Is it possible to use python suds to read a wsdl file from the file system?
38
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?
python soap wsdl suds
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
add a comment |
38
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?
python soap wsdl suds
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
add a comment |
38
38
38
8
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?
python soap wsdl suds
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
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?
python soap wsdl suds
python soap wsdl suds
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
asked Oct 28 '10 at 19:49
Thierry Lam
19k3396136
19k3396136
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
51
try to use url='file:///path/to/file'
edited Jan 28 '14 at 15:19
sherpya
4,04712144
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
5
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
8
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
add a comment |
14
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://stackoverflow.com/a/14298190/622276
- 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")))
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
1
In case someone is using python3, the names have changed:import urllib, osurl = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))
– Sebastian
Jul 1 '17 at 9:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
51
try to use url='file:///path/to/file'
edited Jan 28 '14 at 15:19
sherpya
4,04712144
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
5
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
8
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
add a comment |
51
try to use url='file:///path/to/file'
edited Jan 28 '14 at 15:19
sherpya
4,04712144
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
5
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
8
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
add a comment |
51
51
51
try to use url='file:///path/to/file'
edited Jan 28 '14 at 15:19
sherpya
4,04712144
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
try to use url='file:///path/to/file'
edited Jan 28 '14 at 15:19
sherpya
4,04712144
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
edited Jan 28 '14 at 15:19
sherpya
4,04712144
edited Jan 28 '14 at 15:19
sherpya
4,04712144
edited Jan 28 '14 at 15:19
sherpya
4,04712144
4,04712144
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
answered Oct 28 '10 at 19:51
Gabi Purcaru
23.8k75882
23.8k75882
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
5
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
8
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
add a comment |
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
5
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
8
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
This is the correct answer.
– jathanism
Oct 28 '10 at 22:36
5
5
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
I had to add an extra slash, thanks for the answer.
– Thierry Lam
Nov 1 '10 at 14:38
8
8
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
To add to Thierry's comment, it also has to be an absolute path. (eg. file:///home/admin/service.xml)
– trinth
May 9 '12 at 16:14
add a comment |
14
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://stackoverflow.com/a/14298190/622276
- 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")))
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
1
In case someone is using python3, the names have changed:import urllib, osurl = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))
– Sebastian
Jul 1 '17 at 9:51
add a comment |
14
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://stackoverflow.com/a/14298190/622276
- 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")))
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
1
In case someone is using python3, the names have changed:import urllib, osurl = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))
– Sebastian
Jul 1 '17 at 9:51
add a comment |
14
14
14
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://stackoverflow.com/a/14298190/622276
- 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")))
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
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://stackoverflow.com/a/14298190/622276
- 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")))
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
edited Nov 13 '18 at 3:11
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
answered Feb 20 '15 at 2:11
Josh Peak
1,2131827
1,2131827
1
In case someone is using python3, the names have changed:import urllib, osurl = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))
– Sebastian
Jul 1 '17 at 9:51
add a comment |
1
In case someone is using python3, the names have changed:import urllib, osurl = urllib.parse.urljoin('file:', urllib.request.pathname2url(os.path.abspath("service.xml")))
– Sebastian
Jul 1 '17 at 9:51
1
1
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")))– Sebastian
Jul 1 '17 at 9:51
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")))– Sebastian
Jul 1 '17 at 9:51
add a comment |
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