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I am trying to process a multipart GET call in powershell and then save the zipfile it contains to disk. I can execute this:

$response = Invoke-RestMethod -Uri $reqUrl -Method Get -Headers $headers

and then echo out the filename and contents. In order to save the file, I tried this:

$response = Invoke-RestMethod -Uri $reqUrl -Method Get -Headers $headers -ContentType "multipart/form-data" -Outfile result.zip

This raises an error (Invalid Operation). So I tried this:

$response = Invoke-RestMethod -Uri $reqUrl -Method Get -Headers $headers -ContentType "application/zip" -Outfile result.zip

This creates a file called result.zip which isn't valid. I know that the response is multipart, so while this doesn't raise an error, I am not surprised that the file is invalid because it must contain all of the parts.

I have looked around, but all I find are ways of using Invoke-RestMethod to POST mulitpart content.

This is the error when I try to open the resulting zip file:

enter image description here

I have also tried to decode the result as below, but with the same results.

$B64String = $response.resultObject 
Write-Host "resultObject size: $([Math]::Round($B64String.Length / 1Mb,2)) MiB"
$swDecode = [System.Diagnostics.Stopwatch]::StartNew()
$bytes = [System.Convert]::FromBase64String($B64String)
$swDecode.Stop()
Write-Host "Decoded size: $([Math]::Round($bytes.Count / 1Mb,2)) MiB"
Write-Output $bytes > $($response.fileName)
Mark Chassy
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  • "This creates a file called `result.zip` which isn't valid" - can you expand on this? `result.zip` is a perfectly valid file name in any filesystem I can think of... – Mathias R. Jessen Jan 20 '22 at 17:41
  • I will add more details. – Mark Chassy Jan 21 '22 at 15:21
  • Ahh, you mean the resulting file contents are invalid/corrupt, that makes more sense :) Try using `Invoke-WebRequest -OutFile ...` instead of `Invoke-RestMethod -OutFile` – Mathias R. Jessen Jan 21 '22 at 15:24
  • I don't think the contents are corrupt, I just think that I have not processed them correctly. If I convert the response into JSON, there are three fields: id, fileName, and resultObjet. Using --OutFile does not work because the file will include `id` and `fileName`. But also just using `resultObject` doesn't work either. My question is how should I be encoding/decoding `resultObject` – Mark Chassy Jan 21 '22 at 16:10

1 Answers1

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I found the answer

$response = Invoke-RestMethod -Uri $reqUrl -Method Get -Headers $headers  
$response | ConvertTo-Json
[IO.File]::WriteAllBytes($response.fileName, [System.Convert]::FromBase64String($response.resultObject))
Mark Chassy
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