3

i checked my file with

MsgBox($MB_SYSTEMMODAL, "", "Contents of the file:" & @CRLF & $sFileRead)

so i know i created the file successfully

but when i want to change it to .bat by moving it into a junk .bat file (other people probably have a better way of doing it) i get an error.

$CMD = "move "$sFilePath" " & $jointpath & @CRLF
RunWait(@ComSpec & " /c " & $CMD)

i get 

error: syntax error

here is the whole function:

Func makeJointBat()
    Local Const $sFilePath = _WinAPI_GetTempFileName(@TempDir)
    FileWrite($sFilePath, "you don't need to know what i wrote there")
    Local $hFileOpen = FileOpen($sFilePath, $FO_READ)
    Local $sFileRead = FileRead($hFileOpen)
    FileClose($hFileOpen)
    $CMD = "move "$sFilePath" " & $jointpath & @CRLF
    RunWait(@ComSpec & " /c " & $CMD)
EndFunc   ;==>makeJointBat
burih yidah leib
  • 67

1 Answers1

3

The problem might be that you seem to be opening the file two times: Once when creating it, then again when reading it.

However, you are only closing it once.

Adding a FileClose between the FileWrite and the FileOpen might help with the problem.

harrymc
  • 498,455