如何使用 PowerShell Invoke-RestMethod 发送多部分/表单数据

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:\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() }
}

Answer 2

The problem here was what the API required some additional parameters. Initial request required some parameters to accept raw content and specify filename/size. After setting that and getting back proper link to submit, I was able to use:

Invoke-RestMethod -Uri $uri -Method Post -InFile $filePath -ContentType "multipart/form-data"

Answer 3

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

Answer 4

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)
    }
}