How to execute a command line in a shell script
I'm coding a shell script at this time, and i wanted to know how i could execute a command line in this shell script.
To give you more details, i try to execute ffmpeg with some arguments, it looks like that: ./ffmpeg -an -b 400 -i video.mpg video.m4v.
Thank you for your help.
You just put that line in the script.
Simply include the line in your shell script (without the ./); you need to make sure that the PATH variable contains the path to ffmpeg (usually /usr/local/lib, if I remember correctly).
This will help in most cases (flames welcome):
At the console, type
echo $PATH >> </path/to/your/script/your_script.sh>
then open the script with any text editor, copy the last line to the second line (first will probably read #!/bin/bash), then write at the beginning of the line
If your path is not included in said line, add it to the end with <:/your/path>
ok thanks... but do i really need to set the path???
It should be in usr/local/bin, not lib since it's an executable... but if i put it in the same path that my script, it could work?
If it has no further dependancies, yes. Use
to see which libraries ffmpeg depends on. But I'd say its rather ugly under *NIX to keep binaries in other directories than those specified in file hierarchy standard, see
I need help one more time :D
Something very simple in fact but i do not know how the if else and while tests work in shell bash...
I use a line like that, the same that i use in C-C++...
while (sed -n '$jp' ls_mpg.txt && $ok == 0)
if (sed -n '$jp' ls_mpg.txt == sed -n '$in' ls_mp4.txt) then
$ok = 1
does this ("sed -n '$jp' ls_mpg.txt && $ok == 0") works?
Thank you for your help
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