Hello. В сообщении от Sunday 19 April 2015 05:03:41 Denis Mysenko dus...@mail.ru: > If you use environment variable (like you proposed) or any shell/UNIX 'hack > (eg. -i `cat url.txt`), at the moment of execution a plain-text password > will be inserted anyway. And 'ps' will show what really was launched, > not the string before parsing.
So I suggest exactly not to show plain-text passwords in command line. Not to use variable on command line. Not to use hack -i `cat url.txt`. But give to ffmpeg variable name, and ffmpeg will parse variable themselves. Variant 1. Password on command line. Running: $ export PASS=secret $ ffmpeg rtsp://admin:$PASS@192.168.0.100/stream1.sdp will lead to: $ ps -f -C ffmpeg UID PID PPID C STIME TTY TIME CMD user 1506 1423 88 07:47 pts/11 00:00:08 ffmpeg rtsp://admin:secret@192.168.0.100/stream1.sdp Variant 2. Variable name on command line (name, but not contents). Running: $ export PASS=secret $ ffmpeg 'rtsp://admin:$PASS@192.168.0.100/stream1.sdp' will lead to: $ ps -f -C ffmpeg UID PID PPID C STIME TTY TIME CMD user 1506 1423 88 07:47 pts/11 00:00:08 ffmpeg rtsp://admin: $PASS@192.168.0.100/stream1.sdp In the second variant ffmpeg parameters are quoted and will not be parsed by bash. In the second variant "ps" will not see password. But in the second variant ffmpeg must parse variables at run time. -- Ilya Melnikov Monday 20 April 2015 06:41:40
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