How to get list_blobs to behave like gsutil

Issue

I would like to only get the first level of a fake folder structure on GCS.

If I run e.g.:


gsutil ls 'gs://gcp-public-data-sentinel-2/tiles/'

I get a list like this:

gs://gcp-public-data-sentinel-2/tiles/01/
gs://gcp-public-data-sentinel-2/tiles/02/
gs://gcp-public-data-sentinel-2/tiles/03/
gs://gcp-public-data-sentinel-2/tiles/04/
gs://gcp-public-data-sentinel-2/tiles/05/
gs://gcp-public-data-sentinel-2/tiles/06/
gs://gcp-public-data-sentinel-2/tiles/07/
gs://gcp-public-data-sentinel-2/tiles/08/
gs://gcp-public-data-sentinel-2/tiles/09/
gs://gcp-public-data-sentinel-2/tiles/10/
gs://gcp-public-data-sentinel-2/tiles/11/
gs://gcp-public-data-sentinel-2/tiles/12/
gs://gcp-public-data-sentinel-2/tiles/13/
gs://gcp-public-data-sentinel-2/tiles/14/
gs://gcp-public-data-sentinel-2/tiles/15/
.
.
.

Running code like the following in the Python API give me an empty result:

from google.cloud import storage
bucket_name = 'gcp-public-data-sentinel-2'
prefix = 'tiles/'
storage_client = storage.Client()
bucket = storage_client.get_bucket(bucket_name)
for blob in bucket.list_blobs(max_results=10, prefix=prefix,
                              delimiter='/'):
    print blob.name

If I don’t use the delimiter option I get all the results in the bucket which is not very useful.

Solution

Maybe not the best way, but, inspired by this comment on the official repository:

iterator = bucket.list_blobs(delimiter='/', prefix=prefix)
response = iterator._get_next_page_response()
for prefix in response['prefixes']:
    print('gs://'+bucket_name+'/'+prefix)

Gives:

gs://gcp-public-data-sentinel-2/tiles/01/
gs://gcp-public-data-sentinel-2/tiles/02/
gs://gcp-public-data-sentinel-2/tiles/03/
gs://gcp-public-data-sentinel-2/tiles/04/
gs://gcp-public-data-sentinel-2/tiles/05/
gs://gcp-public-data-sentinel-2/tiles/06/
gs://gcp-public-data-sentinel-2/tiles/07/
gs://gcp-public-data-sentinel-2/tiles/08/
gs://gcp-public-data-sentinel-2/tiles/09/
gs://gcp-public-data-sentinel-2/tiles/10/
...

Answered By – Mangu

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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