is littered with this idiom:
echo >saved.txt cd $NEW_PLACE '&&' command.py --flag $FOO $LOTS $OF $OTHER $VARIABLES
cd $NEW_PLACE && command.py --flag $FOO $LOTS $OF $OTHER $VARIABLES
Obviously updating code in two places is error-prone. Less obvious is that Certain parts need to be quoted in the first line but not the next. Thus, I can not do the update by simple copy-and-paste. If the command includes quotes, it gets even more complicated.
There has got to be a better way! Suggestions?
How about creating a helper function which logs and then executes the command? "$#" will expand to whatever command you pass in.
log() {
echo "$#" >> /tmp/cmd.log
"$#"
}
Use it by simply prepending log to any existing command. It won't handle && or || though, so you'll have to log those commands separately.
log cd $NEW_PLACE && log command.py --flag $FOO $LOTS $OF $OTHER $VARIABLES
are you looking for set -x (or bash -x)? This writes every command to standard out after executing.
use script and you will get archived everything.
use -x for tracing your script, e.g. run them as bash -x script_name args....
use set -x in your current bash (you will get echoed your commands with substitued globs and variables
combine 2 and 3 with the 1
If you just execute the command file immediately after creating it, you will only need to construct the command once, with one level of escapes.
If that would create too many discrete little command files, you could create shell procedures and then run an individual one.
(echo fun123 '()' {
echo echo something important
echo }
) > saved.txt
. saved.txt
fun123
It sounds like your goal is to keep a good log of what your script did so that you can debug it when things go bad. I would suggest using the -x parameter in your shebang like so:
#!/bin/sh -x
# the -x above makes bash print out every command before it is executed.
# you can also use the -e option to make bash exit immediately if any command
# returns a non-zero return code.
Also, see my answer on a previous question about redirecting all of this debug output to a log when --log is passed into your shell script. This will redirect all stdout and stderr. Occasionally, you'll still want to write to the terminal to give the user feedback. You can do this by saving stdout to a new file descriptor and using that with echo (or other programs):
exec 3>&1 # save stdout to fd 3
# perform log redirection as per above linked answer
# now all stdout and stderr will be redirected to the file and console.
# remove the `tee` command if you want it to go just to the file.
# now if you want to write to the original stdout (i.e. terminal)
echo "Hello World" >&3
# "Hello World" will be written to the terminal and not the logs.
I suggest you look into the xargs command. It was made to solve the problem of programtically building up argument lists and passing them off to executables for batch processing
http://en.wikipedia.org/wiki/Xargs

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