Powershell 命令 Set-AzureStorageBlobContent 忽略标志
Powershell command Set-AzureStorageBlobContent ignore flag
我想将新的 blob 上传到 Azure 存储帐户,但只有其中一个不存在。我想知道是否有任何方法可以设置 Set-AzureStorageBlobContent 命令以忽略已经存在的项目,这样我就可以在一次操作中实现它?
I want to upload new blob to Azure storage account, but only of it doesn't already exist.
从 Set-AzureStorageBlobContent, it seems that there are not existing parameters that could be used to skip/ignore the existing blob while uploading the file to blob. As a workaround, we could use Get-AzureStorageBlob cmdlet 的文档中检查 blob 在 运行 Set-AzureStorageBlobContent cmdlet 将文件上传到 blob 之前是否已经存在。
try
{
$blob = Get-AzureStorageBlob -Blob $blobName -Container $containerName -Context $blobContext -ErrorAction Stop
Write-Host "Blob is existing"
}
catch
{
# The blob doesn't exist
# Add logic here
Write-Host "Blob Not Found"
Write-Host "Start Upload Blob"
Set-AzureStorageBlobContent -File $filename -Container $containerName -Blob $blobName -Context $blobContext
}
我想将新的 blob 上传到 Azure 存储帐户,但只有其中一个不存在。我想知道是否有任何方法可以设置 Set-AzureStorageBlobContent 命令以忽略已经存在的项目,这样我就可以在一次操作中实现它?
I want to upload new blob to Azure storage account, but only of it doesn't already exist.
从 Set-AzureStorageBlobContent, it seems that there are not existing parameters that could be used to skip/ignore the existing blob while uploading the file to blob. As a workaround, we could use Get-AzureStorageBlob cmdlet 的文档中检查 blob 在 运行 Set-AzureStorageBlobContent cmdlet 将文件上传到 blob 之前是否已经存在。
try
{
$blob = Get-AzureStorageBlob -Blob $blobName -Container $containerName -Context $blobContext -ErrorAction Stop
Write-Host "Blob is existing"
}
catch
{
# The blob doesn't exist
# Add logic here
Write-Host "Blob Not Found"
Write-Host "Start Upload Blob"
Set-AzureStorageBlobContent -File $filename -Container $containerName -Blob $blobName -Context $blobContext
}