Re: need help with bash command

[Ding Honghui <hhding.gnu@gmail.com>]

tôba wrote:
> Hello,
> 
> Can you help me with a bash command?
> I a directory, I want to extract with a command line for jpg name files.
> I have a lot of:
> 
> icon_3620_0_1.jpg
> icon_3616_0_1.jpg
> icon_3563_0_1.jpg
> 
> I want to extract the number betwen icon_*_0_1.jpg AND I want to use
> the entire filename too in the same command.
> 
> The goal is to create a directory called 3620 and move the
> icon_3620_0_1.jpg file into this directory.
> 
> So, I did like this for test:
> 
> # for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
> icon_$a_0_1.jpg; done
> 
> I hope it should give me:
> 
> icon_3620_0_1.jpg
> icon_3616_0_1.jpg
> icon_3563_0_1.jpg
> 
> like the original
> 
> BUT in my surprise, it shows:
> 
> icon_.jpg
> icon_.jpg
> icon_.jpg
> icon_.jpg
> 
> Why?
> Where is the $a_0_1 gone? Surely I missed something but I cannot
> find it.
> Can you tell me how to reproduce the exact filename so?
> 
> Best regards,
> 
> --
> Tôba
> 
> 

Hello, do you means like that?
for a in `ls | sed -e s'/_/ /g' | awk '{print $2}'`; do echo
icon_${a}_0_1.jpg; done

A better way maybe is:
dhh@mud:~/test/jpg$ touch icon_3563_0_1.jpg  icon_3616_0_1.jpg
icon_3620_0_1.jpg
dhh@mud:~/test/jpg$ ls
icon_3563_0_1.jpg  icon_3616_0_1.jpg  icon_3620_0_1.jpg
dhh@mud:~/test/jpg$ ls | awk -F_ '{print $2}'
3563
3616
3620
dhh@mud:~/test/jpg$