Execute commands in specific location and depending on answer of previous command - linux

I am currently working on a Text-to-speech project and I need to write bash script which will, when it is called, execute two commands. If the first command returns the proper answer (if returns an answer at all), the second command will be called and executed.
My question is, how can I write a script, that executes shell commands in a specific certain file system location?
For example, I need to be in the directory /opt/text/example and execute this command:
sudo ./bin/sample_read -I ../languages/ -I ../languages -v dave -T 2 \
-i /opt/text/example.txt -F 22 -O embedded-pro -o out_file.pcm
and then to wait for the answer, then (if it is good) execute the second command.
The second command is
aplay -f S16_LE -r 22050 -c 1 out_file.pcm

This should help:
pushd /path/to/directory
my_var=$(command1)
if [ "$my_var" == "expected_result" ]; then
command2
fi
popd
You basically run command1 and store its output in my_var. Then you compare the content of $my_var with whatever you're expecting.
Also pushd <path>/popd allow you to move to a directory and back.

Related

bash -- execute command on file change; doubling issue + how to skip loop until command completes

I'm a bash noob, and I am trying to set up a sort of "hot reload" functionality for a project I'm working on using inotifywait. Ubuntu 20.04 if that matters.
Here is what I hoped would have worked:
inotifywait -m -r ../.. -e modify,create,delete |
while read line; do
custom_command
done
I'm having two problems:
Issue #1 is that custom_command takes some time to work, and so if I make more changes to the directory in the meantime, custom command appears to "queue up" custom_command, where really I just want it to keep the most recent one and drop the others.
Issue #2 is that I'm getting some sort of "double output." So for example if I bash auto-exec.sh and auto-exec.sh looks like this:
inotifywait -m -r . -q -e modify,create,delete
Then each time a change registers, I get this as output (not a mistake that it's doubled -- I get two identical lines each time there is a modification):
./ MODIFY auto-exec-testfile.txt
./ MODIFY auto-exec-testfile.txt
I should note I've tried making changes both with Visual Code Studio and gedit, with the same results.
If I modify the bash file like so:
inotifywait -m -r . -q -e modify,create,delete |
while read line; do
echo "$line"
echo "..."
done
I get the following output each time there is a change:
./ MODIFY auto-exec-testfile.txt
...
./ MODIFY auto-exec-testfile.txt
...
If I modify bash_test.sh to the following:
inotifywait -m -r . -q -e modify,create,delete |
while read line; do
echo "help me..."
done
Then I get the following each time a change is made:
help me...
help me...
What happened to the the ./ MODIFY ... line?? Presumably there's something I don't understand about bash, stdout or similar /related concepts here?
And finally, if I change the .sh file to the following:
inotifywait -m -r . -q -q -e modify,create,delete |
while read _; do
echo "help me..."
done
Then I get no output at all. This one I think I understand, because the -q -q means that inotifywait is in "super silent" mode, so there is no log and therefore nothing to trigger the while.
What I'd love to do is just trigger the code once when something changes, and drop all but the most recent execution. I'm not sure doing this using a while is entirely necessary, but I tried inotifywait -m -r . -q -q -e modify,create,delete | echo "help me..", and the script printed "help me..." once at startup, then exited on modification.
Assistance very much appreciated.
EDIT - 20201-Mar-23
I removed -m and create from the inotifywait line, and it appears to work as expected, except that it doesn't stay "up" in monitor mode. So this at least only gives me one entry from inotifywait:
notifywait -r .. -q -e modify,delete |
while read line1; do
echo ${line1}
done
Related:
inotifywait - pause monitoring while executing command
https://unix.stackexchange.com/questions/140679/using-inotify-to-monitor-a-directory-but-not-working-100
inotifywait not performing the while loop in bash script
while inotifywait -e close_write,delete .; do
pkill custom_command
custom_command&
done

How can I feed input within bash [Executed through the Network]

As the title says, within linux how can I feed input to the bash when I do sudo bash
Lets say I have a bash script that reads the name.
The way I execute the script is through sudo using:
cat read-my-name-script.sh | sudo bash
Lets just say this is how I execute the script throught the network.
Now I want to fill the name automatically, is there a way to feed the input. I tried doing this: cat read-my-name-script.sh < name-input-file | sudo bash where the name-input-file is a file for the input that the user will be using to feed the script.
I am new to linux and learning to automate the input and wanted to create a file for input where the user can fill it and feed it to my script.
This is convoluted, but might do what you want.
sudo bash -c "$(cat read-my-name.sh)" <name-input-file
The -c says the next quoted argument are the commands to run (so, read the script as a string on the command line, instead of from a file), and the calling shell interpolates the contents of the file inside the double quotes before the sudo command gets evaluated. So if read-my-name.sh contains
#!/bin/bash
read -p "I want your name please"
then the command gets expanded into
sudo bash -c '#!/bin/bash
read -p "I want your name please"' <name-input-file
(where of course at this time the shell has actually removed the outer double quotes altogether; I put in single quotes in their place instead to show how this would look as actually executable, syntactically valid code).
I think you need that:
while read -r arg; do sudo bash read-my-name-script.sh "$arg";done <name-input-file
So each line of name-input-file will be passed as argument to sudo bash read-my-name-script.sh
If your argslist located on http server, you can do that:
while read -r arg; do sudo bash read-my-name-script.sh "$arg";done < <(wget -q -O- http://some/address/in/internet/name-input-file)
UPD
add [[ -f name-input-file ]] && readarray -t args <name-input-file
to read-my-name-script.sh
and use "${args[#]}" as arguments of command in the script.
For example echo "${args[#]}" or cmd "${args[0]}" "${args[1]}" ... "${args[100]}" in any order.
In this case you can use
wget -q -O- http://some/address/in/internet/read-my-name-script.sh | bash
for run your script with arguments from name-input-file whitout saving script to the local machine

creating bash script to automate task for analyzing multiple files

I don't have a lot of experience with scripting.
I have a directory that contains, among many other files, a set of *.phylip files I need to analyze with a program. I would like to automate this task. I think a loop bash shell script would be appropriate, although I could be wrong.
If I was to perform the analysis manually on one .phylip file, I would use the following command in terminal:
./raxmlHPC-SSE3 -m GTRCAT -y -s uce-5.phylip --print-identical-sequences -p 12345 -n uce-5_result
For the bash shell script, I think it would be close to:
#!/bin/sh
for i in $( ls ); do
./raxmlHPC-SSE3 -m GTRCAT -y -s uce-5.phylip --print-identical-sequences -p 12345 -n test_5 $i
done
The issue I'm aware of, but don't know how to fix, is the -s option, which specifies the input phylip file. Any suggestions on how to modify the script to do what I need done?
Try the below code:
#!/bin/bash
for i in *.phylip
do
./raxmlHPC-SSE3 -m GTRCAT -y -s "$i" --print-identical-sequences -p 12345 -n ${i%.phylip}_result
done
-s option will be passed $i which has the file name with .phylip extension in the current directory.
${i%.phylip}_result replaces the .phylip extension with _result which i guess is what you expect. (Ref: Parameter Substitution)

Passing arguments to a script invoked with bash -c

I'm testing a Bash script I created on GitHub for behavioral correctness (e.g. that it parses options correctly). I want to do this without having to clone the repository locally, so here is how I'm doing it:
curl -sSL https://github.com/jamesqo/gid/raw/master/gid | xargs -0 bash -c
My question is, how can I pass arguments to the script in question? I tried bash -c --help, but that didn't work since it got interpreted as part of the script.
Thanks!
You’re actually over-complicating things by using xargs with Bash’s -c option.
Download the script directly
You don’t need to clone the repository to run the script. Just download it directly:
curl -o gid https://raw.githubusercontent.com/jamesqo/gid/master/gid
Now that it’s downloaded as gid, you can run it as a Bash script, e.g.,
bash gid --help
You can also make the downloaded script executable in order to run it as a regular Unix script file (using its shebang, #!/bin/bash):
chmod +x gid
./gid --help
Use process substitution
If you wanted to run the script without actually saving it to a file, you could use Bash process substitution:
bash <(curl -sSL https://github.com/jamesqo/gid/raw/master/gid) --help
I'll echo Anthony's comments - it makes a lot more sense to download the script and execute it directly, but if you're really set on using the -c option for bash, it's a little bit complicated, the problem is that when you do:
something | xargs -0 bash -c
there's no opportunity to pass any arguments. They all get swallowed as the argument to -c - it essentially gets turned into:
bash -c "$(something)"
so if you place something after the -c in the xargs, it gets before the something. There is no opportunity to put anything after something, as xargs doesn't let you.
If you want to pass arguments, you have to use the substitution position option for xargs, which allows you to place where the argument goes, The option is -J <item>, and the next thing to realize is that the first argument will be $0, so you have to do:
something | xargs -0 -I # bash -c # something <arg1> <arg2>…
I can emulate this with:
echo 'echo hi: ~$0~ ~$1~ ~$2~ ~$3~' | xargs -0 -I # bash -c # something one two three four
which yields:
hi: ~something~ ~one~ ~two~ ~three~

Triple nested quotations in shell script

I'm trying to write a shell script that calls another script that then executes a rsync command.
The second script should run in its own terminal, so I use a gnome-terminal -e "..." command. One of the parameters of this script is a string containing the parameters that should be given to rsync. I put those into single quotes.
Up until here, everything worked fine until one of the rsync parameters was a directory path that contained a space. I tried numerous combinations of ',",\",\' but the script either doesn't run at all or only the first part of the path is taken.
Here's a slightly modified version of the code I'm using
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l '\''/media/MyAndroid/Internal storage'\''' "
Within Backup.sh this command is run
rsync $5 "$path"
where the destination $path is calculated from text in Stamp.
How can I achieve these three levels of nested quotations?
These are some question I looked at just now (I've tried other sources earlier as well)
https://unix.stackexchange.com/questions/23347/wrapping-a-command-that-includes-single-and-double-quotes-for-another-command
how to make nested double quotes survive the bash interpreter?
Using multiple layers of quotes in bash
Nested quotes bash
I was unsuccessful in applying the solutions to my problem.
Here is an example. caller.sh uses gnome-terminal to execute foo.sh, which in turn prints all the arguments and then calls rsync with the first argument.
caller.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh 'long path' arg2 arg3"
foo.sh:
#!/bin/bash
echo $# arguments
for i; do # same as: for i in "$#"; do
echo "$i"
done
rsync "$1" "some other path"
Edit: If $1 contains several parameters to rsync, some of which are long paths, the above won't work, since bash either passes "$1" as one parameter, or $1 as multiple parameters, splitting it without regard to contained quotes.
There is (at least) one workaround, you can trick bash as follows:
caller2.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh '--option1 --option2 \"long path\"' arg2 arg3"
foo2.sh:
#!/bin/bash
rsync_command="rsync $1"
eval "$rsync_command"
This will do the equivalent of typing rsync --option1 --option2 "long path" on the command line.
WARNING: This hack introduces a security vulnerability, $1 can be crafted to execute multiple commands if the user has any influence whatsoever over the string content (e.g. '--option1 --option2 \"long path\"; echo YOU HAVE BEEN OWNED' will run rsync and then execute the echo command).
Did you try escaping the space in the path with "\ " (no quotes)?
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l ''/media/MyAndroid/Internal\ storage''' "

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