问题:我怎样才能让它发挥作用
我尝试使用python请求api将压缩文件发送到服务器。我在文档中看到了这个方法:
r = requests.post(url, files=open('foo.png', 'rb'))
但我所做的不同之处在于,我的压缩文件在内存中,只有一个python对象,没有创建文件的物理压缩版本:
我使用的是zipfile api,创建zip文件的方式如下:
^{pr2}$并尝试以下操作(写入zip文件后):
r = requests.post(url, outFile)
但是它不工作,看起来对象没有被识别为参数。这是堆栈跟踪
Traceback (most recent call last): File
"/Users/abdulahmad/Desktop/upload-script-ve/bin/cogs", line 11, in
<module>
sys.exit(main()) File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/cogs/run.py",
line 396, in main
return run(sys.argv) File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/cogs/run.py",
line 384, in run
return instance() File "/Users/abdulahmad/Desktop/upload-script-ve//src/ctl.py",
line 53, in __call__
handler = uploader(self.url, self.file) File "/Users/abdulahmad/Desktop/upload-script-ve//src/uploader.py",
line 24, in __call__
response = self.session.post(url, files=payload)
#this is where I'm adding the file (the payload)
File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/requests/sessions.py",
line 511, in post
return self.request('POST', url, data=data, json=json, **kwargs) File
"/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/requests/sessions.py",
line 454, in request
prep = self.prepare_request(req) File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/requests/sessions.py",
line 388, in prepare_request
hooks=merge_hooks(request.hooks, self.hooks), File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/requests/models.py",
line 296, in prepare
self.prepare_body(data, files, json) File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/requests/models.py",
line 447, in prepare_body
(body, content_type) = self._encode_files(files, data) File "/Users/abdulahmad/Desktop/upload-script-ve/lib/python2.7/site-packages/requests/models.py",
line 150, in _encode_files
fdata = fp.read() TypeError: read() takes at least 2 arguments (1 given)
实际代码:
inMemoryOutputFile = StringIO()
parentDir, dirToZip = os.path.split(dirPath)
def trimPath(path):
archivePath = path.replace(parentDir, "", 1)
if parentDir:
archivePath = archivePath.replace(os.path.sep, "", 1)
if not includeDirInZip:
archivePath = archivePath.replace(dirToZip + os.path.sep, "", 1)
return os.path.normcase(archivePath)
outFile = zipfile.ZipFile(inMemoryOutputFile, "w",
compression=zipfile.ZIP_DEFLATED)
for (archiveDirPath, dirNames, fileNames) in os.walk(dirPath):
for fileName in fileNames:
filePath = os.path.join(archiveDirPath, fileName)
outFile.write(filePath, trimPath(filePath))
if not fileNames and not dirNames:
zipInfo = zipfile.ZipInfo(trimPath(archiveDirPath) + "/")
outFile.writestr(zipInfo, "")
outFile.close()
return outFile
您需要将}。
StringIO
缓冲区传递给requests
,而不是{ZipFile.read("somefile.txt")
从归档文件中读取未压缩的文件,它不读取压缩的二进制流。这个read
需要1个参数,这就是为什么您得到奇怪的错误消息。在发布之前倒带文件,否则POST
数据将为空。在此示例向您展示了工作流。在
相关问题 更多 >
编程相关推荐