I'm able to post file using curl
curl -X POST -i -F name='barca' -F country='spain' -F
file=#/home/messi/Desktop/barca.png 'http://localhost:8080/new_org/hel/concerts'
Which I can get (file) as
curl -X GET -H 'Accept: image/png' 'http://localhost:8080/new_org/hel/concerts/<id or name of entity>'
But when I tried same thing using requests.post, I got error. Does anybody know why this happen. (Post Error encounter when file pointer is not at last, but when file pointer is at last, I got response 200 but file is not posted)
import requests
url = 'http://localhost:8080/new_org/hel/concerts'
file = dict(file=open('/home/messi/Desktop/barca.png', 'rb'))
data = dict(name='barca', country='spain')
response = requests.post(url, files=file, data=data)
Error: (from usergrid) with response code: 400
{u'duration': 0,
u'error': u'illegal_argument',
u'error_description': u'value is null',
u'exception': u'java.lang.IllegalArgumentException',
u'timestamp': 1448330119021}
https://github.com/apache/usergrid
I believe the problem is that Python is not sending a content-type field for the image that you are sending. I traced through the Usergrid code using a debugger and saw that curl is sending the the content-type for the image and Python is not.
I was able to get this exact code to work on my local Usergrid:
import requests
url = 'http://10.1.1.161:8080/test-organization/test-app/photos/'
files = { 'file': ('13.jpg', open('/Users/dave/Downloads/13.jpg', 'rb'), 'image/jpeg')}
data = dict(name='barca', country='spain')
response = requests.post(url, files=files, data=data)
It is possible that Waken Meng's answer did not work because of the syntax of the files variable, but I'm no Python expert.
I met a problem before when i try to upload image files. Then I read the doc and do this part:
You can set the filename, content_type and headers explicitly:
Here is how I define the file_data:
file_data = [('pic', ('test.png', open('test.png'), 'image/png'))]
r = requests.post(url, files=file_data)
file_data should be a a list: [(param_name, (file_name, file, content_type))]
This works for me, hope can help you.
Related
I have a rest api that offers an upload file post functionality. To test this, I simply did this on my python test code:
fo = open('file.zip', 'rb')
rb_file = {'confile': ('file.zip', fo, 'multipart/form-data')}
resp = requests.post(url,files=rb_file)
fo.close()
This request returns a uuid response {ce9f2d23-8ecd-4c60-9d31-aef0be103d44} that is needed for initiating another post run for a scan.
From Swagger, manually passing this uuid to the scan post request generates the following curl:
curl -X POST "http://....../scan" -H "Content-Type: multipart/form-data" -F "FilID=ce9f2d23-8ecd-4c60-9d31-aef0be103d44"
My question is how to convert this curl to a python request code noting that I no longer have the file on my local machine. The server seems to be expecting a -F rather than a param. How do I pass a file that doesn't exist on my local machine to http request in this case? All I have is the filID. I tried running as param but that doesn't find the resource.
try this !
import requests
headers = {
'Content-Type': 'multipart/form-data',
}
files = {
'FilID': (None, 'ce9f2d23-8ecd-4c60-9d31-aef0be103d44'),
}
response = requests.post('http://....../scan', headers=headers, files=files)
I'm attempting to translate the following curl request to something that will run in django.
curl -X POST https://api.lemlist.com/api/hooks --data '{"targetUrl":"https://example.com/lemlist-hook"}' --header "Content-Type: application/json" --user ":1234567980abcedf"
I've run this in git bash and it returns the expected response.
What I have in my django project is the following:
apikey = '1234567980abcedf'
hookurl = 'https://example.com/lemlist-hook'
data = '{"targetUrl":hookurl}'
headers = {'Content-Type': 'application/json'}
response = requests.post(f'https://api.lemlist.com/api/hooks/', data=data, headers=headers, auth=('', apikey))
Running this python code returns this as a json response
{}
Any thoughts on where there might be a problem in my code?
Thanks!
Adding to what nikoola said, I think you want that whole data line to be as follows so you aren't passing that whole thing as a string. Requests will handle serializing and converting python objects to json for you [EDIT: if you use the json argument instead of data].
source: https://requests.readthedocs.io/en/master/user/quickstart/#more-complicated-post-requests
Instead of encoding the dict yourself, you can also pass it directly
using the json parameter (added in version 2.4.2) and it will be
encoded automatically:
Note, the json parameter is ignored if either data or files is passed.
Using the json parameter in the request will change the Content-Type
in the header to application/json.
data = {"targetUrl":hookurl}
import requests
headers = {
'Content-Type': 'application/json',
}
data = '{"targetUrl":"https://example.com/lemlist-hook"}'
response = requests.post('https://api.lemlist.com/api/hooks', headers=headers, data=data, auth=('', '1234567980abcedf'))
You can visit this url:-
https://curl.trillworks.com/
I am trying to do a curl post query using requests python 2.7, however the API response differently using curl vs. requests lib.
The post query is pretty simple a file and name-value-pair data as the API params.
Below is the curl multipart post request:
curl -uadmin:blabla123 -X POST 127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/6a0ab661-1c43-43ed-b07f-a564f6bcb5ca/children -F filedata=#file1.txt -F name=document__55;nodeType=content
The python 2.7 code is as below:
import requests
from requests.auth import HTTPBasicAuth, HTTPDigestAuth
from config import USER, PASSWD
def createDocument( documentFilename, documentMetadata, targetFolderNodeId):
'''
Uploads a file and its meta-data to the CMIS server under the specified
target folder
'''
with open(documentFilename, 'rb') as file:
files = {'file': file}
# createURL = 'http://127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/{0}/children'.format( targetFolderNodeId )
createURL = 'http://127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/6a0ab661-1c43-43ed-b07f-a564f6bcb5ca/children'
data = {
"name":"document__55",
"nodeType":"cm:content",
}
response = requests.post( createURL, data = data, files = files, auth=HTTPBasicAuth(USER, PASSWD) )
print(response)
print(response.json)
print(response.text)
createDocument('file1.txt', '', '')
Curl returns 200 http code but the script oddly returns 400.
Any help is much appreciate it.
At a first glace the reasons for that script failing vs curl working could be more than one:
nodeType=content in the curl, while "nodeType":"cm:content" in the script
there's an additional comma in the script's payload second line
you're not setting the Content-Type header as multipart/form-data in the script, which curl's -F option actually does
Also, the endpoint should provide a verbose error along with the 400 response. If it does, that could be helpful to identify the error cause.
in fact in the curl I have -F filedata=#file1.txt but in the script I have files = {'file': file} which is not the same so I just had to use files = {'filedata': file} :)
def createDocument( documentFilename, documentMetadata, targetFolderNodeId):
'''
Uploads a file and its meta-data to the CMIS server under the specified
target folder
'''
with open(documentFilename, 'rb') as file:
files = {'filedata': file}
createURL = 'http://127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/{0}/children'.format( targetFolderNodeId )
data = {
"name":"document__77",
"nodeType":"cm:content"
}
response = requests.post( createURL, data = data, files = files, auth=HTTPBasicAuth(USER, PASSWD) )
print(response)
print(response.json)
print(response.text)
print(response.headers)
# print([i for i in dir(response) if 'header' in i])
I have this cURL request that I want to convert into a Python Request code.
The cURL content is
curl -H "X-PrettyPrint: 1"
-F 'json={"title":"PandaTest"};type=application/json'
-F "fileData=#rename.py;type=application/octet-stream"
-X POST https://cs31.salesforce.com/services/data/v39.0/connect/files/users/me
-H 'Authorization: 00Dp000000.....CqqU0.S_5r' --insecure
For more details of the request check the SalesForce docs it contains the HTTP request message - here. Search for the section Upload a file to the Files home.
The Python counter part of it what I've written is
import requests
files = {
"fileData" : open("rename.py", "rb"),
"json" : '{"title":"PandaTest"}'
}
headers = {
'Authorization': 'OAuth 00Dp00000000u....n3ZGuoZK2wYJRCqqU0.S_5r',
"Content-Disposition": "form-data 'fileData'"
}
r = requests.post('https://cs31.salesforce.com/services/data/v39.0/connect/files/users/me/',
data=files, headers=headers)
data = json.loads(r.text)
print data
My request is sent successfully but I get an error Missing expected "fileData" binary parameter.I have a feeling the request which I'm trying to send is not formed correctly. Where exactly did I go wrong?
I have a feeling I'm not handling the 2 -F in the cURL request correctly.
This line:
"fileData" : open("rename.py", "rb"),
is setting "fileName" to the object that is returned when you open a file. If you would like "fileName" to be the actual contents of the file, then change the line to this:
"fileData" : open("rename.py", "rb").read(),
That will read all the bytes in the file and set "fileName" to them.
I have this cURL:
curl -X POST http://user:pass#blabla.com:8080/job/myproject/config.xml --data-binary "#new_config.xml"
I am basically trying to set a new config for a Jenkins installation by changing the pre-existing config.xml file.
I am trying to convert it to something like this in order to use it more flexibly in my code:
url = "http://host:8080/job/myproject/config.xml"
auth = ('user','pass')
payload = {"--data-binary": "#new_config.xml"}
headers = {"Content-Type" : "application/xml"}
r = requests.post(url, auth=auth, data=payload, headers=headers)
I know that I am using incorrectly the payload and the headers.How should I change them?
I run it and I take a 500 responce code.
I read this post , but I am struggling to apply it in my case.
The --data-binary switch means: post the command line argument as the whole POST body, without wrapping in multipart/form-data or application/x-www-form-encoding containers. # tells curl to load the data from a filename; new_config.xml in this case.
You'll need to open the file object to send the contents as the data argument:
url = "http://host:8080/job/myproject/config.xml"
auth = ('user','pass')
headers = {"Content-Type" : "application/xml"}
with open('new_config.xml', 'rb') as payload:
r = requests.post(url, auth=auth, data=payload, headers=headers)
Note that I pass the file object directly into requests; the data will then be read and pushed to the HTTP socket, streaming the data efficiently.