shell scripting

jamrodh singh · 02-27-2015, 11:33 AM

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

suicidaleggroll · 02-27-2015, 11:51 AM

Need some context...

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

rknichols · 02-27-2015, 12:56 PM

${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".