Get response link from HTTP request on 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

Are you trying to get the URL that the 302 redirects to? If so, read the response's Location header instead. — Soc
– Soc, Commented Dec 13, 2019 at 19:28
Why are you using [Net.HttpWebRequest] to make the web request instead of using Invoke-WebRequest? This does not necessarily solve your problem but can guide you in the right direction — Erick Guillen
– Erick Guillen, Commented Dec 13, 2019 at 19:44

vhoang · Accepted Answer · 2019-12-14 01:35:55Z

3

+150

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
    }

answered Dec 14, 2019 at 1:35

vhoang

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