unable to run a remote script with Ansible

[fanvalt]

Hello,

Here is my issue.

I have a script start.sh with the following lines:

#!/bin/ksh

dispatcher.sh --command=start --directory=..

returnCode=$?

exit $returnCode

the both scripts are in the same directory, with the 755 rights.

I run the task this way:

- name: Start

  shell: /bin/ksh start.sh

  args:

    chdir: "/{{ rep_user }}/{{ openhrname }}/bin"

  register: start_op

and I receive the following error:

fatal: [integ]: FAILED! => {"changed": true, "cmd": "/bin/ksh start.sh", "delta": "0:00:00.007132", "end": "2016-05-24 15:13:36.666862", "failed": true, "rc": 127, "start": "2016-05-24 15:13:36.659730", "stderr": "start_openhr.sh[2]: dispatcher.sh: not found [No such file or directory]", "stdout": "", "stdout_lines": [], "warnings": []}

I have tried to run dos2unix on these files before, without success.

I tried also many different ways to write the shell command, without any success:

shell: /bin/ksh ./start.sh

shell: chdir=/{{ rep_user }}/{{ openhrname }}/bin /bin/ksh ./start.sh

shell: /bin/ksh /{{ rep_user }}/{{ openhrname }}/bin/start.sh

  args:

    chdir: "/{{ rep_user }}/{{ openhrname }}/bin"

etc ... 

when I run it manually from the directory where the files are installed, I have no issue.

If I try to run the start.sh shell from another directory manually, the error occurs, which is normal.

Could it be an issue with chdir command ? I run Ansible version 2.1.0

Thanks a lot,

regards

[fanvalt]

Is it normal that when running with the -vvvv options, in the invocation display, the module_name displayed is "command" and not "shell" ?

fatal: [integ]: FAILED! => {"changed": true, "cmd": "/bin/ksh start.sh", "delta": "0:00:00.005499", "end": "2016-05-24 16:43:55.780568", "failed": true, "invocation": {"module_args": {"_raw_params": "/bin/ksh start.sh", "_uses_shell": true, "chdir": "/integuno/openhr/bin", "creates": null, "executable": "/bin/ksh", "removes": null, "warn": true}, "module_name": "command"}, "rc": 127, "start": "2016-05-24 16:43:55.775069", "stderr": "start.sh[2]: dispatcher.sh: not found [No such file or directory]", "stdout": "", "stdout_lines": [], "warnings": []}

[Matt Martz]

The biggest problem looks like all of your scripts assume relative paths.  As a result it looks like things cannot be located.  I'd recommend either adding pathing detection, or using full paths to those files.

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Matt Martz
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[Kai Stian Olstad]

On 24. mai 2016 15:23, fanvalt wrote:
> Hello,
>
> Here is my issue.
> I have a script start.sh with the following lines:
> #!/bin/ksh
> dispatcher.sh --command=start --directory=..
> returnCode=$?
> exit $returnCode
>
> the both scripts are in the same directory, with the 755 rights.
>
> I run the task this way:
> - name: Start
> shell: /bin/ksh start.sh
> args:
> chdir: "/{{ rep_user }}/{{ openhrname }}/bin"
> register: start_op
>
> and I receive the following error:
> fatal: [integ]: FAILED! => {"changed": true, "cmd": "/bin/ksh start.sh",
> "delta": "0:00:00.007132", "end": "2016-05-24 15:13:36.666862", "failed":
> true, "rc": 127, "start": "2016-05-24 15:13:36.659730", "stderr":
> "start_openhr.sh[2]: dispatcher.sh: not found [No such file or directory]",
> "stdout": "", "stdout_lines": [], "warnings": []}

The error message says, start_openhr.sh can't find dispatcher.sh.
So dispatcher.sh is not in path or you should use absolute path.


--
Kai Stian Olstad

[fanvalt]

the script dispatcher.sh is in the directory where start.sh is launched.

I understand your meaning but it is hard to modify scripts that are downloaded from a package and I do not own.

because chdir is supposed to change directory before running a script, it should find all files from that directory when running the script.

[Kai Stian Olstad]

On 24. mai 2016 17:09, fanvalt wrote:
> the script dispatcher.sh is in the directory where start.sh is launched.
> I understand your meaning but it is hard to modify scripts that are
> downloaded from a package and I do not own.
>
> because chdir is supposed to change directory before running a script, it
> should find all files from that directory when running the script.

No it won't, ksh will first check if it find dispatcher.sh built-in
command if not, it will search all directories in the PATH. It will not
check the directory command is started from unless it is in PATH.

--
Kai Stian Olstad