Here I have an image url. filename is image url
def upload(filename, content)
conn = S3Connection(aws_access_key, aws_secret_key)
b = Bucket(conn, bucket_name)
k = Key(b)
k.key = filename..split('/')[::-1][0]
k.set_metadata("Content-Type", 'images/jpeg')
k.set_contents_from_string(content)
k.set_acl("public-read")
It upload things to the S3 but i开发者_JAVA技巧t shows the error:
/tmp/t.jpegcould not be opened, because the associated helper application does not exist. Change the association in your preferences.
I'm pretty sure images/jpeg is a typo. The correct mimetype is:
image/jpeg
k.key = filename..split('/')[::-1][0]
This line has some syntax error. replace .. with .
Check your file permission for the image file you are trying to read. From the error message it seems like you, or at least your program don't have read access to that file.
加载中,请稍侯......
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