linux - Realizing recognition of files/directories in bash -

i want write script in bash, produces list of directory file. necessary mark every line file or directory.

this unfinished attempt:

#!/bin/bash if [ $# -eq 1 ]  if [ -d $1 ]    touch liste.txt   ls -l $1 | grep '^-' >> liste.txt   ls -l $1 | grep '^d' >> liste.txt  fi fi 

now don t know how print in every line "file" or "directory". maybe there more elegant way solve this.

greetings, haniball

thanks pavium,

here finished script:

#!/bin/bash if [ $# -eq 1 ]  if [ -d $1 ]    rm liste.txt   touch liste.txt   ls -l $1 | grep '^-' | sed -e "s/^-/file /g" >> liste.txt   ls -l $1 | grep '^d' | sed -e "s/^d/directory /g" >> liste.txt  fi  more liste.txt fi 

i sure there more elegant solution. maybe grep can thrown out, had restrict output lines match pattern.

greetings, haniball

you can use sed pavium said.

ls -l $1 | grep '^-' | sed 's/^/file: /' >> liste.txt ls -l $1 | grep '^d' | sed 's/^/directory: /' >> liste.txt 

or in 1 command:

ls -l $1 | sed -n -e '/^-/{s/^/file: /p;d;}' -e '/^d/{s/^/directory: /p;d;}' > liste.txt 

or can different:

for f in $1/* ;    if [ -d "$f" ];     echo "directory: $f" >> liste1.txt   else     echo "file: $f" >> liste1.txt   fi done