I want to run two commands together for zkCli.
zkCli addauth digest username:password && zkCli setAcl /zknode-path world:anyone:crdwa
I have already set the ACL value for a zknode, and want to revert it back. But running this command give, authentication is not valid. How to run these two commands in one session?
I managed to run multiple commands in zkCli using heredoc format
(see How does ` cat << EOF` work in bash?)
Insert this snippet into a bash file
TMPVAR="addauth digest username:password\nsetAcl /zknode-path world:anyone:crdwa"
/zookeeper-3.4.10/bin/zkCli.sh <<EOF
$(echo -e ${TMPVAR})
quit
EOF
First, we set TMPVAR with both commands you wanted to execute in a single zkCli session, with a \n delimiter between them
Then we evaluate TMPVAR into STDIN line by line and this will make zkCli execute command after command, and then finally execute quit
Related
In the middle of the script, I have a command that exposes the local port with ssh -R 80:localhost:8080 localhost.run I need to execute this command in the background, parse the output and save it into a variable.
The output returns:
Welcome to localhost.run!
...
abc.lhrtunnel.link tunneled with tls termination, https://abc.lhrtunnel.link
Need to capture this part:
https://abc.lhrtunnel.link
As a result something like this:
...
hostname=$(command)
echo $hostname
...
Try this Shellcheck-clean code:
#! /bin/bash -p
hostname=$(ssh ... \
| sed -n 's/^.*tunneled with tls termination, //p')
declare -p hostname
I'm assuming that you don't really want to background the command that generates the output. You just want to run it in a way that allows its output to be captured and filtered. See How do you run multiple programs in parallel from a bash script? for information about how "background" processes are used for parallel processing.
The -n option to sed means that it doesn't print lines from the input unless explicitly instructed to print.
s/^.*tunneled with tls termination, //p works on input lines that contain anything followed by the string tunneled with tls termination, . It deletes everything on the line up to the end of that string and prints the result, which hopefully will be the URL that you want.
declare -p varname is a much more reliable and useful way to show the value of a variable than using echo.
On Ubuntu 21.04, I installed oracle dtb from docker, works fine I am using it for sql developer but now I need to use it in shell script. I am not sure if it can work this way, or is it some better way. I am trying to run simple script:
#!/bin/bash
SSN_NUMBER="${HOME}/bin/TESTS/sql_log.txt"
select_ssn () {
sudo docker exec -it oracle bash -c "source /home/oracle/.bashrc; sqlplus username/password#ORCLCDB;" <<EOF > $SSN_NUMBER
select SSN from employee
where fname = 'James';
quit
EOF
}
select_ssn
After I run this, nothing happens and I need to kill the session. or
the input device is not a TTY
is displayed
Specifying a here document from outside Docker is problematic. Try inlining the query commands into the Bash command line instead. Remember that the string argument to bash -c can be arbitrarily complex.
docker exec -it oracle bash -c "
source /home/oracle/.bashrc
printf '%s\n' \\
\"select SSN from employee\" \\
\"where fname = \'James\'\;\" |
sqlplus -s username/password#ORCLCDB" > "$SSN_NUMBER"
I took out the sudo but perhaps you really do need it. I added the -s option to sqlplus based on a quick skim of the manual page.
The quoting here is complex and I'm not entirely sure it doesn't require additional tweaking. I went with double quotes around the shell script, which means some things inside those quotes will be processed by the invoking shell before being passed to the container.
In the worst case, if the query itself is static, storing it inside the image in a file when you build it will be the route which offers the least astonishment.
I am logging into a remote server using SSH client. I have written a script that will execute two commands on the server.But, as the first command executes a bash script that calls "bash" command at the end. This results in execution of only one command not the other.
I cannot edit the first script to comment or remove the bash call.
i have written following script:
abc.sh
#!/bin/bash
command1="sudo -u user_abc -H /abc/xyz/start_shell.sh"
command2=".try1.sh"
$command1 && $command2
Only command 1 is getting executed not the second as the "bash" call is creating a new process the second command is not executing.
Solution 1
Since you can execute start_shell.sh you must have read permissions. Therefore, you could copy the script, modify it such that it doesn't call bash anymore, and execute the modified version.
I think this would be the best solution. If you really really really have to use start_shell.sh as is, then you could try one of the following solutions.
Solution 2
Try closing stdin using <&-. An interactive bash session will exit immediately if there is no stdin.
sudo -u user_abc -H /abc/xyz/start_shell.sh <&-; ./try1.sh
Solution 3
Change the order if both commands are independent.
./try1.sh; sudo -u user_abc -H /abc/xyz/start_shell.sh
As an example, I am trying to capture the raw commands that are output by the following script:
https://github.com/adampointer/go-deribit/blob/master/scripts/generate-models.sh
I have tried to following a previous answer:
BASH: echoing the last command run
but the output I am getting is as follows:
last command is gojson -forcefloats -name="${struct}" -tags=json,mapstructure -pkg=${p} >> models/${p}/${name%.*}_request.go
What I would like to do is capture the raw command, in other words have variables such as ${struct}, ${p} and ${p}/${name%.*} replaced by the actual values that were used.
How do I do this?
At the top of the script after the hashbang #!/usr/bin/env bash or #!/bin/bash (if there is any) add set -x
set -x Print commands and their arguments as they are executed
Run the script in debug mode which will trace all the commands in the script: https://stackoverflow.com/a/10107170/988525.
You can do that without editing the script by typing "bash generate-models.sh -x".
I'd like to use lftp in the beginning of a bash script, but how do I exit lftp without stopping the script from processing? I've tried ending the lftp part with "exit", "quit", and "bye", but they all stop the script.
Previously, I split into two scripts and cron'ed them to run in the right order. Is it possible to combine them into one script?
Or more explicitly, use << EOF.
e.g.
#!/bin/bash
#CMP 01.04.2013
#
#<<EOF below is the functional equivalent of hitting .Enter. on your keyboard.
#It allows the rest of the commands to be executed once connected
lftp -e 'mirror -R /home/pi/LocalDirToMirror ~/TargetDir' -u YourUsername,YourPassword ftp://FTP_URL_Location <<EOF
quit 0
EOF
End the LFTP part of the script with EOF, on its own line. That's all you need to do!