My script first sources an API file, the executes a few commands from the file. When the script is executed from the command line, I get "command not found" errors for each of the commands. Here's the script:
#!/bin/bash
## include API file
source /cygdrive/c/path/to/unifi_sh_api
## login
unifi_login
## authorize a client for 30 minutes, limit down/up speed to 2048/1024kbps, quota is 500MB
unifi_authorize_guest "x2:ff:ff:ff:ff:ff" 30 up=1024 down=2048 bytes=500
unifi_logout
This returns "command not found" for the 3 unifi_* commands. There are no other errors.
This is on a Windows server and you can see I'm using Cygwin, so maybe that's part of the problem?
Here's what my sourced file looks like, in case the issue is there: https://dl.ubnt.com/unifi/4.7.6/unifi_sh_api
Any suggestions?
UPDATE:
Definitely using bash.
I did try . instead of source before posting my question here.
Here's the command I'm running in a cmd.exe window:
> C:\cygwin\bin\bash.exe -c '/cygdrive/c/path/to/myscript.sh'
Related
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.
I have written the following code:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
And I am getting error:
array.sh: 3: array.sh: Syntax error: "(" unexpected
From what I came to know from Google, that this might be due to the fact that Ubuntu is now not taking "#!/bin/bash" by default... but then again I added the line but the error is still coming.
Also I have tried by executing bash array.sh but no luck! It prints blank.
My Ubuntu version is: Ubuntu 14.04
Given that script:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
and assuming:
It's in a file in your current directory named array.sh;
You've done chmod +x array.sh;
You have a sufficiently new version of bash installed in /bin/bash (you report that you have 4.3.8, which is certainly new enough); and
You execute it correctly
then that should work without any problem.
If you execute the script by typing
./array.sh
the system will pay attention to the #!/bin/bash line and execute the script using /bin/bash.
If you execute it by typing something like:
sh ./array.sh
then it will execute it using /bin/sh. On Ubuntu, /bin/sh is typically a symbolic link to /bin/dash, a Bourne-like shell that doesn't support arrays. That will give you exactly the error message that you report.
The shell used to execute a script is not affected by which shell you're currently using or by which shell is configured as your login shell in /etc/passwd or equivalent (unless you use the source or . command).
In your own answer, you say you fixed the problem by using chsh to change your default login shell to /bin/bash. That by itself should not have any effect. (And /bin/bash is the default login shell on Ubuntu anyway; had you changed it to something else previously?)
What must have happened is that you changed the command you use from sh ./array.sh to ./array.sh without realizing it.
Try running sh ./array.sh and see if you get the same error.
Instead of using sh to run the script,
try the following command:
bash ./array.sh
I solved the problem miraculously. In order to solve the issue, I found a link where it was described to be gone by using the following code. After executing them, the issue got resolved.
chsh -s /bin/bash adhikarisubir
grep ^adhikarisubir /etc/passwd
FYI, "adhikarisubir" is my username.
After executing these commands, bash array.sh produced the desired result.
I have a bash script start.sh which calls another run.sh, which takes me to another prompt where I have to delete a file file.txt and then exit out of that prompt.
When I call run.sh from inside start.sh, I see the prompt and I believe that it deletes the file.txt but the inner/new prompt waits for me to exit out of it while the script is running - meaning it needs intervention to proceed. How do I avoid it in bash?
In Python I can use Popen and get it going but not sure about bash.
EDIT: I would rather like to know what command to provide to exit out of the shell (generated from running run.sh") so I can go back to the prompt where "start.sh" was started.
Etan: To answer your question
VirtualBox:~/Desktop/ > ./start
company#4d6z74d:~$ ->this is the new shell
company#4d6z74d:~$ logout ---> I did a "Control D here" so the script could continue.
Relevant part of start.sh which:
/../../../../run.sh (this is the one that takes us to the new $ prompt)
echo "Delete file.txt "
rm -f abc/def/file.txt
You can run run.sh in the background using &. In start.sh, you would invoke the script via /path/run.sh &. Now, start.sh will exit without waiting for run.sh to finish (which is running in the background).
I'm trying to create a job in Jenkins that will execute a simple shell script. However, it seems that Jenkins isn't actually doing anything with my script. No matter what value I put into the Execute Shell Command section, Jenkins always says that it passes. Even if I put in a bogus filename like "RandomBogusFilename.sh" it'll say the job was a success.
Can anyone point out what I'm doing wrong and how I can get Jenkins to actually use my shell script?
The shell script, the job config, and the console output are all shown below. I'm currently trying to do this on a Windows Server 2008 R2 Standard machine.
Thanks.
My .sh file
File Name: surveyToolRequest.sh
File Location: /jobs/Jeff Shell Script Test/workspace
Description:
Hit a web address and retrieve the HTTP Response. Then print out the HTTP Response.
#!/bin/bash
response_code=$(curl -s -o /dev/null -w "%{http_code}" http://SOME-WEBSITE.COM)
echo "The response code is " $response_code
My Jenkins Job Config
Jenkins Console Output
I played with this and found that it worked if I specified the path to the script. If the script is in your job's workspace directory,
./surveyToolRequest.sh
should work as Jenkins looks for files relative to the root of the workspace.
It's more common to just put the contents of the script file directory into the job configuration; that way you can see what the job is doing and you'll avoid problems like this one.
You should use run "Execute windows batch command" and not "Execute Shell"
I know this has been asked many times, but I can find a suitable answer in my case.
I croned a backup script using rsync and would like to see all output, errors or not, from the all script commands. I must write the command inside the script itself, and do not want to see output in my shell.
I have been trying with no success. Below part of the script.
#!/bin/bash
.....
BKLOG=/mnt/backup_error_$now.txt
# Log everything to log file
# something like
exec 2>&1 | tee $BKLOG
# OR
exec &> $BKLOG
I have been adding at the script beginig all kinds of exec | tee $BKLOG with adding &>, 2>&1at various part of the command line, but all failed. I either get an empty log file or incomplete. I need to see on log file what rsync has done, and the error if script failed before syncing.
Thank you for help. My shell is zsh, so any solution in zsh is welcomed.
To redirect all the stdout/stderr to a file place this line on top of your script:
BKLOG=/mnt/backup_error_$now.txt
exec &> "$BKLOG"