Re: variable in loop

2011-01-03 Thread Roberto Ragusa
Rodolfo Alcazar Portillo wrote: > On Sun, 2011-01-02 at 14:27 -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

Re: variable in loop

2011-01-02 Thread Rodolfo Alcazar Portillo
On Sun, 2011-01-02 at 14:27 -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

Re: variable in loop

2011-01-02 Thread Chris Tyler
On Sun, 2011-01-02 at 14:27 -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..

Re: variable in loop

2011-01-02 Thread Sam Sharpe
On 2 January 2011 22:27, 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. Because {1.

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? -- users mailing list users@