Simple Batch Job

[ptestfall]

Hi,

the simple thing I want to do is a dos command:

echo text > $1

and convert the autoit-file to an exe-file and give the name of the text file as a parameter like e.g. "test.exe textfile.txt".

But if I use autoit with

Run("cmd /c echo text > $1")

and convert it to a exe-file I cannot give a parameter upon execution or it is ignored and the result is a file named $1 with "text" in it.

Please help me with this little problem.

Thanks

pt

[Don N]

would an input box work...run the exe and asks for the textfile name

or can it just use a default one? lets say text.txt?

[ptestfall]

An input box is unwanted and a default filename as well.

The command for execution should be:

test.exe textfile.txt

Where test.exe is a compiled autoit script and textfile.txt is just an example.

In the end I want to end a complete path like "C:\Program Files\Installation\textfile.txt".

[Xenobiologist]

Hi,

something like:

If $cmdLine[0] = 2 Then
;Run("cmd /c echo " &  $cmdLine[1] & ">" & $cmdLine[2])
    Run(@ComSpec & " /c " & 'echo '&  $cmdLine[1] & ">" & $cmdLine[2], "", @SW_HIDE)
Else
    MsgBox(64, "Error", "Needs two parameters" & @CRLF & @CRLF & _
    "1. text" & @CRLF & _
    "2. filename.extension")
EndIf

So long,

Mega

[Geert]

Or

If $CmdLine[0] > 0 Then
    FileWriteLine($CmdLine[1], "text")
EndIf

[ptestfall]

Thanks very much for your great and quick help