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Python BundleService.read_file方法代码示例

本文整理汇总了Python中apps.web.bundles.BundleService.read_file方法的典型用法代码示例。如果您正苦于以下问题:Python BundleService.read_file方法的具体用法?Python BundleService.read_file怎么用?Python BundleService.read_file使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在apps.web.bundles.BundleService的用法示例。


在下文中一共展示了BundleService.read_file方法的6个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: get

# 需要导入模块: from apps.web.bundles import BundleService [as 别名]
# 或者: from apps.web.bundles.BundleService import read_file [as 别名]
 def get(self, request, uuid, path):
     user_id = self.request.user.id
     service = BundleService()
     try:
         content_type = BundleFileContentApi._content_type(path)
         return StreamingHttpResponse(service.read_file(uuid, path), content_type=content_type)
     except Exception as e:
         return Response(status=service.http_status_from_exception(e))
开发者ID:csrampey,项目名称:codalab,代码行数:10,代码来源:views.py

示例2: BundleDownload

# 需要导入模块: from apps.web.bundles import BundleService [as 别名]
# 或者: from apps.web.bundles.BundleService import read_file [as 别名]
def BundleDownload(request, uuid):
    service = BundleService(request.user)

    local_path, temp_path = service.download_target(uuid, return_zip=True)
    item = service.get_bundle_info(uuid)

    response = StreamingHttpResponse(service.read_file(uuid, local_path), content_type="zip")
    response['Content-Disposition'] = 'attachment; filename="%s.zip"' % item['metadata']['name']
    return response
开发者ID:samarjeet,项目名称:codalab,代码行数:11,代码来源:views.py

示例3: get

# 需要导入模块: from apps.web.bundles import BundleService [as 别名]
# 或者: from apps.web.bundles.BundleService import read_file [as 别名]
 def get(self, request, uuid, path):
     # user_id = self.request.user.id
     service = BundleService(self.request.user)
     try:
         content_type, _encoding = mimetypes.guess_type(path)
         if not content_type:
             content_type = 'text/plain'
         return StreamingHttpResponse(service.read_file(uuid, path), content_type=content_type)
     except Exception as e:
         tb = traceback.format_exc()
         log_exception(self, e, tb)
         return Response({"error": smart_str(e)}, status=500)
开发者ID:abmnv,项目名称:codalab,代码行数:14,代码来源:worksheet_views.py

示例4: get

# 需要导入模块: from apps.web.bundles import BundleService [as 别名]
# 或者: from apps.web.bundles.BundleService import read_file [as 别名]
 def get(self, request, uuid, path):
     user_id = self.request.user.id
     service = BundleService(self.request.user)
     try:
         content_type = BundleFileContentApi._content_type(path)
         return StreamingHttpResponse(service.read_file(uuid, path), content_type=content_type)
     except Exception as e:
         logging.error(self.__str__())
         logging.error(smart_str(e))
         logging.error("")
         logging.debug("-------------------------")
         tb = traceback.format_exc()
         logging.error(tb)
         logging.debug("-------------------------")
         return Response(status=service.http_status_from_exception(e))
开发者ID:prags,项目名称:codalab,代码行数:17,代码来源:views.py

示例5: get

# 需要导入模块: from apps.web.bundles import BundleService [as 别名]
# 或者: from apps.web.bundles.BundleService import read_file [as 别名]
 def get(self, request, uuid, path):
     # user_id = self.request.user.id
     service = BundleService(self.request.user)
     try:
         content_type, _encoding = mimetypes.guess_type(path)
         if not content_type:
             content_type = 'text/plain'
         return StreamingHttpResponse(service.read_file(uuid, path), content_type=content_type)
     except Exception as e:
         logging.error(self.__str__())
         logging.error(smart_str(e))
         logging.error('')
         logging.debug('-------------------------')
         tb = traceback.format_exc()
         logging.error(tb)
         logging.debug('-------------------------')
         return Response(status=service.http_status_from_exception(e))
开发者ID:TPNguyen,项目名称:codalab,代码行数:19,代码来源:worksheet_views.py

示例6: BundleDownload

# 需要导入模块: from apps.web.bundles import BundleService [as 别名]
# 或者: from apps.web.bundles.BundleService import read_file [as 别名]
def BundleDownload(request, uuid):
    service = BundleService(request.user)

    local_path, temp_path = service.download_target(uuid, return_zip=True)

    return StreamingHttpResponse(service.read_file(uuid, local_path), content_type="zip")
开发者ID:riccitensor,项目名称:codalab,代码行数:8,代码来源:views.py


注:本文中的apps.web.bundles.BundleService.read_file方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。