Question

我正在尝试在Powershell中创建等效于

的命令

curl -u username:abcd -i -F name=files -F filedata=@employees.csv https://myservice.com/v1/employees/csv

我需要在请求中输入文件名。因此在Powershell中

$FilePath = 'employees.csv'
$FieldName = 'employees.csv'
$ContentType = 'text/csv'
$username = "user"
$password = "..."

$FileStream = [System.IO.FileStream]::new($filePath, [System.IO.FileMode]::Open)
$FileHeader = [System.Net.Http.Headers.ContentDispositionHeaderValue]::new('form-data')
$FileHeader.Name = $FieldName
$FileHeader.FileName = Split-Path -leaf $FilePath
$FileContent = [System.Net.Http.StreamContent]::new($FileStream)
$FileContent.Headers.ContentDisposition = $FileHeader
$FileContent.Headers.ContentType = [System.Net.Http.Headers.MediaTypeHeaderValue]::Parse($ContentType)

$MultipartContent = [System.Net.Http.MultipartFormDataContent]::new()
$MultipartContent.Add($FileContent)

$base64AuthInfo = [Convert]::ToBase64String([Text.Encoding]::ASCII.GetBytes("$($username):$($password)" ))
$Response = Invoke-WebRequest -Headers @{Authorization = "Basic $base64AuthInfo" } -Body $MultipartContent -Method 'POST' -Uri 'https://myservice.com/v1/employees/csv'

是否有更好（更短）的方法，所以我在Content Disposition中有一个文件名？

$body = get-content employees.csv -raw
$base64AuthInfo = [Convert]::ToBase64String([Text.Encoding]::ASCII.GetBytes("user:pass" ))
Invoke-RestMethod -Headers @{Authorization = "Basic $base64AuthInfo" } -uri url -Method Post -body $body -ContentType 'text/csv
# a flag -ContentDispositionFileName would be great

Answer 1

猜测端点将接受什么，但这是您在curl中的powershell请求的示例：

$u, $p = 'username', 'password'
$b64 = [Convert]::ToBase64String([Text.Encoding]::ASCII.GetBytes("${u}:$p"))
$invokerestmethodParams = @{
    'Uri'             = 'https://myservice.com/v1/employees/csv'
    'Method'          = 'POST'
    'Headers'         = @{ Authorization = "Basic $b64" }
    'InFile'          = 'C:\path\to\employees.csv'
    'SessionVariable' = 's' # use $s to view content headers, etc.
}
$output = Invoke-RestMethod @invokerestmethodParams

使用Powershell上传文件

1 个答案: