I made simple script. file name is sutest.
#!/bin/bash
cd ~/Downloads/redis-4.0.1/src
./redis-server
echo "uid is ${UID}"
echo "user is ${USER}"
echo "username is ${USERNAME}"
I runed script.$ . sutest
But, script code is stopped at ./redis-server.
So I can't see echo messages.
I want to make this kind of script files. How can I do that??
I would be appreciate your help.
Let's say more general case.
myscript1 file executes process like redis-server above.
another myscript2 file executes process like redis-server above.
another myscript3 file executes process like redis-server above.
How can I run three script files simultaneously??
I want to do job in ssh connection.
To make the matter worse, If I can't use screen or tmux??
Add a '&' char at the end of the row
./redis-server &
this char permits to run in backgroud the job, and the script continues.
Just do the echos first:
cd ~/Downloads/redis-4.0.1/src
echo "uid is ${UID}"
echo "user is ${USER}"
echo "username is ${USERNAME}"
exec ./redis-server
The use of exec is a small trick (which you can omit if you prefer): it replaces the shell script with redis-server, so the shell script is no longer running at all. Without exec, you end up with the shell script waiting around for redis-server to finish, which is unnecessary if the script will do nothing further.
If you don't like that for some reason, you can keep the original order:
cd ~/Downloads/redis-4.0.1/src
./redis-server & # run in background
echo "uid is ${UID}"
echo "user is ${USER}"
echo "username is ${USERNAME}"
wait # optional
Related
I have 1 bash script that runs another bash script, however the first bashscript isn't waiting for the second one to complete before proceeding, how can I force it to wait?
For example:
#!/bin/bash
# first.sh
#call to secondary script
sh second.sh
echo "second.sh has completed"
echo "continuing with the rest of first.sh..."
The way it is now, it will run second.sh, and continue on, without waiting for second.sh to complete.
AS I use scheme like this in few scripts - just calling second scripts in the same shell-copy using source.
In script-1:
source script2.sh
or:
. script2.sh
So - no one command in script-1 will not be proceeded till script2.sh will end all it's tasks.
Little example.
First script:
$ cat script-1.sh
#!/bin/bash
echo "I'm sccript $0."
echo "Runnig script-2..."
source script-2.sh
echo "script-2.sh finished!"
Second script:
$ cat script-2.sh
#bin/bash
echo "I'm script-2. Running wait operation..."
sleep 2
echo "I'm ended my task."
How it works:
$ ./script-1.sh
I'm sccript ./script-1.sh.
Runnig script-2...
I'm script-2. Running wait operation...
I'm ended my task.
script-2.sh finished!
Normally it does; something else is happening. Are you sure that the other script isn't running something in the background instead? You can try using wait regardless.
You can simply add the command wait after you execute the second script, it will wait for all process that you launch from your principal script
You can even recuperate the PID of your second script using the command echo $! directly after you call the second script, and then pass this PID as an argument to the wait command
try using bash second.sh and check your second.sh and make sure you don't have programs that run in the background
Another way to do it $(second.sh)
I want to log into server based on user's choice so I wrote bash script. I am totally newbie - it is my first bash script:
#!/bin/bash
echo -e "Where to log?\n 1. Server A\n 2. Server B"
read to_log
if [ $to_log -eq 1 ] ; then
echo `ssh user#ip -p 33`
fi
After executing this script I am able to put a password but after nothing happens.
If someone could help me solve this problem, I would be grateful.
Thank you.
The problem with this script is the contents of the if statement. Replace:
echo `ssh user#ip -p 33`
with
ssh user#ip
and you should be good. Here is why:
Firstly, the use of back ticks is called "command substitution". Back ticks have been deprecated in favor of $().
Command substitution tells the shell to create a sub-shell, execute the enclosed command, and capture the output for assignment/use elsewhere in the script. For example:
name=$(whoami)
will run the command whoami, and assign the output to the variable name.
the enclosed command has to run to completion before the assignment can take place, and during that time the shell is capturing the output, so nothing will display on the screen.
In your script, the echo command will not display anything until the ssh command has completed (i.e. the sub-shell has exited), which never happens because the user does not know what is happening.
You have no need to capture the output of the ssh command, so there is no need to use command substitution. Just run the command as you would any other command in the script.
I'm novice to running bash script. (you can suggest me, if title I've given is incorrect.)
I want to run a jar file using bash script in loop. Then it should write the output of jar command into some file.
Bash file datagenerate.sh
#!/bin/bash
echo Total iterations are 500
for i in {1..500}
do
the_output="$(java -jar data-generator.jar 10 1 mockData.csv data_200GB.csv)"
echo $the_output
echo Iteration $i processed
done
no_of_lines="$(wc -l data_200GB.csv)"
echo "${no_of_lines}"
I'm running above script using command nohup sh datagenerate.sh > datagenerate.log &. As I want to run this script in background, so that even I log out from ssh it should keep running & output should go into datagenerate.log.
But when I ran above command and hit enter or close the terminal it ends the process. Only Total iterations are 500 is getting logged into output file.
Let me know what I'm missing. I followed following two links to create above shell script: link-1 & link2.
nohup sh datagenerate.sh > datagenerate.log &
nohup should work this way without using screen program, but depending on your distro your sh shell might be linked to dash.
Just make your script executable:
chmod +x datagenerate.sh
and run your command like this:
nohup ./datagenerate.sh > datagenerate.log &
You should check this out:
https://linux.die.net/man/1/screen
With this programm you can close your shell while a command or script is still running. They will not be aborted and you can pick the session up again later.
In my bash script, I do:
ssh me9#some_mad_server.com;
cd ~/apple;
echo "Before Exit"
exit
echo "After Exit"
I never see Before Exit or After Exit. I can understand why I may not see Before Exist as my script at that stage is in another console. But I am confused if the Exit mean my script ends and hence why After Exit never gets logged.
Any help appreciated.
To execute a series of commands on a remote host, you need to pass them to ssh on the command line, not execute them after the ssh call. Like this:
ssh me9#some_mad_server.com '
cd ~/apple
echo "Before Exit"
'
echo "After Exit"
This uses a multiline string to pass multiple commands. An exit is implicit when the end of the string is reached.
Importantly, the commands in the quoted string are executed on the remote host, while the final echo is executed on the local server. I've indented the remote commands for clarity.
You can use screen for this.
screen -d -m <command>
Use screen -r to attach to that screen again
screen -r <screen ID>
I have 1 bash script that runs another bash script, however the first bashscript isn't waiting for the second one to complete before proceeding, how can I force it to wait?
For example:
#!/bin/bash
# first.sh
#call to secondary script
sh second.sh
echo "second.sh has completed"
echo "continuing with the rest of first.sh..."
The way it is now, it will run second.sh, and continue on, without waiting for second.sh to complete.
AS I use scheme like this in few scripts - just calling second scripts in the same shell-copy using source.
In script-1:
source script2.sh
or:
. script2.sh
So - no one command in script-1 will not be proceeded till script2.sh will end all it's tasks.
Little example.
First script:
$ cat script-1.sh
#!/bin/bash
echo "I'm sccript $0."
echo "Runnig script-2..."
source script-2.sh
echo "script-2.sh finished!"
Second script:
$ cat script-2.sh
#bin/bash
echo "I'm script-2. Running wait operation..."
sleep 2
echo "I'm ended my task."
How it works:
$ ./script-1.sh
I'm sccript ./script-1.sh.
Runnig script-2...
I'm script-2. Running wait operation...
I'm ended my task.
script-2.sh finished!
Normally it does; something else is happening. Are you sure that the other script isn't running something in the background instead? You can try using wait regardless.
You can simply add the command wait after you execute the second script, it will wait for all process that you launch from your principal script
You can even recuperate the PID of your second script using the command echo $! directly after you call the second script, and then pass this PID as an argument to the wait command
try using bash second.sh and check your second.sh and make sure you don't have programs that run in the background
Another way to do it $(second.sh)