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• How can I pass a variable containing spaces as an argument in a bash script ?

  8 décembre 2017, par Bill D

  I am trying to pass a variable containing a filename with spaces as an argument to ffmpeg in a bash script, and I am not having success.

  varToPass="file name with spaces.mp4"
  
  
  
  ffmpeg -i "$varToPass" -ss "${array[j]}" -t "${anotherArray[j]}" output.mp4

  I have tried passing the argument as "$varToPass" with double quotes as suggested by this answer and others, but I’m still unable to get it to work.

  I’ve also tried putting escape characters in the string like this :

  varToPass=`echo $varToPass | sed 's/ /\ /'`

  But still no luck.

  Any suggestions for how this can be done ?

• ffmpeg getting syntax error when run inside shell script only.. Why ? [duplicate]

  10 décembre 2017, par bmck2006

  This question already has an answer here :

  • Unexpected ’(’ in bash [duplicate]
    
    2 answers
    

  OS : Raspbian

  I’m currently working on a bash script that will merge all .AVI videos in the current directory, to one .AVI file. I’ve found the following to work perfectly for me, when run directly in the terminal :

  ffmpeg -f concat -safe 0 -i <(find . -name '*.avi' -printf "file '$PWD/%p'\n") -c copy "$(date -d '-1 day' +'%F')".avi

  This works perfectly, and I find a new .AVI file listed in the current directory (with a date name from yesterday).

  Now, when I add this same command to my script (merge.sh), I get the following output :

  /home/pi/Documents/MotionScripts/merge.sh: 11: 
  
  /home/pi/Documents/MotionScripts/merge.sh: Syntax error: "(" unexpected

  This is beyond frustrating. And yes, ffmpeg command from earlier is on line 11. The script navigates to the correct directory where the multiple .avi files are stored.

  Here is the entire script :

  #!/bin/bash
  
  
  
  #This script combines multiple AVI files within the current directory, into one long video
  
  
  
  cd /media/myBook/Security/yesterday
  
  
  
  sudo mv * /home/pi/Videos/tempconversion
  
  
  
  cd /home/pi/Videos/tempconversion
  
  
  
  ffmpeg -f concat -safe 0 -i <(find . -name '*.avi' -printf "file '$PWD/%p'\n") -c copy "$(date -d '-1 day' +'%F')".avi
  
  
  
  sudo mv "$(date -d '-1 day' +'%F')".avi /media/myBook/Security/yesterday
  
  
  
  sudo rm -f *

  Why am I getting the syntax error only when the same command is being issued within an executable script ?

  RESOLVED ! Shouldn’t have used ’sh’ to run the script. Make sure the script is executable (sudo chmod +x merge.sh) and run by entering the full path.

• When passing arguments with php shell script eval isn't properly executing the script

  19 décembre 2017, par Aleksandra Lozanovska

  This is the part of my php script :

  $result ='ffmpeg -re -i /home/neotel/sample.mp4 -s 320x240 -vf setdar=4:3 -r 
  
  25 -c:v libx264 -b:v 2500k -c:a aac -b:a 128k -f mpegts 
  
  udp://192.168.88.211:5000' 
  
  $FILE1 = escapeshellarg($result);
  
  $output = shell_exec("sh testpetar.sh $FILE1 2>&1");

  This is my shell script :

  #!/usr/bin/env bash
  
  FILE1=$1
  
  FILE2="nohup "
  
  FILE3=" > log.txt"
  
  FILE="$FILE2$FILE1$FILE3"
  
  eval $FILE

  I am trying to pass the $results strings as the $FILE1 argument and then concatenate it with the strings $FILE2 and $FILE3 to form a whole ffmpeg command and execute it as a script in shell. When i execute it manually through Putty it is working properly, but when i am passing the $results command from php the eval function is not working.