从 Powershell 上的 HTTP 请求获取响应链接答案

【问题标题】：Get response link from HTTP request on Powershell从 Powershell 上的 HTTP 请求获取响应链接
【发布时间】：2019-12-14 09:35:55
【问题描述】：

以下 Powershell 脚本运行 Google 搜索存储在我的硬盘中的图像。

我怎样才能获得进入结果页面的链接？是否可以导航到其上显示的不同网页？

我尝试$request.Links | Select href 尝试获取链接列表，但没有成功。我也尝试将Write-Output $respStream 添加到代码中，但它没有运行。

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

【问题讨论】：

您是否尝试获取302 重定向到的URL？如果是这样，请改为阅读响应的 Location 标头。
你为什么使用[Net.HttpWebRequest]而不是Invoke-WebRequest来发出网络请求？这不一定能解决您的问题，但可以引导您朝着正确的方向前进

标签： powershell httprequest

【解决方案1】：

正如@soc 所提到的，您不需要流来拉取 url，移动到的位置在响应标头中：

    $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
    }

【讨论】：