Matching File Input Request Example

I am trying to build a matching request using csv file using python request library. I am getting <Response [500]> What I am doing wrong here?

My example csv input file contains just one column with the header name and 1 row, like this:

name

Facebook

There is my code:

request_url = "https://api-eit.refinitiv.com/permid/match/file"

headers = { 'Content-Type': 'multipart/form-data',

'x-ag-access-token': my_apikey,

'x-openmatch-numberOfMatchesPerRecord': '1',

'x-openmatch-dataType': 'Organization'}

files = {'file': open('C:/My_Folder/input_file.csv') }

response = requests.post(url=request_url, headers=headers, files=files)

r = response.json()

print(r)

----------

{'error': {'status': {'code': '500',

'errorCode': 'Server Error',

'errorDescription': 'java.lang.NullPointerException'}}}

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Accepted answers

zoya faberov

Hello @EA_MZ ,

In Python, by specifying "files" as parameter, the request is already 'Content-Type': 'multipart/form-data':

This discussion thread may be helpful.

It seems, that if you have it twice, Requests module gets it slightly off, resulting in the invalid submission.

It is mandatory.

All comments

zoya faberov

Hello @EA_MZ ,

Please see Record Matching API Swagger documentation. You can test with your file, to see if it is formatted correctly.

Via link "templates can be found here" you can download the required format templates, per record type of search, and test with this file. Next you can follow the same format for your entries file, that will ensure the valid format of the submitted file.

EA_MZ

Hi Zoya, I tested my file through the website and it works without problems, however it doesn't work when submitted through api using python requests.

It is a very basic input, just one row with a header 'Name'.

zoya faberov

Hi @EA_MZ ,

Try

import requests

request_url = "https://api-eit.refinitiv.com/permid/match/file&quot;

headers = { #'Content-Type': 'multipart/form-data',

'x-ag-access-token': 'YOURTOKEN',

'x-openmatch-numberOfMatchesPerRecord': '1',

'x-openmatch-dataType': 'Organization'}


files = {'file': open('.\exampleRM.csv') }

response = requests.post(request_url, files=files, headers=headers)

r = response.json()

print('Response:')
print(r)

works on my side

EA_MZ

Thanks Zoya! Disabling the 'Content-Type': 'multipart/form-data' in the header solved the issue! Do you know what is the cause of this issue?

The 'Content-Type': 'multipart/form-data' parameter is presented in the manual as mandatory, so it was confusing.

Image Caption

zoya faberov

Hello @EA_MZ ,

In Python, by specifying "files" as parameter, the request is already 'Content-Type': 'multipart/form-data':

This discussion thread may be helpful.

It seems, that if you have it twice, Requests module gets it slightly off, resulting in the invalid submission.

It is mandatory.

Jirapongse

@EA_MZ

You can refer to the OpenPermID Python library.

The source code is available on GitHub.

               files = {'file': open(filename)}
                response = requests.post(
                    url,
                    headers=headers,
                    files = files,
                    timeout=self.__timeout__)

1627360551087.png

EA_MZ

they omit the 'Content-Type': 'multipart/form-data' parameter in the openpermid module

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