If you try to launch GUI apps from a service in Vista, you'll have lots of trouble. As a security feature, Vista mediates the interaction of services with the desktop using 'Interactive Services Detection'.
That means, if you are running PHP as a module of an Apache service, you won't be able to launch GUI apps using any method. This kind of thing just won't work:
$WshShell = new COM("WScript.Shell");
$oExec = $WshShell->Run("notepad.exe", 7, false);
So, if you want to use Apache/PHP as a proxy for launching GUI apps, you'll need to run Apache as a console application.
First, if Apache is already installed as a service, you'll need to set it's startup type to "manual" using the services snap-in. (%SystemRoot%\system32\services.msc) Search for Services in the start menu search box.
Then add a shortcut to C:\apache\bin\httpd.exe (or wherever Apache is installed) to your Startup folder, and set that shortcut to start minimized. You can use an app like TrayIt! to force Apache down into the system tray.
Then use any of the methods outlined on the PHP website and you will be able to open a Windows application from PHP and see it's GUI.
System program execution
- Introduction
- Installing/Configuring
- Predefined Constants
- Program execution Functions
- escapeshellarg — Escape a string to be used as a shell argument
- escapeshellcmd — Escape shell metacharacters
- exec — Execute an external program
- passthru — Execute an external program and display raw output
- proc_close — Close a process opened by proc_open and return the exit code of that process
- proc_get_status — Get information about a process opened by proc_open
- proc_nice — Change the priority of the current process
- proc_open — Execute a command and open file pointers for input/output
- proc_terminate — Kills a process opened by proc_open
- shell_exec — Execute command via shell and return the complete output as a string
- system — Execute an external program and display the output
Joe Engel ¶
4 years ago
