# ldd /usr/bin/ffmpeg
linux-vdso.so.1 => (0x00007ffffc1fe000)
libavfilter.so.0 => not found
libpostproc.so.51 => not found
libswscale.so.0 => not found
libavdevice.so.52 => not found
libavformat.so.52 => not found
libavcodec.so.52 => not found
libavutil.so.49 => not found
libm.so.6 => /lib/x86_64-linux-gnu/libm.so.6 (0x00007fdd18259000)
libpthread.so.0 => /lib/x86_64-linux-gnu/libpthread.so.0 (0x00007fdd1803a000)
libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007fdd17c75000)
/lib64/ld-linux-x86-64.so.2 (0x00007fdd18583000)
I am trying to grep only the names left from the "=>" symbol.
It works with echo easily:
echo linux-vdso.so.1 | grep -oP "^[a-z0-9.]*"
linux-vdso.so.1
But when I perform the same RegEx onto the output of ldd it does display anything:
ldd /usr/bin/ffmpeg | grep -oP "^[a-z0-9.]*"
So I thought, maybe I have to include some whitespace
ldd /usr/bin/ffmpeg | grep -oP "^([a-z0-9.]|\w)*"
But, this did not work and so I do not know further...
awk '{ print $1 }'
? – jordanm Apr 02 '15 at 17:21