How to do String manipulations using Substring function in Shell

anilvvnn · #1 (**permalink**) 08-18-2009

Hi,

I have a scenario to just plug out the file name from the following location path.

/opt/project/data/int/holdFiles/csv195687.csv

So, how do I get just file name which is "csv195687.csv" from the above line using awk/shell scripting? Can we use indexOf and Substring in awk to get last indexOf "/" and then end of String to get file name?

Thanks.

malcomex999 · #2 (**permalink**) 08-18-2009

if u made a little search, you could have get what u wanted by urself....

Code:

 
str="/opt/project/data/int/holdFiles/csv195687.csv"
echo ${str##*/}
csv195687.csv

anilvvnn · #3 (**permalink**) 08-18-2009

Thanks for quick reply. It is a cool answer. From next time onwards, I will definitely search the forum before I ask the question.

durden_tyler · #4 (**permalink**) 08-18-2009

Code:

$
$ str="/opt/project/data/int/holdFiles/csv195687.csv"
$ basename $str
csv195687.csv
$
$ echo $str | sed 's/.*\///'
csv195687.csv
$
$ echo $str | awk -F/ '{print $NF}'
csv195687.csv
$
$ echo $str | perl -lne '/.*\/(.*)/ && print $1'
csv195687.csv
$

tyler_durden

anilvvnn · #5 (**permalink**) 08-18-2009

Thanks Tyler. Hmmm...lot of options.

leobsd · #6 (**permalink**) 2 Weeks Ago

str="/opt/project/data/int/holdFiles/csv195687.csv"
basename $str

cfajohnson · #7 (**permalink**) 2 Weeks Ago

Quote:

Originally Posted by anilvvnn

Hmmm...lot of options.

The only one that makes sense for dealing with a string is:
Code:
echo ${str##*/}
Save external commands for dealing with files.

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