Bash expression to list files beginning and ending with a pattern

Question

In my shell bash, I have to select files beginning by ab or xyz and don't end by .jpg or .gif

here is what i did but it doesn't work:

$ echo ab*[!.jpg] ab*[!.gif] xyz*[!.jpg] xyz*[!.gif]

Answer 1

With bash's extended glob syntax:

$ touch {ab,xyz}1234.{jpg,gif,txt,doc}

$ shopt -s extglob    
$ echo @(ab|xyz)!(*@(.jpg|.gif))
ab1234.doc ab1234.txt xyz1234.doc xyz1234.txt

The exclamation point is for negation, and the @ symbol is for or.

References:

• Using OR patterns in shell wildcards
• How can I use inverse or negative wildcards when pattern matching in a unix/linux shell?

Answer 2

Using grep:

ls | grep -E '^ab|^xyz' | grep -E -v '\.jpg$|\.gif$'

-v is to inverse the match

Answer 3

Hi you can try with below command:-

ls {ab*,xyz*}.* | sed '/.jpg/d;/.gif/d'

IF you store the output into a file:-

ls {ab*,xyz*}.* | sed '/.jpg/d;/.gif/d' > shortedFile.txt

How will it work? ls {ab*,xyz*}.* command will list out all the files begin with ab and xyz and redirect the output to sed command by using | (pipe) and sed command will remove file name ended with .jpg and gif.

Hope this will help your