如何在python中发送多部分/表单数据请求?怎么发文件,我懂,但是怎么用这种方法发表单数据就不懂了。
当前回答
为了澄清上面的例子,
即使不需要上传任何文件,也需要使用files参数发送多部分表单POST请求。
文件= {}
不幸的是,这行不通。
您将需要放入一些虚拟值,例如。
files={"foo": "bar"}
当我试图上传文件到Bitbucket的REST API时,我遇到了这个问题,不得不写这个讨厌的东西,以避免可怕的“不支持的媒体类型”错误:
url = "https://my-bitbucket.com/rest/api/latest/projects/FOO/repos/bar/browse/foobar.txt"
payload = {'branch': 'master',
'content': 'text that will appear in my file',
'message': 'uploading directly from python'}
files = {"foo": "bar"}
response = requests.put(url, data=payload, files=files)
:O=
其他回答
您需要使用网站HTML中上传文件的name属性。例子:
autocomplete="off" name="image">
你看到name="image">?你可以在上传文件的网站的HTML中找到它。您需要使用它来上传Multipart/form-data文件
脚本:
import requests
site = 'https://prnt.sc/upload.php' # the site where you upload the file
filename = 'image.jpg' # name example
这里,在image的位置,用HTML添加上传文件的名称
up = {'image':(filename, open(filename, 'rb'), "multipart/form-data")}
如果上传需要点击按钮进行上传,可以这样使用:
data = {
"Button" : "Submit",
}
然后启动请求
request = requests.post(site, files=up, data=data)
完成,文件上传成功
这是你需要上传一个大的单一文件作为多部分formdata的python片段。在服务器端运行NodeJs Multer中间件。
import requests
latest_file = 'path/to/file'
url = "http://httpbin.org/apiToUpload"
files = {'fieldName': open(latest_file, 'rb')}
r = requests.put(url, files=files)
对于服务器端,请检查multer文档:https://github.com/expressjs/multer 这里的字段single('fieldName')用于接受单个文件,如下所示:
var upload = multer().single('fieldName');
我试图用python 3中的请求模块发送一个请求到URL_server。 这对我来说很管用:
# -*- coding: utf-8 *-*
import json, requests
URL_SERVER_TO_POST_DATA = "URL_to_send_POST_request"
HEADERS = {"Content-Type" : "multipart/form-data;"}
def getPointsCC_Function():
file_data = {
'var1': (None, "valueOfYourVariable_1"),
'var2': (None, "valueOfYourVariable_2")
}
try:
resElastic = requests.post(URL_GET_BALANCE, files=file_data)
res = resElastic.json()
except Exception as e:
print(e)
print (json.dumps(res, indent=4, sort_keys=True))
getPointsCC_Function()
地点:
URL_SERVER_TO_POST_DATA =我们要发送数据的服务器 报头=发送的报头 file_data =发送的参数
通过在POST请求中指定files参数,请求的Content-Type被自动设置为multipart/form-data(后面跟着用于分隔multipart负载中的每个主体部分的边界字符串),无论您只发送文件,还是同时发送form-data和文件(因此,在这种情况下,不应该尝试手动设置Content-Type)。然而,如果只发送表单数据,则Content-Type将自动设置为application/x-www-form-urlencoded。
You can print out the Content-Type header of the request to verify the above using the example given below, which shows how to upload multiple files (or a single file) with (optionally) the same key (i.e., 'files' in the case below), as well as with optional form-data (i.e., data=data in the example below). The documentation on how to POST single and multiple files can be found here and here, respectively. In case you need to upload large files without reading them into memory, have a look at Streaming Uploads. For the server side—in case this is needed—please have a look at this answer, from which the code snippet below has been taken, and which uses the FastAPI web framework.
import requests
url = 'http://127.0.0.1:8000/submit'
files = [('files', open('a.txt', 'rb')), ('files', open('b.txt', 'rb'))]
#file = {'file': open('a.txt','rb')} # to send a single file
data ={"name": "foo", "point": 0.13, "is_accepted": False}
r = requests.post(url=url, data=data, files=files)
print(r.json())
print(r.request.headers['content-type'])
import requests
# assume sending two files
url = "put ur url here"
f1 = open("file 1 path", 'rb')
f2 = open("file 2 path", 'rb')
response = requests.post(url,files={"file1 name": f1, "file2 name":f2})
print(response)
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