how to use splunk appserver receive and save files

[hellosplunkit]

Hi Splunker,

Today, I refer to this link https://wiki.splunk.com/Community:40GUIDevelopment  to develop the interface. I use the python library requests to upload the file, and use the CherryPy of Splunk to receive the file. But I can't receive and save the file. I printed the parameter content, but I didn't receive any parameter. Here is my code。

I use the python library requests to upload the file

    def forwardFile(self, filePath, fileName, url):
        files = {'ufile': open(os.path.join(filePath, fileName), 'rb')}
        data = {'xxx':'aaa','xxdd':'bbb'}
        headers = {}
        headers["content-type"] = 'multipart/form-data'
        r = requests.post(url, files=files,data=data ,headers=headers, verify=False)
        logger.info(r)

use the CherryPy of Splunk to receive the file

    @route('/:action=receive')
    @expose_page(must_login=False, methods=['GET','POST'])
    @cherrypy.expose
    def receive(self, **kwargs):
        logger.info("111")
        logger.info(kwargs)
        logger.info("222")

then i see the file web_service.log 

2020-10-15 23:21:07,472 INFO	[5f8868e36d107834250] <string>:68 - hello world:http://127.0.0.1
2020-10-15 23:21:07,501 INFO	[5f8868e37f107d00ad0] <string>:145 - 111
2020-10-15 23:21:07,501 INFO	[5f8868e37f107d00ad0] <string>:146 - {'action': 'receive'}
2020-10-15 23:21:07,501 INFO	[5f8868e37f107d00ad0] <string>:147 - 222
2020-10-15 23:21:07,507 INFO	[5f8868e36d107834250] <string>:79 - <Response [200]>

How to get the value of ufile ？

then i can save it.

[feelcool]

hello,hellosplunkit  

I have the same problem. Have you solved the problem yet??

[richgalloway]

That wiki page is very old (4 major versions!) so I would be very careful about using anything you read there.  CherryPy is deprecated in Splunk 8 so that's another reason to try a different approach.

Please describe what you want to accomplish and perhaps someone will have a modern way to do it.