Is it possible to use python suds to read a wsdl file from the file system?

Question

From suds documentation, I can create a Client if I have a url for the WSDL.

from suds.client import Client
url = 'http://localhost:7080/webservices/WebServiceTestBean?wsdl'
client = Client(url)

I currently have the WSDL file on my file system. Is it possible to use suds to read the WSDL file from my file system instead of hosting it on a web server?

Kissable

Answer 1

try to use url='file:///path/to/file'

Sarraute

To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml) – Barberabarberry

Answer 2

Oneliner

# Python 3
import urllib, os 
url = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))

This is a more complete one liner that will:

• let you specify just the local path,
• get you the absolute path,
• and then format it as a file-url.

Based upon:

• the comments in the accepted answer and
• this https://mcmap.net/q/329239/-convert-a-filename-to-a-file-url
• and thanks to user Sebastian the updated Python 3 implementation since we should avoid writing legacy python at this time.

Original for reference

# Python 2 (Legacy Python)
import urlparse, urllib, os

url = urlparse.urljoin('file:', urllib.pathname2url(os.path.abspath("service.xml")))

Unwrap

In case someone is using python3, the names have changed: import urllib, os url = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml"))) – Flapper

that solved my problem that appeared when switching from python 3.8.x to 3.9.2 – Liminal

Answer 3

Using pathlib:

from pathlib import Path
url = Path('resources/your_definition.wsdl').absolute().as_uri()
    

Quartziferous