<p>Answers to some parts of your question are:</p>

<ul>
<li><p>"I want to find a softlink (to another file) in the $PATH environment variable on Linux managed nodes"</p>

<p>Use "shell" module instead of lookup. </p>

<blockquote>
  <p><a href="https://docs.ansible.com/ansible/latest/user_guide/playbooks_lookups.html#lookups-and-variables" rel="nofollow noreferrer">Lookup</a> occur on the local computer, not on the remote computer.</p>
</blockquote></li>
<li><p>"I want to extract the path of the file name for the file that is found. How do I do this?"</p>

<p>Use "files_found.files" to see the complete path. Checkout the return values section in <a href="https://docs.ansible.com/ansible/latest/modules/find_module.html#return-values" rel="nofollow noreferrer">find</a> module documentation.</p></li>
<li><p>As far as i understood, you want to do something like this</p></li>
</ul>

<p>Try out the following:</p>

<pre><code>- name: finding files matching "goodfile" pattern
  find:
     paths: "{{ some_path }}"
     patterns: "goodfile"
     file_type: link
  register: files_found

- name : get the complete path
  set_fact:
    files_found_path: "{{ files_found.files }}"

 - debug:
     var: files_found_path
</code></pre>

CC BY-SA 4.0
2018-12-16T06:30:45.370Z
2018-12-16T06:46:32.877Z
2018-12-16T06:46:32.877Z
dheerendra
<p>This has been reworded, expanded upon and clarified via an update late on 11/9/18.  I request it be upvoted now that it has enough code.</p>

<p>I want to find a softlink (to another file) in the $PATH environment variable on Linux managed nodes.</p>

<p>I have an Ansible playbook that uses a find module.  I have a string of potential directories to look in.  This string is the value assigned to the variable "result_from_prev".  My playbook looks like this:</p>

<pre><code> - debug:
     msg: "{{ item }}: {{ lookup('env', 'PATH') }}"
   with_dict: {a, 1}
   register: prevar
 - set_fact:
     midvar: "{{ prevar | dictsort | join('') }}"
 - set_fact:
     result_from_prev: "{{ midvar.split(':') }}"
 - find:
     paths: "{{ item }}"
     patterns: "goodfile"
     file_type: link
   register: files_found
   loop: "{{ result_from_prev }}"
 - set_fact: 
     jsonvar: "{{ files_found | to_json }}"
 - set_fact: 
     var: "{{ files_found | list }}"
 - debug:
     msg: "{{ item }}"
   loop: "{{ files_found | json_query('files[*].path' }}"
</code></pre>

<p>I want to extract the path of the file name for the file that is found.  How do I do this?</p>

<p>The above gives me an error about invalid data being passed to loop.  A list must be passed. I think that files_found is a dictionary.  How would I use wantlist=True here?  I tried, but I had no luck.</p>

<p>I tried using jsonvar instead of "files_found" in the loop.  But I got the same error about invalid data being passed to 'loop.'</p>

<p>I tried using listvar instead of files_found.  But I got the same error about invalid data being passed to 'loop.'</p>

<p>I tried using "with_items" instead of "loop".  But this prints out nothing for the msg.</p>

<p>In the JSON that the jsonvar variable can print out, there is a "matched" key.  The value is "0" for each directory path that the file was not found and the value is "1" if the file was found.  But I do not know how to extract just the full path of the file found.  I'd like it to be clean like this: </p>

<blockquote>
  <p>/tmp/subdirectory/goodfile</p>
</blockquote>

<p>The potential locations of where the file will be will change.  I have a framework that works: I have the potential files in a list in a variable named "result_from_prev".</p>

<p>I have tried using the "when" conditional and the item.matched notation to analyze the value of specific keys.  These efforts do not work for me.  I have tried putting the JSON into a file or using the "join" feature to create a long string.  But I have not had luck parsing this long string.  I have also tried using the "lookup" keyword with set_fact to assign a variable to that directory path that was found with the find module.  I tried to start over completely with the stat module instead of the find module.  But I found this to be difficult to use a list of potential paths whereas the searching by the file's name is supported with the find module.  I want to avoid using the "shell" module.  Nothing I have tried works.  </p>

<p>I expect the find module to return only successfully found files with a supported syntax that makes extracting the directory path relatively easy.  This is not what I have experienced.  How do I return just the full path of the file named "goodfile" that is matched with the find module above?</p>

CC BY-SA 4.0
2018-11-09T19:36:45.093Z
I down-rated the question because it does not [Include just enough code to allow others to reproduce the problem](https://stackoverflow.com/help/how-to-ask)
Vladimir Botka
CC BY-SA 4.0
2018-11-10T02:20:29.823Z
Can it be up-rated now?
Ranjit
CC BY-SA 4.0
2018-11-12T16:40:40.037Z
What could be done to improve this?
Ranjit
<ansible><linux>
How do I return just the full path of a file found with the find module in Ansible?
Ranjit
# How do I return just the full path of a file found with the find module in Ansible?

(Asked by: username_0 on 2018-11-09)

This has been reworded, expanded upon and clarified via an update late on 11/9/18.  I request it be upvoted now that it has enough code.
I want to find a softlink (to another file) in the $PATH environment variable on Linux managed nodes.
I have an Ansible playbook that uses a find module.  I have a string of potential directories to look in.  This string is the value assigned to the variable "result_from_prev".  My playbook looks like this:
 - debug:
     msg: "{{ item }}: {{ lookup('env', 'PATH') }}"
   with_dict: {a, 1}
   register: prevar
 - set_fact:
     midvar: "{{ prevar | dictsort | join('') }}"
 - set_fact:
     result_from_prev: "{{ midvar.split(':') }}"
 - find:
     paths: "{{ item }}"
     patterns: "goodfile"
     file_type: link
   register: files_found
   loop: "{{ result_from_prev }}"
 - set_fact: 
     jsonvar: "{{ files_found | to_json }}"
 - set_fact: 
     var: "{{ files_found | list }}"
 - debug:
     msg: "{{ item }}"
   loop: "{{ files_found | json_query('files[*].path' }}"

I want to extract the path of the file name for the file that is found.  How do I do this?
The above gives me an error about invalid data being passed to loop.  A list must be passed. I think that files_found is a dictionary.  How would I use wantlist=True here?  I tried, but I had no luck.
I tried using jsonvar instead of "files_found" in the loop.  But I got the same error about invalid data being passed to 'loop.'
I tried using listvar instead of files_found.  But I got the same error about invalid data being passed to 'loop.'
I tried using "with_items" instead of "loop".  But this prints out nothing for the msg.
In the JSON that the jsonvar variable can print out, there is a "matched" key.  The value is "0" for each directory path that the file was not found and the value is "1" if the file was found.  But I do not know how to extract just the full path of the file found.  I'd like it to be clean like this: 

/tmp/subdirectory/goodfile

The potential locations of where the file will be will change.  I have a framework that works: I have the potential files in a list in a variable named "result_from_prev".
I have tried using the "when" conditional and the item.matched notation to analyze the value of specific keys.  These efforts do not work for me.  I have tried putting the JSON into a file or using the "join" feature to create a long string.  But I have not had luck parsing this long string.  I have also tried using the "lookup" keyword with set_fact to assign a variable to that directory path that was found with the find module.  I tried to start over completely with the stat module instead of the find module.  But I found this to be difficult to use a list of potential paths whereas the searching by the file's name is supported with the find module.  I want to avoid using the "shell" module.  Nothing I have tried works.  
I expect the find module to return only successfully found files with a supported syntax that makes extracting the directory path relatively easy.  This is not what I have experienced.  How do I return just the full path of the file named "goodfile" that is matched with the find module above?

**Tags:** ansible, linux

> **Comments:**
> **username_1**: I down-rated the question because it does not [Include just enough code to allow others to reproduce the problem](https://stackoverflow.com/help/how-to-ask)  
> **username_0**: Can it be up-rated now?  
> **username_0**: What could be done to improve this?  

---

## 1 answers

### Answer by username_2 on 2018-12-16. Score: 1

Answers to some parts of your question are:

"I want to find a softlink (to another file) in the $PATH environment variable on Linux managed nodes"
Use "shell" module instead of lookup. 

Lookup occur on the local computer, not on the remote computer.

"I want to extract the path of the file name for the file that is found. How do I do this?"
Use "files_found.files" to see the complete path. Checkout the return values section in find module documentation.
As far as i understood, you want to do something like this

Try out the following:
- name: finding files matching "goodfile" pattern
  find:
     paths: "{{ some_path }}"
     patterns: "goodfile"
     file_type: link
  register: files_found

- name : get the complete path
  set_fact:
    files_found_path: "{{ files_found.files }}"

 - debug:
     var: files_found_path


---

