Running compiled exe on command line with log path as parameter

[TheSaint]

$logFile = _FileCreate($CmdLine[1] & $CmdLine[2])

Are the two CmdLine elements parts of one path or two completely separate paths?

Either way, you are missing a separator or joiner between ... either a space or backwards slash.

So perhaps either -->

$logFile = _FileCreate($CmdLine[1] & " " & $CmdLine[2])

or

$logFile = _FileCreate($CmdLine[1] & "" & $CmdLine[2])

And what is $logFile supposed to be to you?

In reality it is a success or failure return value (1 or 0) from the _FileCreate function, not a file.

Edited August 27, 2013 by TheSaint

[FredMellink]

Hello John,

This is the command line:

"C:Program Files (x86)AutoIt3AutoIt3.exe" C:SnapsAutoITfrednotepad_exercise20130826.au3 C:FredTestNotepadLog.txt"

[FredMellink]

I like to run "mytestscript.exe C:myLogPathmytestlog.csv" and the script should create the path and file and append logging from the test steps. I like to know the easiest way to establish this. Thanks you all wizkids to help me

[hannes08]

Hello Fred,

if you use FileOpen("<yourfile>", 9) it automatically creates your directories and the file if it doesn't exist.

This information can be found in the helpfile under FileOpen() as well. Take your time and look at the description of the parameters in the helpfile it is really usefull.

[JohnOne]

FileWrite($CmdLine[1], '')

Will create path and file.

[hannes08]

Well, I never really understood why you'll need a function to create a new empty file. :-D

[FredMellink]

The following little test I run:

#include <WindowsConstants.au3>
#include <Date.au3>
#include <File.au3>

Global $answer, $timer, $Secs, $Mins, $Hour, $Time1, $Time2

If $CmdLine[0] = 0 Then
Exit MsgBox(0, 'Command Line Info', 'No arguments passed')
Else
MsgBox(0, 'Command Line Info', 'There are ' & $CmdLine[0] & ' arguments passed')
MsgBox(0, 'Command Line Info', 'The first argument is ' & $CmdLine[1])

$logFile = FileOpen($CmdLine[1],9)

MsgBox(0, "Check logfile creation", 'The logfile is: ' & $logFile)
_FileWriteLog($logFile, "version20130826")
EndIf

And I get the message "The first argument is C:FredTestNotepadLog.txt" and :The logfile is: 1

using commandline: "C:Program Files (x86)AutoIt3AutoIt3.exe" C:SnapsAutoITfrednotepad_exercise20130826.au3 C:FredTestNotepadLog.txt"

[FireFox]

And what's wrong with that?

The FileOpen function returns a file handle as stated in the helpfile.

Br, FireFox.

[FredMellink]

Yes, so how to result a file?? By using

$logFile = FileOpen(" & $CmdLine[1] & ",9)     ?

commandline execution using $CmdLine is new to me.

Edited August 27, 2013 by FredMellink

[JohnOne]

What is it you are expecting/wanting $logfile to contain?

[FredMellink]

I want to add logging output of each test step using _FileWriteLog($logFile, "test step" & $i & "result")

[FireFox]

Your first syntax was correct:

$logFile = FileOpen($CmdLine[1], 9)

PS: to post your code, use the autoit tags.

[JohnOne]

If the file is not being created and written to then you might be trying to write to a location which needs admin rights.

Put this in code

#RequireAdmin

Anywhere, but top is usual.

[FredMellink]

Thanks John... and all others, Now it works