Using powershell Invoke-RestMethod to GET multipart content

Question

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:

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)

"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
– Mathias R. Jessen, Commented Jan 20, 2022 at 17:41
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
– Mathias R. Jessen, Commented Jan 21, 2022 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
– Mark Chassy, Commented Jan 21, 2022 at 16:10

Mark Chassy · Accepted Answer · 2022-01-27 10:20:16Z

1

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))

answered Jan 27, 2022 at 10:20

Mark Chassy

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