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:\Temp\test.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() }
}

I found this post and changed it a bit

$fileName = "..."
$uri = "..."

$currentPath = Convert-Path .
$filePath="$currentPath\$fileName"

$fileBin = [System.IO.File]::ReadAlltext($filePath)
$boundary = [System.Guid]::NewGuid().ToString()
$LF = "`r`n"
$bodyLines = (
    "--$boundary",
    "Content-Disposition: form-data; name=`"file`"; filename=`"$fileName`"",
    "Content-Type: application/octet-stream$LF",
    $fileBin,
    "--$boundary--$LF"
) -join $LF

Invoke-RestMethod -Uri $uri -Method Post -ContentType "multipart/form-data; boundary=`"$boundary`"" -Body $bodyLines

Update Powershell 7+

Invoke-RestMethod and Invoke-WebRequest now both accept a -Form parameter that takes a hashtable like this:

$Uri = 'https://api.contoso.com/v2/profile'
$Form = @{
    firstName  = 'John'
    lastName   = 'Doe'
    email      = '[email protected]'
    avatar     = Get-Item -Path 'c:\Pictures\jdoe.png'
    birthday   = '1980-10-15'
    hobbies    = 'Hiking','Fishing','Jogging'
}
$Result = Invoke-WebRequest -Uri $Uri -Method Post -Form $Form

You can check your current version with $PSVersionTable

See Also

• powershell invoke-restmethod multipart/form-data
• multipart/form-data file upload with PowerShell

For anyone wondering (like Jelphy) whether David's answer can be used with cookies/credentials, the answer is yes.

First set the session with Invoke-WebRequest:

Invoke-WebRequest -Uri "$LoginUri" -Method Get -SessionVariable 'Session'

Then POST to the Login URL, which stores the authentication cookie in $Session:

$Response = Invoke-WebRequest -Uri "$Uri" -Method Post -Body $Body -WebSession $Session

The steps above are the standard way to deal with session in Powershell. But here is the important part. Before creating the HttpClient, create an HttpClientHandler and set it's CookieContainer property with the cookies from the session:

$ClientMessageHandler = New-Object System.Net.Http.HttpClientHandler
$ClientMessageHandler.CookieContainer = $Session.Cookies

Then pass this object to the HttpClient constructor

$Client = [System.Net.Http.HttpClient]::new($ClientMessageHandler)

Voila, you now have an HttpClient with session cookies set automatically via Invoke-WebRequest. The rest of David's example should work (copied here for completeness):

$MultipartFormData = 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)
$MultipartFormData.Add($FileContent, $FieldName, $FileName)

$Result = $Client.PostAsync($url, $content).Result
$Result.EnsureSuccessStatusCode()

I had many files to upload with each request, so I factored out this last bit into a lambda function:

function Add-FormFile {
    param ([string]$Path, [string]$Name)

    if ($Path -ne "")
    {    
        $FileStream = [System.IO.File]::OpenRead($Path)
        $FileName = [System.IO.Path]::GetFileName($Path)
        $FileContent = [System.Net.Http.StreamContent]::new($FileStream)
        $MultipartFormData.Add($FileContent, $Name, $FileName)
    }
}