how to make $CmdLine[] work with quotes?

[E1M1]

If you compile it and then execute with argument abcd then it displays you abcd but if you execute it with argument "abcd" then it still shows you abcd with out quotes. is there way to get command line arg with arg?

Problem is that when I make program that takes multiple args I need quotes too. I want to make something that would accept such commandline arg: -i "C:\some dir\file.asm" -o "C:\some other path\file.o" -d

#include <Array.au3>

If $cmdline[0] > 0 Then
$input = _ArrayToString($CmdLine, " ", 1)
MsgBox(0,0,$input)
EndIf

I know that it's possible since ffmpeg works fine with quotes in command line arg compilers also accept quotes. in command line argument.

Does anyone know how I could make my program accept quotes in commandline argument?

[Jos]

You can add the double quotes around the Filename Commandline value yourself ...so don't understand your issue here?

Edited August 17, 2011 by Jos

[E1M1]

yea but if I have -i "\path to\somefile -a" -o something then it won't work that way because you don't know if file name is somefile -a of if -a is an other arg.

or of you want to give other executable (exe file) as arg then it also wouldnt work because you cant tell if you want to give args to executable file or if you want to give args to your autoit script.

I mean -i "file.exe -a" -d If i'd add quotes manually I cant tell if a is argument for file.exe of if it's arg for my script.

Is there any workaround with dllstructs or with some other way (like some function to get raw input))? or is this Operating System related problem?

Edited August 17, 2011 by E1M1

[Jos]

Don't understand, isn't the -i parameter always followed by a file name?

I have something like this in Autoit3Wrapper which works perfectly

For $x = 1 To $CMDLINE[0]
    $T_Var = $CMDLINE[$x]
    Select
        Case $T_Var = "-i"
            $x = $x + 1
            $ScriptFile_In = $CMDLINE[$x]
        Case $T_Var = "-o"
            $x = $x + 1
            $ScriptFile_Out = $CMDLINE[$x]
        Case $T_Var = "-d"
            ; do what you need for -D
        Case Else
            ; other/wrong option
    EndSelect
Next

Jos

[BrewManNH]

You can't have a space in the command line argument or the program will assume the next item is a new argument. If you're looking for the -i and then looking in the next array element for the parameter for -i then it will show the whole string enclosed in the parentheses as one array element, this "\path to\somefile -a" shows up in a single array element. I think your use of _arraytostring is what is confusing the matter.

[UEZ]

Well, command line parsing is sometimes very complicated.

#AutoIt3Wrapper_Change2CUI=y
#include <Array.au3>
_ArrayDisplay($CmdLine)

If you compile the script above and start it with the parameters -i "file.exe -a" -d you can see that after -i the parameter file.exe -a is given.

You can do some string manipulation to strip out the parameter for file.exe.

I hope it helps you.

Br,

UEZ

Edited August 17, 2011 by UEZ

[E1M1]

thanks. I found that single quote works. Double quote seems to be OS related problem as it also didn't work with C++ program.

so -i 'file.exe -a' -d works well

Thank you Jos for your help. I appreciate that.

edit @UEZ that's really good find. I helps alot. thank you.

Edited August 17, 2011 by E1M1

[JohnOne]

Is there any workaround with dllstructs or with some other way (like some function to get raw input))? or is this Operating System related problem?

$CmdLineRaw possibly.

[ProgAndy]

Here are the rules the cmdline-array uses to parse the commandline:

http://msdn.microsoft.com/de-de/library/17w5ykft.aspx

[E1M1]

$CmdLineRaw possibly.

Perfect!