使用 python 请求发布带有 json 数据的文件

问题描述 投票:0回答:1

我有接受带有 json 数据的多个文件的 API -

curl -X POST \
  http://localhost:25965/v1/import \
  -H 'Content-Type: application/json' \
  -H 'Postman-Token: xxxxxxxx-xxxx-xxxx-xxxx-ba66a9b8d6cb' \
  -H 'cache-control: no-cache' \
  -H 'content-type: multipart/form-data; boundary=----WebKitFormBoundary7MA4YWxkTrZu0gW' \
  -F 'files=@C:\Users\user\File1.xlsx' \
  -F 'files=@C:\Users\user\File2.xlsx' \
  -F 'files=@C:\Users\user\File3.xlsx' \
  -F 'values=[
   {
        "field1": "Value11",
        "field2": "Value21",
        "field3": "File1.xlsx"
    },
   {
        "field1": "Value21",
        "field2": "Value22",
        "field3": "File2.xlsx"
    },
   {
        "field1": "Value31",
        "field2": "Value32",
        "field3": "File3.xlsx"
    }
]'

我正在尝试将此请求转换为Python。 对于请求,我收到错误需要解包的值太多(预期为 2)

def PostFiles(self, apiName, values, files):
        url = self.ApiUrl + apiName 
        params = {}
        for file in files:
            params.update({'files':file})
        response = requests.post(url, 
                                 files=params,
                                 data=values,
                                 headers={
                                     'Content-Type':'application/json'},
                                 auth=HTTPKerberosAuth(delegate=True))
        return response

我也尝试过 requests_toolbelt 但不起作用。

def PostFiles(self, apiName, values, files):
        url = self.ApiUrl + apiName  
        params = {}
        for file in files:
            params.update({'files':file})
        params.update({'values':json.dumps(values)})

        multipart_data = MultipartEncoder(params)

        headers = { 'Content-Type':multipart_data.content_type}
        response = requests.post(url, 
                                 data=multipart_data,
                                 headers=headers,
                                 auth=HTTPKerberosAuth(delegate=True))
        return response

这是我的函数调用 -

 files = [open(join(directory, filename), 'rb')]
    #files = [('files',(filename, open(join(directory, filename), 'rb'),'application/vnd.ms-excel'))]
    values = [{
             'field1': 'value11',
             'field2' : 'value21',
             'field3' : 'File1.xlsx'
        }]
    response = PostFiles('import', values, files)
    print(response)

有与此主题相关的帖子,但我找不到任何使用 json 发布多个文件的内容。

python api python-requests
1个回答
0
投票

我按照在单个请求中发布 JSON 和文件修改了我的代码并按预期工作。

def PostFiles(self, apiName, values, files):
        url = self.ApiUrl + apiName 
        params = {
            'values': (None, json.dumps(values), 'application/json')
            }
        for file in files:
            params.update({'files':file})
        response = requests.post(url, 
                                 files=params,
                                 auth=HTTPKerberosAuth(delegate=True))
        result = response.json()
        if response.status_code != 200 :
            raise Exception(result)
        return json.dumps(result)
© www.soinside.com 2019 - 2024. All rights reserved.