Which I don't understand!

[angelo2449]

With this code working great, I have to do something that bob I understand!



   [Stampa1]



   if FileExists("C:\CalCMill\Varie\RigheConMille.dat") then

       open "C:\CalCMill\Varie\RigheConMille.dat" for input as #1

   else

       Notice "Attenzione" + chr$(13) + "Devi prima eseguire > Calcola <"

       wait

   end if



   Percorso$ = "C:\CalcMill\Soggetti\" + Soggetto$ + "\" + EdiX$ + "\Elaborati\Elaborato1.doc"

   print "Percorso$ xxxxxxxx  "; Percorso$



   if FileExists(Percorso$) then

       kill Percorso$

   end if



   open Percorso$ for output as #2



       gosub [Copertina]



       print #2, "Edificio       Piano   Unità  Scala   Int.  Sub.       Tipo         Millesimi"

       print #2, "------------------------------------------------------------------------------"

       print #2, ""



       while eof(#1) = 0

           line input #1, StringaIn$



           Pos1 = instr(trim$(StringaIn$), ",", 1)

           Pos2 = instr(trim$(StringaIn$), ",", Pos1 + 1)

           Pos3 = instr(trim$(StringaIn$), ",", Pos2 + 1)

           Pos4 = instr(trim$(StringaIn$), ",", Pos3 + 1)

           Pos5 = instr(trim$(StringaIn$), ",", Pos4 + 1)

           Pos6 = instr(trim$(StringaIn$), ",", Pos5 + 1)

           Pos7 = instr(trim$(StringaIn$), ",", Pos6 + 1)



           Edificio$ = left$(StringaIn$, Pos1 - 1)

           Piano$ = mid$(StringaIn$, Pos1 + 1, Pos2 - Pos1 - 1)

           Unita$ = mid$(StringaIn$, Pos2 + 1, Pos3 - Pos2 - 1)

           Scala$ = mid$(StringaIn$, Pos3 + 1, Pos4 - Pos3 - 1)

           Interno$ = mid$(StringaIn$, Pos4 + 1, Pos5 - Pos4 - 1)

           SubAlterno$ = mid$(StringaIn$, Pos5 + 1, Pos6 - Pos5 - 1)

           Tipo$ = mid$(StringaIn$, Pos6 + 1, Pos7 - Pos6 - 1)

           Mille$ = mid$(StringaIn$, Pos7 + 1)



           print Edificio$

           print Piano$

           print Unita$

           print Scala$

           print Interno$

           print SubAlterno$

           print Tipo$

           print Mille$



           print #2, Edificio$ + space$(15 - len(Edificio$)) +_

                     Piano$ + space$(10 - len(Piano$)) +_

                     Unita$ + space$(7 - len(Unita$)) +_

                     Scala$ + space$(7 - len(Scala$)) +_

                     Interno$ + space$(6 - len(Interno$)) +_

                     SubAlterno$ + space$(7 - len(SubAlterno$)) +_

                     Tipo$ + space$(15 -len(Tipo$)) +_

                     using("#####.##", val(Mille$))

       wend



   close #1

   close #2



   Percorso1$ = "C:\CalcMill\Soggetti\" + Soggetto$ + "\\" + chr$(34) + EdiX$ + chr$(34) + "\Elaborati\Elaborato1.doc"

   Run "C:\Program Files (x86)\OpenOffice 4\program\swriter.exe " + Percorso1$



   wait



Why do I have to insert a double backslash in the definition of > Percorso1$ < for the program to work?
If I remove a backslash, the path is not recognized!
Thank You



[Walt Decker]

if FileExists("C:\CalCMill\Varie\RigheConMille.dat") then

open "C:\CalCMill\Varie\RigheConMille.dat" for input as #1

I assume you are having trouble here. FileExists() is not an LB function, it is a User32.dll function. So you will have to use CALLDLL for the query.  Look up CALLDLL in the help.

[alincon]

From LB help file:

This also makes it practical to use to check for file existence. If you want to know if a file c:\config.bak exists, you could try...

files "c:\", "config.bak", info$()

If val(info$(0, 0)) > 0, then the file exists.

[angelo2449]

My question is related to this line of program where there are the two backslash of which I do not understand the meaning :

Percorso1$ = "C:\CalcMill\Soggetti\" + Soggetto$ + "\\" + chr$(34) + EdiX$ + chr$(34) + "\Elaborati\Elaborato1.doc"

[tsh73]

We don't know what Soggetto$ and EdiX$ is.
How about just print it

Percorso1$ = "C:\CalcMill\Soggetti\" + Soggetto$ + "\\" + chr$(34) + EdiX$ + chr$(34) + "\Elaborati\Elaborato1.doc"
print Percorso1$
Run "C:\Program Files (x86)\OpenOffice 4\program\swriter.exe " + Percorso1$

and show us?

[Chris Iverson]

Do you have an example of a full string that Percorso1$ would be? Try adding a PRINT or a NOTICE to display the value just before the program is run.

TL;DR down below, but I suspect the command-line string gets double-parsed by a string parser in the program, and in most parsers nowadays(indeed, most programming languages), the backslash in a string is a special character - an escape character. You use backslashes when you want to put special characters in a string.

For example, if you want to add chr$(13) + chr$(10) to the end of a string "test" in many programming languages, you could just do "test\r\n".

Another example, you can include a quote in a string by escaping it. Like, to get the output string test"quote, in LB you'd do "test" + chr$(34) + "quote", in these languages you could do "test\"quote".

Because the \ is special, that indicates something special about the next character(s), you need a way to indicate that you really intend for a backslash to be there.

And you do that by escaping the backslash itself, so you get two backslashes. In such a language, "C:\new folder" would be interpreted as "C:" + chr$(10) + "ew folder". To clarify that you want the backslash there, you'd do "C:\\new folder", which would parse to "C:\new folder".


TL;DR, it's likely not an LB thing, but a whatever-program-you're-calling thing.

[angelo2449]

Hi Guys,

Soggetto$ = "COSMOPOLITANSpA"
EdiX$ = "Palazina D"

In the first writing of the code, the line looked like this:
Percorso1$ = "C:\CalcMill\Soggetti\" + Soggetto$ + "\" + EdiX$ "\Elaborati\Elaborato1.doc"
But the system did not recognize the path due to the presence of a blank in 'Palazzina D'.
So I added> chr $ (34) <to encapsulate 'Palazzina D'.
I had to enter > \\ < for the program to work!
Thank You.

[Chris Iverson]

If those are the variables, that means the string that gets built is this:

C:\CalcMill\Soggetti\COSMOPOLITANSpA\\"Palazina D"\Elaborati\Elaborato1.doc

In which case, I think I was correct about the \ being an escape character, but wrong about which one. I think the first "\" is used as the path separator, and the second is used to escape the quote character, so it doesn't think the parameter ended there. If it's splitting the parameters based on the quote, then without that escape character, it might be interpreting the command line as this:

Param 1: C:\CalcMill\Soggetti\COSMOPOLITANSpA\
Param 2: Palazina D
Param 3: \Elborati\Elborato1.doc

Instead of this:
Param 1: C:\CalcMill\Soggetti\COSMOPOLITANSpA\Palazina D\Elborati\Elborato1.doc

Out of curiosity, try removing both the double-backslash and the quotes around EdiX$, but try quoting the entire Percorso1$ string.

Like this:

Percorso1$ = chr$(34) + "C:\CalcMill\Soggetti\" + Soggetto$ + "\" + EdiX$ "\Elaborati\Elaborato1.doc" + chr$(34)

This will result in the string being built as:

"C:\CalcMill\Soggetti\COSMOPOLITANSpA\Palazina D\Elaborati\Elaborato1.doc"

Which is the standard for path parameters that include spaces.

[Walt Decker]

I would help greatly if you supplied source code.  It is very difficult to debug an app without it.

[Brandon Parker]

Jan 13, 2021 14:44:24 GMT -5 Walt Decker said:

if FileExists("C:\CalCMill\Varie\RigheConMille.dat") then

open "C:\CalCMill\Varie\RigheConMille.dat" for input as #1

I assume you are having trouble here. FileExists() is not an LB function, it is a User32.dll function. So you will have to use CALLDLL for the query.  Look up CALLDLL in the help.

This is just a response to clear something up. I would assume that the OP is actually using a user-defined function similar to the one shown below. Using the "Files" command and a double-dimension array LB can peruse the file structure with ease. One simply has to understand which parameters are placed where in the array when the "Files" command returns.


   Function fileExists(path$, filename$)
       Files path$, filename$, Info$()
       fileExists = Val(Info$(0, 0))
       ReDim Info$(0, 0)
   End Function


{:0)

Brandon Parker