@WorkingDir Not Working (for me!)

[MichaelPracht]

I've created and compiled a script that lives in Windows\System32. When I run it, I want to create a log file in the folder where the script is executed (so, I run it from c:\ but it lives in c:\Windows\System32 and I want it to create c:\script.log). The problem is that @WorkingDir and @ScriptDir returns "C:\Windows\System32" instead of the directory where the script was executed. Anybody know how I can set a variable to the folder path where the script is actually called from? If I just execute:

FileWriteLine("test.log", "Output goes here.")

...then it creates the log file in the folder where the script lives (C:\Windows\System32), instead of where I am when I execute the script.

[Varian]

Just use

FileChangeDir("C:\")

before you start the output to file

[MichaelPracht]

Just use

FileChangeDir("C:\")

before you start the output to file

But I won't know where the script is being run from. So, for example the script may be run from G:\folder\mystuff and I would want to create the log file G:\folder\mystuff\script.log

[willichan]

How are you executing the script?

[MichaelPracht]

From a command window. The script is in C:\Windows\System32, which is in my PATH, but I may execute it from C:\thisfolder\thatfolder\righthere

[trancexx]

Then pass command line parameter with desired location and add parsing in your script.

Regardless of the solution you seem to be confused with the terms.

[MichaelPracht]

Confused with what terms?

If you mean: runme -path "c:\thispath"

well, I'd rather not force the user to reenter the path they're already in. I assumed the script could know where it is being run from.

[Varian]

Then run it this way:

RunWait("C:\Windows\System32\YourProgram.exe", @ScriptDir)

You'll have to provide the full executable path since your working folder is different

[trancexx]

Confused with what terms?

With this...

I assumed the script could know where it is being run from.

It knows, but you don't.

@WorkingDir and @ScriptDir are just fine. What you want is something else and that's not related to your script but to the process that starts them.

Edited November 3, 2010 by trancexx

[MichaelPracht]

Ok, let me try to explain this again. I have a compiled script, let’s say it’s called “files-list.exe”, and its purpose is to write a list of the files in the folder where it’s executed from into a text file (called “files-list.txt”) that the user can then edit or print.

The compiled script has been placed in C:\WINDOWS\System32 (which is in the user’s PATH) so that it can be called from any folder on the user’s system.

The user starts a command prompt. They navigate to “C:\Documents and Settings\All Users\Desktop” and then they type “files-list”. The script executes. If the script tries to create files-list.txt in @WorkingDir, the script is created at C:\WINDOWS\System32\files-list.txt. If it tries to create the text file in @ScriptDir, it creates C:\WINDOWS\System32\files-list.txt. I’ve not found a way to tell the script to create the file in <currentdir>\files-list.txt. The answer is probably very simple, but so far it eludes me.

[Richard Robertson]

Is your script receiving elevated permissions? If so, it will run with a changed working directory.

[JoHanatCent]

Try:

$CurPath = EnvGet("cd")

[MichaelPracht]

It looks like @WorkingDir will not return the actual current/working directory unless your script is run from an elevated command prompt. Adding #requireadmin to the beginning of the script is not sufficient enough. This behavior isn’t specified on the website (http://www.autoitscript.com/autoit3/docs/macros.htm).

So, my problem now has become: how do I detect that the script was not run from an elevated command prompt, so I can instruct the user? If they run the script from a non-elevated command prompt and it tries to create “files-list.txt” in @WorkingDir, it will attempt to create “C:\WINDOWS\System32\files-list.txt”. Whether the script succeeds or fails to write there, the user will be looking in their current folder, thinking that the script didn’t work. I don’t want to tell the user to look in “C:\WINDOWS\System32” because (aside from being a hassle) I want them to be able to place the compiled script in ANY folder in their PATH.

Ideas?

BTW, $CurPath = EnvGet("cd") returns an empty string regardless of whether I run from an elevated or non-elevated command prompt (I'm running on 32-bit Windows 7 Enterprise.)

[Richard Robertson]

You should use the Au3Wrapper command of #AutoIt3Wrapper_Res_requestedExecutionLevel=requireAdministrator

That might solve it.

[Juvigy]

C:\Windows\System32, which is in my PATH, but I may execute it from C:\thisfolder\thatfolder\righthere

And how would you do that ? No mater from where you execute it , the exe is still in C:\Windows\System32 right ? @WorkingDir will change only if you make a shortcut with another workingdir path. I think it would be best to do:

1. FileWriteLine(@MyDocumentsDir & "test.log", "Output goes here.")

or

2. FileWriteLine(@UserProfileDir & "test.log", "Output goes here.")

or something like that

[MichaelPracht]

The beginning of my script now looks like this:

#RequireAdmin

#AutoIt3Wrapper_Res_requestedExecutionLevel=requireAdministrator

However, @WorkingDir still returns "C:\WINDOWS\System32", if not run from an elevated command prompt.

[MichaelPracht]

And how would you do that ? No mater from where you execute it , the exe is still in C:\Windows\System32 right ? @WorkingDir will change only if you make a shortcut with another workingdir path. I think it would be best to do:

1. FileWriteLine(@MyDocumentsDir & "test.log", "Output goes here.")

or

2. FileWriteLine(@UserProfileDir & "test.log", "Output goes here.")

or something like that

I just don't want the user to have to navigate somewhere else for the output. I want to create it in the same folder they're already in.

[BrewManNH]

If you change your program to accept command line arguments, you can run it from a batch file placed anywhere in your system's path, such as the Windows\System32. In your batch file that you use to run your program, add the command line argument %CD% and that will be translated to the current folder that you are running the batch file from.

Example batch file:

dir "%cd%"

This will give you a directory listing of all files in the current directory.

[JoHanatCent]

Try:

RunWait(@ComSpec & " /c %SystemDrive%&&cd %temp%&&echo %cd%>temp.tmp", "", @SW_hide); create temp file to save %cd%
$file = FileOpen(@TempDir & "\temp.tmp", 0)
; Check if file opened for reading OK
If $file = -1 Then
    MsgBox(0, "Error", "Unable to open file." & $file)
    Exit
EndIf
; Read in lines of text until the EOF is reached
While 1
    $line = FileReadLine($file)
    If @error = -1 Then ExitLoop
    MsgBox(0, "Current DIR:", $line)
WEnd
FileClose($file)
FileDelete(@TempDir & "\temp.tmp")

[MichaelPracht]

Try:

RunWait(@ComSpec & " /c %SystemDrive%&&cd %temp%&&echo %cd%>temp.tmp", "", @SW_hide); create temp file to save %cd%
$file = FileOpen(@TempDir & "\temp.tmp", 0)
; Check if file opened for reading OK
If $file = -1 Then
    MsgBox(0, "Error", "Unable to open file." & $file)
    Exit
EndIf
; Read in lines of text until the EOF is reached
While 1
    $line = FileReadLine($file)
    If @error = -1 Then ExitLoop
    MsgBox(0, "Current DIR:", $line)
WEnd
FileClose($file)
FileDelete(@TempDir & "\temp.tmp")

This returns "C:\WINDOWS\System32", so no advantage here that I can see (but thanks for the suggestion).