Thanks Jon, it's working now

On Wed, Oct 9, 2019 at 3:38 AM 'J Hawkesworth' via Ansible Project <
ansible-project@googlegroups.com> wrote:

> You would probably need to write like this:
>
> - win_shell: 'dir PackageDeploy*.* /b > {{directory}}\configurations\temp'
>    args:
>       executable: cmd
>       chdir: '{{directory}}\configurations\jmxterm'
>    register: list_out
>
> But its probably much better to use win_find module
> https://docs.ansible.com/ansible/latest/modules/win_find_module.htmld as
> the results will be easier to make use of than the contents of your
> list_out registered variable.
>
> - name: find PackageDeploy files
>   win_find:
>      paths: '{{directory}}\configurations\jmxterm'
>      patterns: 'PackageDeploy*.*'
>   register: package_deploy_info
>
> Hope this helps,
>
> Jon
>
> On Tuesday, October 8, 2019 at 9:31:48 AM UTC+1, Uppara venkat wrote:
>>
>> Hi all,
>> I'm trying to list the files with the name of  "PackageDeploy" using as
>> below but getting all other files also in the directory.
>>
>> - win_shell: 'dir PackageDeploy*.* /b
>> {{directory}}\configurations\jmxterm > {{directory}}\configurations\temp'
>>       args:
>>          executable: cmd
>>       register: list_out
>>
>>
>> Could you please suggest the right method for list the file with name ?
>>
>>
>> Best regards,
>> Venkat
>>
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-- 






Thanks& Regards,
venkatesulu.U
8147041026

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