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I am trying to call the PHP file by, passing the request_number in the localhost URL using Powershell

In HTML:

<a href='workflow_execution_progress.php?view_id=".$row['request_number']."' title='Click to see the progress of workflow'>

I referred to this link but not sure to modify it with parameters. Executing php script on powershell

Update: My PowerShell

$PhpExe  = "C:\path\to\php\install\dir\php.exe"
$PhpFile = "C:\path\to\script.php"
$PhpArgs = '-f "{0}"' -f $PhpFile  //args like view_id = 1 / 2 /3 (dynamically)
$PhpOutput = & $PhpExe $PhpArgs
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  • If you call a local PHP file through powershell, then there's not "url" involve (since you're not making a network request). Commented Oct 14, 2020 at 11:59
  • yup, it's a local file now. will deploy after the completion of the project. My question is how to pass dynamic id to the PHP file using PowerShell script @MagnusEriksson Commented Oct 14, 2020 at 12:03
  • Does this answer your question? Pass variable to php script running from command line Commented Oct 14, 2020 at 12:34
  • @MagnusEriksson Sorry i am not getting how i can add it in my powershellcode $PhpExe = "C:\path\to\php\install\dir\php.exe" $PhpFile = "C:\path\to\script.php" $PhpArgs = '-f "{0}"' -f $PhpFile Commented Oct 14, 2020 at 12:53

1 Answer 1

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You can use $argv to get an array of arguments passed to the script.

https://www.php.net/manual/en/reserved.variables.argv.php

php script.php arg1 arg2 arg3

<?php
var_dump($argv);
?>

array(4) {
  [0]=>
  string(10) "script.php"
  [1]=>
  string(4) "arg1"
  [2]=>
  string(4) "arg2"
  [3]=>
  string(4) "arg3"
}

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4 Comments

Sorry i am not getting how i can add it in my powershellcode $PhpExe = "C:\path\to\php\install\dir\php.exe" $PhpFile = "C:\path\to\script.php" $PhpArgs = '-f "{0}"' -f $PhpFile
Maybe I did not understand your problem. I am not sure if it is about how to get the argument passed to the script, or how to pass arguments ?
Need to pass arguments to the PHP file using powershell script. - how to pass arguments
I don't get it. I read your update but why not just place arguments after $Phpfile ? $PhpArgs = '-f "{0}"' -f $PhpFile 'view_id=1'

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