How to extract a word from a string in bash?

Question

linux

Good afternoon habrayudi.
Faced such a problem - it is necessary to select the first word from the end (before the slash) from the string, for example, "/etc/somewhere" into another variable.
How to implement this in bash?

Thanks in advance

Answer 1

If it is stored in a variable, then:

T=/etc/somewhere ; echo ${T##*/}

If there is a stream from a file, then sed, awk. It is also possible in the while read LINE loop, but this is inefficient (sed and awk together will have time to work out 5-10 times each =) )

Answer 2

Try sed 's|.*/||'
Example:
[email protected]:~$ echo /etc/somewhere | sed 's|.*/||'
somewhere

Answer 3

echo /etc/somewhere | grep -E -o '[^/]+$'

Answer 4

[email protected]:~$ basename /etc/crontab
crontab

Answer 5

vvv=`basename /etc/crontab`; echo $vvv;

Answer 6

echo "/etc/somewhere" | awk -F"/" '{print $3}'

Answer 7

$ foo=/etc/some/where $ bar=${foo##*/} $ echo $bar where $