Script to return Code 200

PowerShell Help
1

Hi everyone,

i need a script that return code 200 from site, that i must login with username and password. I write a script the accept credential, but script show me all content of site. For example:

$uri = “http://example.com”
$user = ‘user’
$pass = ‘password’ | ConvertTo-SecureString -AsPlainText -Force

$cred = New-Object Management.Automation.PSCredential ($user, $pass)

Invoke-WebRequest -Uri $uri -Credential $cred

$test = Invoke-WebRequest -Uri $uri -Credential $cred

If ($test.StatusCode-eq 200) {
Write-Host “site is Running”
}

When i started, the script show me following:

"StatusCode : 200
StatusDescription :
Content : <!DOCTYPE HTML PUBLIC “-//W3C//DTD HTML 4.01//EN”
“http://www.w3.org/TR/html4/strict.dtd”>
<html>
<head>
<title>test</title>
<style type=“text/css”>

  • {
    p…
    RawContent : HTTP/1.1 200
    Transfer-Encoding: chunked
    Content-Type: text/html
    Date: Tue, 23 Feb 2021 16:04:03 GMT

<!DOCTYPE HTML PUBLIC “-//W3C//DTD HTML 4.01//EN”
“http://www.w3.org/TR/html4/strict.dtd”…
Forms : {}
Headers : {[Transfer-Encoding, chunked], [Content-Type, text/html], [Date, Tue, 23 Feb 2021 16:04:03 GMT]}
Images : {}
InputFields : {}
Links : {}
ParsedHtml : mshtml.HTMLDocumentClass
RawContentLength : 9120

site is Running"

The result must be only “site is Running” without any other information.

Thank you.

 

2
Ivcho,
... when you post code, error messages, sample data or console output format it as code, please.
Here you can read how that works: Guide to Posting Code.
You can go back and edit your existing post. You don't have to create a new one. ; - )
Thanks in advance.
You run the command Invoke-WebRequest twice. Once just as it is and once with a variable assignment. I think the second time should be enough:
$uri = 'http://example.com&#8221'
$user = 'user'
$pass = 'password' | ConvertTo-SecureString -AsPlainText -Force
$cred = New-Object Management.Automation.PSCredential ($user, $pass)

$test = Invoke-WebRequest -Uri $uri -Credential $cred
If ($test.StatusCode -eq 200) {
    'site is Running'
}