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Old 02-27-2015, 12:33 PM   #1
jamrodh singh
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Registered: Feb 2015
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shell scripting


can some1 please tell me ....i am writing script to display type of file....and i m not able to get what "filename##*\." means?....please be brief in your explanation..thank you
the program is like:
for filename in$(ls)
do
ext=${filename##*\.}
case '$ext' in

Last edited by jamrodh singh; 02-27-2015 at 12:55 PM.
 
Old 02-27-2015, 12:51 PM   #2
suicidaleggroll
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Need some context...

edit: thanks for the edit, looks like rknichols answered it.

Last edited by suicidaleggroll; 02-27-2015 at 02:41 PM.
 
Old 02-27-2015, 01:56 PM   #3
rknichols
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${filename##*\.} is the value of $filename with all leading characters up to and including the last "." character deleted. (The backslash escape on the "." is superfluous.)

${varname#*.} drops all leading characters up through the first "." (shortest string that matches "*.") The double "##" looks for the longest match.

It's all in the bash manpage under "Parameter Expansion".
 
  


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