I have written code on my backend (hosted on Elastic Beanstalk) to retrieve a file from an S3 bucket and save it back to the bucket under a different name. I am using boto3 and have created an s3 client called 's3'.
bucketname is the name of the bucket, keyname is name of the key. I am also using the tempfile module
tmp = tempfile.NamedTemporaryFile()
with open(tmp.name, 'wb') as f:
s3.download_fileobj(bucketname, keyname, f)
s3.upload_file(tmp, bucketname, 'fake.jpg')
I was wondering if my understanding was off (still debugging why there is an error) - I created a tempfile and opened and saved within it the contents of the object with the keyname and bucketname. Then I uploaded that temp file to the bucket under a different name. Is my reasoning correct?
The upload_file() command is expecting a filename (as a string) in the first parameter, not a file object.
Instead, you should use upload_fileobj().
However, I would recommend something different...
If you simply wish to make a copy of an object, you can use copy_object:
response = client.copy_object(
Bucket='destinationbucket',
CopySource='/sourcebucket/HappyFace.jpg',
Key='HappyFaceCopy.jpg',
)
Related
I was trying to open a file/image in python/django and upload it to s3 but I get different errors depending on what I try. I can get it to work when I send the image using the front end html form but not when opening the file on the back end. I get errors such as "'bytes' object has no attribute 'file'" Any ideas how to open an image and upload it to s3? I wasn't sure if I was using the correct upload function, but it worked when I received the file from an html form instead of opening it directly.
image = open(fileURL, encoding="utf-8")
S3_BUCKET = settings.AWS_BUCKET
session = boto3.Session(
aws_access_key_id=settings.AWS_ACCESS_KEY_ID,
aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY,
)
s3 = session.resource('s3')
s3.Bucket(S3_BUCKET).put_object(Key='folder/%s' % fileName, Body=image)
Thanks.
The open command return a file object. Therefore Body=image does not contain the actual contents of the object.
Since you want to upload an existing object, you could use:
Key = 'folder/' + fileName
s3.Object(S3_BUCKET, Key).upload_file(fileURL)
For a project, I'm trying to get an uploaded image file, stored in a bucket. I'm trying to have Python save a copy temporarily, just to perform a few tasks on this file (read, decode and give the decoded file back as JSON). After this is done, the temp file needs to be deleted.
I'm using Python 3.8, if that helps at all.
If you want some snippets of what I tried, I'm happy to provide :)
#edit
So far, I tried just downloading the file from the bucket, which works. But I can't seem to figure out how to temporarily save it to just decode (I got an API that will decode the image and get data from that file). This is the code for downloading
def download_file_from_bucket(blob_name, file_path, bucket_name):
try:
bucket = storage_client.get_bucket(bucket_name)
blob = bucket.blob(blob_name)
with open(file_path, 'wb') as f:
storage_client.download_blob_to_file(blob, f)
except Exception as e:
print(e)
return False
bucket_name = 'white-cards-with-qr'
download_file_from_bucket('My first Blob Image', os.path.join(os.getcwd(), 'file2.jpg'), bucket_name)
for object store in cloud environment, you can sign your object to give access for ones who don't have account for that object, you may read this for google cloud
You can use the tempfile library. This is a really basic snippet. You can also name the file or read it after writing it.
import tempfile
temp = tempfile.TemporaryFile()
try:
temp.write(blob)
finally:
temp.close()
I have a json file placed in s3. The s3 url is similar to the below one:
https://s3-eu-region-1.amazonaws.com/dir-resources/sample.json
But in pyspark when pass the same, It is not reading the file.
path = "https://s3-eu-region-1.amazonaws.com/dir-resources/sample.json"
df=spark.read.json(path)
But I am able to download it through browser.
Assuming that dir-resources is the name of your bucket, you should be able to access to the file with the following URI:
path = "s3://dir-resources/sample.json"
In some cases, you may have to use the s3n protocol instead:
path = "s3n://dir-resources/sample.json"
This code writes json to a file in s3,
what i wanted to achieve is instead of opening data.json file and writing to s3 (sample.json) file,
how do i pass the json directly and write to a file in s3 ?
import boto3
s3 = boto3.resource('s3', aws_access_key_id='aws_key', aws_secret_access_key='aws_sec_key')
s3.Object('mybucket', 'sample.json').put(Body=open('data.json', 'rb'))
I'm not sure, if I get the question right. You just want to write JSON data to a file using Boto3? The following code writes a python dictionary to a JSON file.
import json
import boto3
s3 = boto3.resource('s3')
s3object = s3.Object('your-bucket-name', 'your_file.json')
s3object.put(
Body=(bytes(json.dumps(json_data).encode('UTF-8')))
)
I don't know if anyone is still attempting to use this thread, but I was attempting to upload a JSON to s3 trying to use the method above, which didnt quite work for me. Boto and s3 might have changed since 2018, but this achieved the results for me:
import json
import boto3
s3 = boto3.client('s3')
json_object = 'your_json_object here'
s3.put_object(
Body=json.dumps(json_object),
Bucket='your_bucket_name',
Key='your_key_here'
)
Amazon S3 is an object store (File store in reality). The primary operations are PUT and GET. You can not add data into an existing object in S3. You can only replace the entire object itself.
For a list of available operations you can perform on s3 see this link
http://docs.aws.amazon.com/AmazonS3/latest/API/RESTObjectOps.html
An alternative to Joseph McCombs answer can be achieved using s3fs.
from s3fs import S3FileSystem
json_object = {'test': 3.14}
path_to_s3_object = 's3://your-s3-bucket/your_json_filename.json'
s3 = S3FileSystem()
with s3.open(path_to_s3_object, 'w') as file:
json.dump(json_object, file)
I'm attempting to save an image to S3 using boto. It does save a file, but it doesn't appear to save it correctly. If I try to open the file in S3, it just shows a broken image icon. Here's the code I'm using:
# Get and verify the file
file = request.FILES['file']
try:
img = Image.open(file)
except:
return api.error(400)
# Determine a filename
filename = file.name
# Upload to AWS and register
s3 = boto.connect_s3(aws_access_key_id=settings.AWS_KEY_ID,
aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY)
bucket = s3.get_bucket(settings.AWS_BUCKET)
f = bucket.new_key(filename)
f.set_contents_from_file(file)
I've also tried replacing the last line with:
f.set_contents_from_string(file.read())
But that didn't work either. Is there something obvious that I'm missing here? I'm aware django-storages has a boto backend, but because of complexity with this model, I do not want to use forms with django-storages.
Incase you don't want to go for django-storages and just want to upload few files to s3 rather then all the files then below is the code:
import boto3
file = request.FILES['upload']
s3 = boto3.resource('s3', aws_access_key_id=settings.AWS_ACCESS_KEY, aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY)
bucket = s3.Bucket('bucket-name')
bucket.put_object(Key=filename, Body=file)
You should use django-storages which uses boto internally.
You can either swap the default FileSystemStorage, or create a new storage instance and manually save files. Based on your code example I guess you really want to go with the first option.
Please consider using django's Form instead of directly accessing the request.