Bash – How to Grep by First Character

bashgrep

I'm trying to list all files and select only certain from them.
ls -1 return me the list

abc 
abcd
rm_0818-051752-753-06

after that I want to select only by this criteria to return the 3rd line so I do:

ls -1 | grep -E "^rm_.*"

but I get nothing.

I'm trying to recognize the first character of 3rd line in this way:

var=$(ls -1 | grep -E "rm_")
echo $var //returns rm_0818-051752-753-06
echo ${var:0:1} //returns some strange symbol 001B in square

Can you explain me this behavior and how could I grep by first character ^? Thanks

Best Answer

ls -1 | grep -E "^rm_.*" looks good and should work. Possible reason why it doesn't work is alias bound to ls command in your profile.
To ensure, try /bin/ls -1 | grep -E "^rm_.*"

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Command-Line Grep – Params to Return Only File Names

   -l, --files-with-matches
          Suppress normal output; instead print the  name  of  each  input
          file  from  which  output would normally have been printed.  The
          scanning will stop on the first  match.   (-l  is  specified  by
          POSIX.)

from grep manpage

Bash – How to decode a list of base64-encoded file names

Just add the base64 encoding of newline (Cg==) after each file name and pipe the whole thing to base64 -d:

find . -name "*_*" -printf "%f\n" |
  sed -n 's/_....-..-..\.pdf$/Cg==/p' |
  base64 -d

With your approach, that would have to be something like:

find . -name "*_*" -printf "%f\0" |
  sed -zn 's/_....-..-..\.pdf$//p' |
  xargs -r0 sh -c '
    for i do
      echo "$i" | base64 -d
    done' sh

as you need a shell to create those pipelines. But that would mean running several commands per file which would be quite inefficient.

Best Answer

Related Solutions

Command-Line Grep – Params to Return Only File Names

Bash – How to decode a list of base64-encoded file names

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