Delete everything before last / in bash
Using sed
for this is vast overkill -- bash has extensive string manipulation built in, and using this built-in support is far more efficient when operating on only a single line.
s=/home/rtz11/files/testfiles/547/prob547455_01
basename="${s##*/}"
echo "$basename"
This will remove everything from the beginning of the string greedily matching */
. See the bash-hackers wiki entry for parameter expansion.
If you only want to remove everything prior to the last /
, but not including it (a literal reading of your question, but also a generally less useful operation), you might instead want if [[ $s = */* ]]; then echo "/${s##*/}"; else echo "$s"; fi
.
awk '{print $NF}' FS=/ input-file
The 'print $NF' directs awk to print the last field of each line, and assigning FS=/ makes forward slash the field delimeter. In sed, you could do:
sed 's@.*/@@' input-file
which simply deletes everything up to and including the last /
.
Meandering but simply because I can remember the syntax I use:
cat file | rev | cut -d/ -f1 | rev
Many ways to skin a 'cat'. Ouch.