Simple YAML to XML

Script converter of YAML to XML. Written in Python 3

Example

sample: file
list:
    example 1
    example 2
items:
    - name: example1
      price: 2

    - name: example2
      price: 3

Result in

<list>example 1 example 2
</list>
<sample>file</sample>
<items>
  <price>2</price>
  <name>example1</name>
</items>
<items>
  <price>3</price>
  <name>example2</name>
</items>

Information

• This is a simple script that parses an YAML file to XML
• This is testing and sampling purposes but can be changed to production ready
• This should be run in an isolated python environment (virtual env)

Requirements

• Python 3.6+
• Virtual env

Usage

This can be executed inside any python environment 3.6+ but I would recommend creating a virtual environment for this purpose and not to break any system

For the user of virtualenv wrapper (my preference), please follow the instructions below:

1. In the command line execute pip install virtualenvwrapper-win and linux users pip install virtualenvwrapper
2. Create your virtual env mkvirtualenv _name_at_your_choice
3. Install the requirements inside pip install -r requirements.txt
4. Execute the application python ./app.py with the required parameters

E.g.: python ./app.py -l <location>/<something>.yml -d <localtion>/<something>.xml

Parameters

• -l - Location of the file to be parsed (required)
• -d - Destination of the file with the file name after parsing (optional)
• -t - Whether elements get a data type attribute (defaulting to True. E.g.: <element type="str">)
• -s - Whether the output should be on the screen (defaulting to False)

You can run the result only on the screen or just destination or both

• Outputting to a file only python ./app.py -l resources/test.yml -d resources/result.xml -t false -s false
• Outputting to a file and screen python ./app.py -l resources/test.yml -d resources/result.xml -t false
• Outputting only to screen python ./app.py -l resources/test.yml -t false