Get response link from HTTP request on Powershell - powershell

Question

The following Powershell script runs a Google search of an image stored within my hard drive.
How can I get the link which is followed to get to the results page? Is it possible to navigate to the different webpages displayed on it?
I've tried $request.Links | Select href to try and get a list of the links, but it didn't work. I've also tried to add Write-Output $respStream to the code, but then it doesn't run.
Set-ExecutionPolicy Bypass -scope Process -Force
function Get-GoogleImageSearchUrl
{
param(
[Parameter(Mandatory = $true)]
[ValidateScript({ Test-Path $_ })]
[string] $ImagePath
)
# extract the image file name, without path
$fileName = Split-Path $imagePath -Leaf
# the request body has some boilerplate before the raw image bytes (part1) and some after (part2)
# note that $filename is included in part1
$part1 = #"
-----------------------------7dd2db3297c2202
Content-Disposition: form-data; name="encoded_image"; filename="$fileName"
Content-Type: image/jpeg
"#
$part2 = #"
-----------------------------7dd2db3297c2202
Content-Disposition: form-data; name="image_content"
-----------------------------7dd2db3297c2202--
"#
# grab the raw bytes composing the image file
$imageBytes = [Io.File]::ReadAllBytes($imagePath)
# the request body should sandwich the image bytes between the 2 boilerplate blocks
$encoding = New-Object Text.ASCIIEncoding
$data = $encoding.GetBytes($part1) + $imageBytes + $encoding.GetBytes($part2)
# create the HTTP request, populate headers
$request = [Net.HttpWebRequest] ([Net.HttpWebRequest]::Create('http://images.google.com/searchbyimage/upload'))
$request.Method = "POST"
$request.ContentType = 'multipart/form-data; boundary=---------------------------7dd2db3297c2202' # must match the delimiter in the body, above
$request.ContentLength = $data.Length
# don't automatically redirect to the results page, just take the response which points to it
$request.AllowAutoredirect = $false
# populate the request body
$stream = $request.GetRequestStream()
$stream.Write($data, 0, $data.Length)
$stream.Close()
# get response stream, which should contain a 302 redirect to the results page
$respStream = $request.GetResponse().GetResponseStream()
# pluck out the results page link that you would otherwise be redirected to
(New-Object Io.StreamReader $respStream).ReadToEnd() -match 'HREF\="([^"]+)"' | Out-Null
$matches[1]
}
$url = Get-GoogleImageSearchUrl "C:\Users\Path\filename.jpeg"
Start-Process $url

Answer 1

As mentioned by #soc, you shouldn't need the stream to pull the url, the moved to location is in the response header:
$request = [Net.HttpWebRequest] ([Net.HttpWebRequest]::Create('http://images.google.com/searchbyimage/upload'))
$request.AllowAutoredirect = $false
...
$response = $request.GetResponse()
if ($response.StatusCode -eq 302) {
$redirect_url = $response.Headers["Location"]
write-host $redirect_url
}