Re: variable in loop

2011-01-10 Thread Brian Ryans
Quoting Karl E. Jorgensen on 2011-01-02 17:22:20: > for i in `seq 1 $a` That `seq 1 $a` could be trimmed to `seq $a`. Unsure of portability. -- _ Brian Ryans 8B2A 54C4 E275 8CFD 8A7D 5D0B 0AD0 B014 C112 13D0 . ( ) ICQ 43190205 | Mail/Jabber/Yahoo/MSN: brianlry...@gmail.com ..: X A

Re: variable in loop

2011-01-02 Thread Andrew McGlashan
Hi, S Mathias wrote: $ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}') )); echo "$a $ASDF"$(for i in {1..$a}; do printf "."; done) 65 hello. $ Why doesn't it print: 65 hello. What am i missing? Don't know,

Re: variable in loop

2011-01-02 Thread François TOURDE
Le 14976ième jour après Epoch, S. Mathias écrivait: > What am i missing? Saying "Hello, this post is Off Topic, but can someone help me, please", perhaps? -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of "unsubscribe". Trouble? Contact listmas...@lists.debian.o

Re: variable in loop

2011-01-02 Thread Karl E. Jorgensen
On Sun, Jan 02, 2011 at 02:27:19PM -0800, S Mathias wrote: > $ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}') )); echo > "$a $ASDF"$(for i in {1..$a}; do printf "."; done) > 65 hello. > $ > > > Why doesn't it print: > 65 hello...

variable in loop

2011-01-02 Thread S Mathias
$ ASDF=hello; a=0; a=$(( 70 - $(echo $ASDF | awk '{print length}') )); echo "$a $ASDF"$(for i in {1..$a}; do printf "."; done) 65 hello. $ Why doesn't it print: 65 hello. What am i missing? -- To UNSUBSCRIBE, email t