declare a variable to ffmpeg-CodePudding

In PowerShell, i want declare a variable to ffmpeg:

## DOSSIERS DE TRAVAIL
$Ingest = "C:\INGEST\" ## DOSSIER WATCHE
$outPath = "C:\TRANSCODE\" ## DOSSIER DE DESTINATION, A definir
$TEMP = "C:\temp\" ## dossier temporaire de récupération des essences (de-wrapp fichier ingesté) 
 

$MAPaudio = ""
$Filesaudio = ""


for ($i = 1; $i -le $nombreAudios; $i  ) {
    $TEMP_AUDIO = $TEMP   "audio"   $i   ".wav"
    ##Extraction de l'audio et conversion en 48000hz (Obligatoire pour le MXF)
    $MAPaudio  = " -map "   $i   ":a "
    $Filesaudio  = " -i "   $TEMP_AUDIO
}

    
$TEMP_VIDEO = $TEMP   "video.mxf"

write-host $MAPaudio
write-host $Filesaudio
##Encapsulage de la vidéo
$result = " -i $TEMP_VIDEO $Filesaudio -c copy -map 0:v:0 $MAPaudio -y c:\transcode\test.mxf".tostring()
  
ffmpeg "$result"

I have an error:

Output #0, mxf, to ' -i C:\temp\video.mxf  -i C:\temp\audio1.wav -i C:\temp\audio2.wav -i C:\temp\audio3.wav -i C:\temp\audio4.wav -c copy -map 0:v:0  -map 1:a  -map 2:a  -map 3:a  -map 4:a  -y c:\transcode\test.mxf':

Output file #0 does not contain any stream

Yet, if i give the variable to ffmpeg without powershell, it works !!! I don't understand why, ffmpeg doesn't want my variable.

CodePudding user response：

You are effectively calling ffmpeg like this:

ffmpeg " -i $TEMP_VIDEO $Filesaudio -c copy -map 0:v:0 $MAPaudio -y c:\transcode\test.mxf"

Note the double quotes that are automatically added by PowerShell, when you pass a variable that contains whitespace, as an argument to a native command!

As a result, ffmpeg sees only a single argument, that it doesn't understand.

If you want to store multiple arguments in a variable, it has to be an array instead of a single string. PowerShell unrolls and quotes (as necessary) the individual arguments, when passing the arrays to the command.

To fix the code:

$TEMP = "C:\temp" ## dossier temporaire de récupération des essences (de-wrapp fichier ingesté) 
 
$MAPaudio = @()    # create empty array
$Filesaudio = @()  # create empty array

for ($i = 1; $i -le $nombreAudios; $i  ) {
    $TEMP_AUDIO = Join-Path $TEMP "audio$i.wav"
    ##Extraction de l'audio et conversion en 48000hz (Obligatoire pour le MXF)

    # Append two elements to the argument array
    $MAPaudio  = "-map", "${i}:a"

    # Append two elements to the argument array
    $Filesaudio  = "-i", $TEMP_AUDIO
}
    
$TEMP_VIDEO = Join-Path $TEMP "video.mxf"

# Finally pass the arguments separately, instead of a single string
ffmpeg -i $TEMP_VIDEO $Filesaudio -c copy -map 0:v:0 $MAPaudio -y 'c:\transcode\test.mxf'

CodePudding user response：

I test this:

   ##Récupération et formatage du nom de la vidéo
$name=[string]$_
$nameFileToEncode=$name.split('.')[-2]
$nameFileToEncode=$nameFileToEncode.replace(" ","_")

##Path complet de la Video de sortie
$outVideo="$outPath" "\" "$nameFileToEncode" ".mxf"


$MAPaudio=""
$Filesaudio=""


for($i=1;$i -le $nombreAudios;$i  )
{
$TEMP_AUDIO=$TEMP "audio" $i ".wav"
##Extraction de l'audio et conversion en 48000hz (Obligatoire pour le MXF)
$MAPaudio  = " -map " $i ":a "
$Filesaudio  =" -i "  $TEMP_AUDIO
}


$TEMP_VIDEO=$TEMP "video.mxf"

write-host $MAPaudio
write-host $Filesaudio
##Encapsulage de la vidéo

&ffmpeg -i $TEMP_VIDEO $Filesaudio -c copy -map 0:v:0 $MAPaudio -y 'c:\transcode\test.mxf'

but i have this error:

Output #0, wav, to ' -i C:\temp\audio1.wav -i C:\temp\audio2.wav -i C:\temp\audio3.wav -i C:\temp\audio4.wav':
Output file #0 does not contain any stream

Curiously, it says that is a output then this is inputs...