CODEWITHSUNDEEP
regexbashshell
manisha
manisha

Reputation: 695

Explanation of a regex

What does the following regex mean?

fspec="/exp/home1/abc.txt"
fname="${fspec##*/}"

I know what it does, but not how it does? Fetching fname is not clear to me.

Please explain.

Upvotes: 3

Views: 62

Answers (2)

Fidel
Fidel

Reputation: 1037

Below is the explanation from bash documentation.

${parameter#word}
${parameter##word}
The  word  is  expanded  to  produce  a pattern just as in pathname
expansion.  If the pattern matches the beginning of  the  value  of
parameter,  then  the result of the expansion is the expanded value
of parameter with the shortest matching pattern (the ``#'' case) or
the longest matching pattern (the ``##'' case) deleted.

As per the above explanation, In your example word=*/ which means zero (or) any number of characters ending with /.

bash-3.2$fspec="/exp/home1/abc.txt"
bash-3.2$echo "${fspec##*/}"   # Here it deletes the longest matching pattern
# (i.e) /exp/home1/
# Output is abc.txt

bash-3.2$echo "${fspec#*/}"    # Here it deletes the shortest matching patter
#(i.e) /
# Output is exp/home1/abc.txt
bash-3.2$

Upvotes: 1

fedorqui
fedorqui

Reputation: 289745

The ${var##*/} syntax strips everything up to the last /.

$ fspec="/exp/home1/abc.txt"
$ echo "${fspec##*/}"
abc.txt

In general, ${string##substring} strips the longest match of $substring from front of $string.

For further reference you can for example check Bash String Manipulation with several explanations and examples.

Upvotes: 4

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