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.
${i%.phylip}_result replaces the .phylip extension with _result which i guess is what you expect. (Ref: Parameter Substitution)
Related
I'm a bash noob, and I am trying to set up a sort of "hot reload" functionality for a project I'm working on using inotifywait. Ubuntu 20.04 if that matters.
Here is what I hoped would have worked:
inotifywait -m -r ../.. -e modify,create,delete |
while read line; do
custom_command
done
I'm having two problems:
Issue #1 is that custom_command takes some time to work, and so if I make more changes to the directory in the meantime, custom command appears to "queue up" custom_command, where really I just want it to keep the most recent one and drop the others.
Issue #2 is that I'm getting some sort of "double output." So for example if I bash auto-exec.sh and auto-exec.sh looks like this:
inotifywait -m -r . -q -e modify,create,delete
Then each time a change registers, I get this as output (not a mistake that it's doubled -- I get two identical lines each time there is a modification):
./ MODIFY auto-exec-testfile.txt
./ MODIFY auto-exec-testfile.txt
I should note I've tried making changes both with Visual Code Studio and gedit, with the same results.
If I modify the bash file like so:
inotifywait -m -r . -q -e modify,create,delete |
while read line; do
echo "$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 ${line1}
done
Related:
inotifywait - pause monitoring while executing command
https://unix.stackexchange.com/questions/140679/using-inotify-to-monitor-a-directory-but-not-working-100
inotifywait not performing the while loop in bash script
while inotifywait -e close_write,delete .; do
pkill custom_command
custom_command&
done
As the title says, within linux how can I feed input to the bash when I do sudo bash
Lets say I have a bash script that reads the name.
The way I execute the script is through sudo using:
cat read-my-name-script.sh | sudo bash
Lets just say this is how I execute the script throught the network.
Now I want to fill the name automatically, is there a way to feed the input. I tried doing this: cat read-my-name-script.sh < name-input-file | sudo bash where the name-input-file is a file for the input that the user will be using to feed the script.
I am new to linux and learning to automate the input and wanted to create a file for input where the user can fill it and feed it to my script.
This is convoluted, but might do what you want.
sudo bash -c "$(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]}" ... "${args[100]}" in any order.
In this case you can use
wget -q -O- http://some/address/in/internet/read-my-name-script.sh | bash
for run your script with arguments from name-input-file whitout saving script to the local machine
I am currently working on a Text-to-speech project and I need to write bash script which will, when it is called, execute two commands. If the first command returns the proper answer (if returns an answer at all), the second command will be called and executed.
My question is, how can I write a script, that executes shell commands in a specific certain file system location?
For example, I need to be in the directory /opt/text/example and execute this command:
sudo ./bin/sample_read -I ../languages/ -I ../languages -v dave -T 2 \
-i /opt/text/example.txt -F 22 -O embedded-pro -o out_file.pcm
and then to wait for the answer, then (if it is good) execute the second command.
The second command is
aplay -f S16_LE -r 22050 -c 1 out_file.pcm
This should help:
pushd /path/to/directory
my_var=$(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 $my_var with whatever you're expecting.
Also pushd <path>/popd allow you to move to a directory and back.
I am not sure this is possible, but I've often thought that and some solutions have amazed me. Is it possible to create the equivalent of the following script without creating two processes (in this case, it is clear two processes are created because there is a pipe):
#!/bin/bash
EVENTS="CREATE,CLOSE_WRITE,DELETE,MODIFY,MOVED_FROM,MOVED_TO"
inotifywait -e "$EVENTS" -m -r ~/Desktop/testing | \
while true; do
read TMP
echo "${TMP}" >> ~/Desktop/eventlog
done
Note that inside the while loop I do want to have access to the event.
It seems to me that a pipe is necessary because we need to write with one process and read from another. But maybe there exists a trick?
In bash 4.2, you can set the lastpipe option to allow the while loop to run in the current shell instead of a separate process.
shopt -s lastpipe
inotifywait -e "$EVENTS" -m -r ~/Desktop/testing |
while true; do
read TMP
echo "${TMP}" >> ~/Desktop/eventlog
done
(You don't need an explicit line continuation after the |, since bash knows that a line cannot end with that character.)
Is it correct to assume that you are looking to avoid a secondary script so that variables modified in the for loop dont lose their value once the loop is done?
In that case you can just swap and so something like
while read TMP; do
echo "${TMP}" >> ~/Desktop/eventlog
done < <(inotifywait -e "$EVENTS" -m -r ~/Desktop/testing)
but if you are concerned about flexibility of your code flow
you can redirect to a file handle and then read from that handle whenever
comment below if you want me to fish out en example
if it's something else - please add detail as to what you are actually looking to do
I am very new to bash scripting and in Ubuntu\Debian package system.
Today I am studying the content of this preinst file that the script executes before that package is unpacked from its Debian archive (.deb) file.
My fist doubt is about a line containing this:
!:-
Probably it is a stupid question but, using Google, I can't find an answer.
Insert the last command without the last argument (bash)
/usr/sbin/ab2 -f TLS1 -S -n 1000 -c 100 -t 2 http://www.google.com/
then
!:- http://www.stackoverflow.com/
is the same as
/usr/sbin/ab2 -f TLS1 -S -n 1000 -c 100 -t 2 http://www.stackoverflow.com/