curl - How to send multipart/form-data with PowerShell Invoke-RestMethod

Question

I'm trying to send a file via Invoke-RestMethod in a similar context as curl with the -F switch.

Curl Example

curl -F FileName=@"/path-to-file.name" "https://uri-to-post"

In powershell, I've tried something like this:

$uri = "https://uri-to-post"
$contentType = "multipart/form-data"
$body = @{
    "FileName" = Get-Content($filePath) -Raw
}

Invoke-WebRequest -Uri $uri -Method Post -ContentType $contentType -Body $body
}

If I check fiddler I see that the body contains the raw binary data, but I get a 200 response back showing no payload has been sent.

I've also tried to use the -InFile parameter with no luck.

I've seen a number of examples using a .net class, but was trying to keep this simple with the newer Powershell 3 commands.

Does anyone have any guidance or experience making this work?

Answer 1

The accepted answer won't do a multipart/form-data request, but rather a application/x-www-form-urlencoded request forcing the Content-Type header to a value that the body does not contain.

One way to send a multipart/form-data formatted request with PowerShell is:

$ErrorActionPreference = 'Stop'

$fieldName = 'file'
$filePath = 'C:Tempest.pdf'
$url = 'http://posttestserver.com/post.php'

Try {
    Add-Type -AssemblyName 'System.Net.Http'

    $client = New-Object System.Net.Http.HttpClient
    $content = New-Object System.Net.Http.MultipartFormDataContent
    $fileStream = [System.IO.File]::OpenRead($filePath)
    $fileName = [System.IO.Path]::GetFileName($filePath)
    $fileContent = New-Object System.Net.Http.StreamContent($fileStream)
    $content.Add($fileContent, $fieldName, $fileName)

    $result = $client.PostAsync($url, $content).Result
    $result.EnsureSuccessStatusCode()
}
Catch {
    Write-Error $_
    exit 1
}
Finally {
    if ($client -ne $null) { $client.Dispose() }
    if ($content -ne $null) { $content.Dispose() }
    if ($fileStream -ne $null) { $fileStream.Dispose() }
    if ($fileContent -ne $null) { $fileContent.Dispose() }
}