Windows Vista “Could not open input file”

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I get this strange error when I try to run a php file from anywhere in the system:

Could not open input file: drush.php.

…except for its own directory. When I try invoke it from its own directory, no problem — everything works!

Is there a permissions issue here? I looked under security tab in the properties for the file, but every user was given all the permissions that are available.

So I don't get why Windows is not able to open this file from any other directory except where it is located.

Best Answer

It could be a "working directory" issue. Have you tried making a shortcut to it on the desktop, then right-click, properties, and change "start in" to the folder where the php file is?

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Windows – How to find what process is holding a file open in Windows

I've had success with Sysinternals Process Explorer. With this, you can search to find what process(es) have a file open, and you can use it to close the handle(s) if you want. Of course, it is safer to close the whole process. Exercise caution and judgement.

To find a specific file, use the menu option Find->Find Handle or DLL... Type in part of the path to the file. The list of processes will appear below.

If you prefer command line, Sysinternals suite includes command line tool Handle, that lists open handles.

Examples

c:\Program Files\SysinternalsSuite>handle.exe |findstr /i "e:\" (finds all files opened from drive e:\"
c:\Program Files\SysinternalsSuite>handle.exe |findstr /i "file-or-path-in-question"

Windows – Useful Command-line Commands on Windows

A little known one is

getmac

It shows the MAC address(es) of your network adapter(s).

Screenshot of running getmac from a Windows commandline window.

Best Answer

Related Solutions

Windows – How to find what process is holding a file open in Windows

Windows – Useful Command-line Commands on Windows

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