我正在开发一个Django web应用程序,它接收PDF文件,并对PDF的每个页面执行一些图像处理。我得到一个PDF,我需要将每个页面保存到我的。我正在使用pdf2image的convert_from_path()为PDF中的每一页生成一个枕头图像列表。现在,我想把这些图像保存到Google存储库中,但是我想不出答案。
我已经成功地在本地保存了这些枕头图像,但是我不知道如何在云中实现这一点。
fullURL = file.pdf.url
client = storage.Client()
bucket = client.get_bucket('name-of-my-bucket')
blob = bucket.blob(file.pdf.name[:-4] + '/')
blob.upload_from_string('', content_type='application/x-www-form-urlencoded;charset=UTF-8')
pages = convert_from_path(fullURL, 400)
for i,page in enumerate(pages):
blob = bucket.blob(file.pdf.name[:-4] + '/' + str(i) + '.jpg')
blob.upload_from_string('', content_type='image/jpeg')
outfile = file.pdf.name[:-4] + '/' + str(i) + '.jpg'
page.save(outfile)
of = open(outfile, 'rb')
blob.upload_from_file(of)发布于 2019-04-04 05:43:43
所以从不使用by商店开始吧。他们试图摆脱它,让人们使用云存储。首先设置云存储
我使用的是webapp2,而不是Django,但我相信您可以搞清楚。另外,我不使用枕头图像,所以你必须打开你要上传的图片。然后这样做(假设您试图发布数据):
import cloudstorage as gcs
import io
import StringIO
from google.appengine.api import app_identity在get和post在它自己的部分之前
def create_file(self, filename, Dacontents):
write_retry_params = gcs.RetryParams(backoff_factor=1.1)
gcs_file = gcs.open(filename,
'w',
content_type='image/jpeg',
options={'x-goog-meta-foo': 'foo',
'x-goog-meta-bar': 'bar'},
retry_params=write_retry_params)
gcs_file.write(Dacontents)
gcs_file.close()在你的HTML中
<form action="/(whatever yoururl is)" method="post"enctype="multipart/form-data">
<input type="file" name="orders"/>
<input type="submit"/>
</form>在邮政
orders=self.request.POST.get(‘orders)#this is for webapp2
bucket_name = os.environ.get('BUCKET_NAME',app_identity.get_default_gcs_bucket_name())
bucket = '/' + bucket_name
OpenOrders=orders.file.read()
if OpenOrders:
filename = bucket + '/whateverYouWantToCallIt'
self.create_file(filename,OpenOrders)发布于 2019-04-03 09:52:27
因为您已经在本地保存了这些文件,那么它们就可以在运行web应用程序的本地目录中使用。
您可以做的只是迭代该目录的文件,然后一个一个地将它们上传到。
这里是一个示例代码:
您将需要这个库:
谷歌云存储
Python代码:
#Libraries
import os
from google.cloud import storage
#Public variable declarations:
bucket_name = "[BUCKET_NAME]"
local_directory = "local/directory/of/the/files/for/uploading/"
bucket_directory = "uploaded/files/" #Where the files will be uploaded in the bucket
#Upload file from source to destination
def upload_blob(source_file_name, destination_blob_name):
storage_client = storage.Client()
bucket = storage_client.get_bucket(bucket_name)
blob = bucket.blob(destination_blob_name)
blob.upload_from_filename(source_file_name)
#Iterate through all files in that directory and upload one by one using the same filename
def upload_files():
for filename in os.listdir(local_directory):
upload_blob(local_directory + filename, bucket_directory + filename)
return "File uploaded!"
#Call this function in your code:
upload_files()注意:我在应用程序中测试了代码,它对我起了作用。了解它是如何工作的,并根据您的需要对其进行修改。我希望这能帮上忙。
https://stackoverflow.com/questions/55483775
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