如果其他基于 PS 中 invoke-webrequest 的输出 [英] if else based on output of invoke-webrequest in PS

问题描述

有人可以帮我只捕获下面 invoke-webrequest 的 StatusCode，以便我可以确定站点是启动 (200) 还是关闭(任何其他代码).我认为本质上需要一个 if else 语句.

Can someone help me capture only the StatusCode of invoke-webrequest below so that I can determine if a site is up (200) or down (any other code). I think essentially an if else statement will be needed.

注意:我不需要 invoke-webrequest 的输出.这是我现在运行的代码:.\websiteCheck.ps1 https://google.com

NOTE: I don't need the output of invoke-webrequest. This is the code I'm running now with: .\websiteCheck.ps1 https://google.com

$url = $args[0]

[Net.ServicePointManager]::SecurityProtocol = "Tls11"

invoke-webrequest -uri $url -DisableKeepAlive -UseBasicParsing -Method head

返回的信息比我需要的要多得多:

which returns a lot more info to work with than I need:

StatusCode        : 200
StatusDescription : OK
Content           :
RawContent        : HTTP/1.1 200 OK
                X-XSS-Protection: 1; mode=block
                X-Frame-Options: SAMEORIGIN
                Cache-Control: private, max-age=0
                Content-Type: text/html; charset=UTF-8
                Date: Wed, 20 Feb 2019 05:55:14 GMT
                Expires: ...
Forms             :
Headers           : {[X-XSS-Protection, 1; mode=block], [X-Frame-Options, SAMEORIGIN], [Cache-Control, private,
                max-age=0], [Content-Type, text/html; charset=UTF-8]...}
Images            : {}
InputFields       : {}
Links             : {}
ParsedHtml        :
RawContentLength  : 0

我只想做这样的事情:

如果 StatusCode = 200 写主机站点已启动"否则写主机网站已关闭"

if StatusCode = 200 write-host "The site is up" else write-host "The site is down"

这必须基于 invoke-webrequest 而不是类似 [Net.HttpWebRequest] $req = [Net.WebRequest]::Create($url) 的东西，否则它不能与执行多次重定向的 url 一起正常工作.

This must be based on invoke-webrequest and not something like [Net.HttpWebRequest] $req = [Net.WebRequest]::Create($url) because otherwise it doesn't work properly with url's that do multiple redirects.

推荐答案

有更多的状态码可以返回.请参阅HTTP 状态代码.

There are more status codes that can be returned. See HTTP Status Codes.

那么也许更细粒度的脚本会是更好的选择?

So maybe a bit more fine-grained script would be a better choice?

$uri = $args[0]
[Net.ServicePointManager]::SecurityProtocol = [Net.SecurityProtocolType]::Tls12

try{
    $status = [int](Invoke-WebRequest $uri -UseBasicParsing -DisableKeepAlive -Method Head).StatusCode
    switch ($status) {
        {$_ -ge 100 -and $_ -lt 200}  { Write-Output "The site is up. Statuscode: $status"; break }
        {$_ -ge 200 -and $_ -lt 300}  { Write-Output "The site is up. Statuscode: $status"; break }
        {$_ -ge 300 -and $_ -lt 400}  { Write-Output "The site is redirected. Statuscode: $status"; break }
        {$_ -ge 400 -and $_ -lt 500}  { Write-Output "Client error. Statuscode: $status"; break }
        {$_ -ge 500 -and $_ -lt 600}  { Write-Output "Server error. Statuscode: $status"; break }
        default { Write-Output "The site returned an unhandled status code. Statuscode: $status"}
    }
}
catch {
    Write-Error "An error occurred on Invoke-WebRequest.`r`n$($_.Exception.Message)"
}

您还可以使用 [System.Net.WebRequest] 代替 Invoke-WebRequest.类似的东西:

Instead of the Invoke-WebRequest you could also use [System.Net.WebRequest]. Something like:

$uri = $args[0]
[Net.ServicePointManager]::SecurityProtocol = [Net.SecurityProtocolType]::Tls12

try{
    # Web request
    $res = ([System.Net.WebRequest]::Create($uri)).GetResponse()
}
catch {
    $res = $_.Exception.Response
}
$status = [int]$res.StatusCode
switch ($status) {
    {$_ -ge 100 -and $_ -lt 200}  { Write-Output "The site is up. Statuscode: $status"; break }
    {$_ -ge 200 -and $_ -lt 300}  { Write-Output "The site is up. Statuscode: $status"; break }
    {$_ -ge 300 -and $_ -lt 400}  { Write-Output "The site is redirected. Statuscode: $status"; break }
    {$_ -ge 400 -and $_ -lt 500}  { Write-Output "Client error. Statuscode: $status"; break }
    {$_ -ge 500 -and $_ -lt 600}  { Write-Output "Server error. Statuscode: $status"; break }
    default { Write-Output "An unhandled error occurred. Statuscode: $status"}
}
# Dispose response if available
if($res){ $res.Dispose() }

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