Rapidshare API with PowerShell

Question

$FreeUploadServer = Invoke-RestMethod -Uri "http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=nextuploadserver"
Invoke-RestMethod -Uri "http://rs$FreeUploadServer.rapidshare.com/cgi-bin/rsapi.cgi?sub=upload&login=43533592&password=password&filecontent=C:\libs\test.txt" 

I have no clue why that isnt working, its something to do with the filecontent parameter but i have read the entire documentation about the API and i cant figure it out.

API Documentation

Answer 1

Make sure you use the -Method Post parameter/arg combo on the second Invoke-RestMethod call. Also, take a look at this SO response to a similar question about using CURL. After looking at the question and answer, it appears that RapidShare requires filling in POST form fields. In this case you need to specify the -Body parameter. I think you'll use a PowerShell hashtable where each entry corresponds to a form field name/value pair. Looks like the one field needed is the filecontent field. Presumably the value is the file's contents.

Also, when you use POST, you'll have to convert the GET query parameters to POST form fields e.g.:

$url = "http://rs$FreeUploadServer.rapidshare.com/cgi-bin/rsapi.cgi"
$fields = @{sub='upload';login='43533592';password='password';filename='test.txt';
            filecontent=([IO.File]::ReadAllText('c:\libs\test.txt'))}

Invoke-RestMethod -Uri $url -Body $fields -Method Post

If [IO.File]::ReadAlltext() doesn't cut it perhaps try [IO.File]::ReadAllBytes() instead.

Answer 2

Turns out there is a tiny section in the API Documentation that says if you use the POST method all parameters need to be passed in the body. I wrote a function that uploads to RapidShare in PowerShell, i hope this helps people in the future cause this was VERY annoying.

function Upload-RapidShare([string]$File,[string]$Name)
{
     $FreeUploadServer = Invoke-RestMethod -Uri "http://api.rapidshare.com/cgi-bin/rsapi.cgi?sub=nextuploadserver"
     Invoke-RestMethod "http://rs$FreeUploadServer.rapidshare.com/cgi-bin/rsapi.cgi" -Body  "sub=upload&login=USERNAME&password=PASSWORD&filename=$Name&filecontent=$File" -Method Post
}