Question

I am trying to upload a file to Windows Azure blob storage. From my understanding, you can upload files into something similar to a subdirectory. I want to upload a file to a blob such that the file essentially resides in /TestContainer/subDirectory1/subDirectory2/file.png

// Setup the Windows Aure blob client
CloudStorageAccount storageAccount = CloudStorageAccount.FromConfigurationSetting("BlobStorage");
CloudBlobClient client = storageAccount.CreateCloudBlobClient();

// Retrieve the TestContainer container from blob storage
CloudBlobContainer container = client.GetContainerReference("TestContainer");
if (container.CreateIfNotExist())
  container.SetPermissions(new BlobContainerPermissions { PublicAccess = BlobContainerPublicAccessType.Blob });

// Setup the blob
CloudBlob blob = container.GetBlobReference("");

// Create the meta data for the blob
NameValueCollection metadata = new NameValueCollection();
metadata["id"] = fileID.ToString();
blob.Metadata.Add(metadata);

// Store the blob 
byte[] bytes = GetFileBytes();
blob.UploadByteArray(bytes);

How do I upload a file with a directory structure? The link here mentions there’s a way to do it. However, it doesn’t show you how to do it.

Thank you!

Answer 1

There are a couple of ways. The easier way is to just use / characters as @makerofthings7 already mentioned. You can also use a CloudBlobDirectory object if you prefer. Here’s an example showing both.

CloudBlobContainer testContainer = blobClient.GetContainerReference("testcontainer");

//Upload using a CloudBlobDirectory object
var dir = testContainer.GetDirectoryReference("UsingCloudDirectory/foo/bar/baz/");
var blobRef = dir.GetBlockBlobReference("BlobByDir.bin");

using (MemoryStream ms = new MemoryStream(new byte[] { 0x0 }))
{
    blobRef.UploadFromStream(ms);
}

//Upload using the filename without a CloudBlobDirectory
var blobRef2 = testContainer.GetBlockBlobReference("UsingBlobName/foo/bar/baz/BlobByName.bin");
using (MemoryStream ms = new MemoryStream(new byte[] { 0x0 }))
{
    blobRef2.UploadFromStream(ms);
}