Hi,
I have been scratching my head for bash parameter substitution.
I understand ## in following examples but I can not understand %%.
Is this how bash 2.0 supposed to work?
$ XXX="123456123456"
$ echo $XXX
123456123456
$ echo ${XXX%%*1}
123456123456
$ echo ${XXX%%*6}
$ echo ${XXX%*1}
123456123456
$ echo ${XXX%*6}
12345612345
$ echo ${XXX##*1}
23456
$ echo ${XXX#*1}
23456123456
??????????????????????????
--
~\^o^/~~~ ~\^.^/~~~ ~\^*^/~~~ ~\^_^/~~~ ~\^+^/~~~ ~\^:^/~~~ ~\^v^/~~~
+ Osamu Aoki <[EMAIL PROTECTED]>, GnuPG-key: 1024D/D5DE453D +
+ My debian quick-reference, http://www.aokiconsulting.com/quick/ +
