I wrote a script with mbratch's help.
I run as below;
./scriptname folder1
However, I see neither error nor results and I'm not sure what's wrong.
sh -x ./scriptname folder1
+ MAIN
+ check_directories
+ [ -d ]
This works fine for me:
Note: updated to support additional options.
opt="$1"
folder1="$2"
folder2="$3"
case "$opt" in
-d)
if [ -d "${folder1}" ] && [ -d "${folder2}" ] ; then
for i in "${folder1}/*" ; do
echo "test <$i>"
if [ -f "${folder2}/${i##*/}" ] ; then
echo "<${i##*/}>"
else
echo "! <${i##*/}>"
fi
done
fi
;;
# Example option
-h)
# Print help and exit.
;;
# Default case
*)
echo "Unknown option '$opt'" >&2
exit 1
;;
esac
Try replacing ~/ with $HOME/, and be sure to set folder1 and folder2 before using them. Note also that this will break if your directory or file names include spaces. In that case, use find; check the find man page for details.
Is that the entirety of your script? The variables $folder1 are never defined anywhere. By default, the program will take the first two chunks of text in as $1 and $2, so use those variables instead.
Put some echo statements in there to see what variables have what values.
You have a for loop already, so go with that. However you might have to put the part where you are getting a file list inside of $() to have it assigned to a variable, and then loop over it.
Do a quick search on "Looping through files in bash" and you will find a good template for the for loop.
Related
I'm trying to print all directories/subdirectories from a given start directory.
for i in $(ls -A -R -p); do
if [ -d "$i" ]; then
printf "%s/%s \n" "$PWD" "$i"
fi
done;
This script returns all of the directories found in the . directory and all of the files in that directory, but for some reason the test fails for subdirectories. All of the directories end up in $i and the output looks exactly the same.
Let's say I have the following structure:
foo/bar/test
echo $i prints
foo/
bar/
test/
While the contents of the folders are listed like this:
./foo:
file1
file2
./bar:
file1
file2
However the test statement just prints:
PWD/TO/THIS/DIRECTORY/foo
For some reason it returns true for the first level directories, but false for all of the subdirectories.
(ls is probably not a good way of doing this and I would be glad for a find statement that solves all of my issues, but first I want to know why this script doesn't work the way you'd think.)
As pointed out in the comments, the issue is that the directory names include a :, so -d is false.
I guess that this command gives you the output you want (although it requires Bash):
# enable globstar for **
# disabled in non-interactive shell (e.g. a script)
shopt -s globstar
# print each path ending in a / (all directories)
# ** expands recursively
printf '%s\n' **/*/
The standard way would either to do the recursion yourself, or to use find:
find . -type d
Consider your output:
dir1:
dir1a
Now, the following will be true:
[ -d dir1/dir1a ]
but that's not what your code does; instead, it runs:
[ -d dir1a ]
To avoid this, don't attempt to parse ls; if you want to implement recursion in baseline POSIX sh, do it yourself:
callForEachEntry() {
# because calling this without any command provided would try to execute all found files
# as commands, checking for safe/correct invocation is essential.
if [ "$#" -lt 2 ]; then
echo "Usage: callForEachEntry starting-directory command-name [arg1 arg2...]" >&2
echo " ...calls command-name once for each file recursively found" >&2
return 1
fi
# try to declare variables local, swallow/hide error messages if this fails; code is
# defensively written to avoid breaking if recursing changes either, but may be faulty if
# the command passed as an argument modifies "dir" or "entry" variables.
local dir entry 2>/dev/null ||: "not strict POSIX, but available in dash"
dir=$1; shift
for entry in "$dir"/*; do
# skip if the glob matched nothing
[ -e "$entry" ] || [ -L "$entry" ] || continue
# invoke user-provided callback for the entry we found
"$#" "$entry"
# recurse last for if on a baseline platform where the "local" above failed.
if [ -d "$entry" ]; then
callForEachEntry "$entry" "$#"
fi
done
}
# call printf '%s\n' for each file we recursively find; replace this with the code you
# actually want to call, wrapped in a function if appropriate.
callForEachEntry "$PWD" printf '%s\n'
find can also be used safely, but not as a drop-in replacement for the way ls was used in the original code -- for dir in $(find . -type d) is just as buggy. Instead, see the "Complex Actions" and "Actions In Bulk" section of Using Find.
This a small bash program that is tasked with looking through a directory and counting how many files are in the directory. It's to ignore other directories and only count the files.
Below is my bash code, which seems to fail to count the files specifically in the directory, I say this because if I remove the if statement and just increment the counter the for loop continues to iterate and prints 4 in the counter (this is including directories though). With the if statement it prints this to the console.
folder1 has files
Looking at other questions I think the expression in my if statement is right and I am getting no compilation errors for syntax or another problems.
So I just simply dumbfounded as to why it is not counting the files.
#!/bin/bash
folder=$1
if [ $1 = empty ]; then
folder=empty
counter=0
echo $folder has $counter files
exit
fi
for d in $(ls $folder); do
if [[ -f $d ]]; then
let 'counter++'
fi
done
echo $folder has $counter files
Thank you.
Your entire script could be very well simplified as below with enhancements made. Never use output of ls programmatically. It should be used only in the command-line. The -z construct allows to you assert if the parameter following it is empty or non-empty.
For looping over files, use the default glob expansion provided by the shell. Note the && is a short-hand to do a action when the left-side of the operand returned a true condition, in a way short-hand equivalent of if <condition>; then do <action>; fi
#!/usr/bin/env bash
[ -z "$1" ] && { printf 'invalid argument passed\n' >&2 ; exit 1 ; }
shopt -s nullglob
for file in "$1"/*; do
[ -f "$file" ] && ((count++))
done
printf 'folder %s had %d files\n' "$1" "$count"
I need a shell script that will monitor all the folders given in the command
and will notify the user if a certain file will be created inside them (the name of
the file will be read from keyboard).
I am allowed to use simple commands, so not inotify...
this is what i managed to do so far:
#!/bin/bash
echo "Please enter a file you want to monitor: "
read file_monitor
#this is an infinite while
while [ 1 ] ; do
#using test -e to search for the file
test -e $file_monitor && echo "The file has been created!"
sleep 5
done
I have to find a way to stop the while when the file has been created, and also to search for the file in the folders given in the command line. Can someone help me, please?
To exit the loop, use break:
test -e $file_monitor && echo "The file has been created!" && break
I would prefer to break first, and echo after the loop, or as #mkemp6 suggested, directly use the test as the condition for the loop.
To check the folders, simply loop through them, and check the file in each one.
break [n]
Exit from within a for, while, until, or select loop. If n is specified, break n
levels. n must be ≥ 1. If n is greater than the number of enclosing loops, all
enclosing loops are exited.
while ! test -e "$file_monitor"; do sleep 5; done
But you are much better off using something like inotify to monitor the appropriate directories.
#!/bin/bash
arr=$#
for i in $arr; do
if [ ! -d $i ]
then echo "The parameter $i is no a directory!"
exit 1
fi
done
echo -n "Please give file you want to monitor: "
read file_monitor
a=1
while [ $a -eq 1 ]; do
for i in $arr; do
cd $i
test -e $file_monitor && echo "The file has been created" && a=$((a+1))
cd ..
done
done
So this is what I have managed to do.
I'm writing a script in bash. It will receive from 2 to 5 arguments. For example:
./foo.sh -n -v SomeString Type Directory
-n, -v and Directory are optional.
If script doesn't receive argument Directory it will search in current directory for a string.
Otherwise it will follow received path and search there. If this directory doesn't exist it will send a message.
The question is: Is there a way to check if the last arg is a path or not?
You can get last argument using variable reference:
numArgs=$#
lastArg="${!numArgs}"
# check if last argument is directory
if [[ -d "$lastArg" ]]; then
echo "it is a directory"
else
echo "it is not a directory"
fi
you can use this:
#!/bin/bash
if [[ -d ${!#} ]]
then
echo "DIR EXISTS"
else
echo "dosen't exists"
fi
First, use getopts to parse the options -n and -v (they will have to be used before any non-options, but that's not usually an issue).
while getopts nv opt; do
case $opt in
n) nflag=1 ;;
v) vflag=1 ;;
*) printf >&2 "Unrecognized option $opt\n"; exit 1 ;;
esac
done
shift $((OPTIND-1))
Now, you will have only your two required arguments, and possibly your third optional argument, in $#.
string_arg=$1
type_arg=$2
dir_arg=$3
if [ -d "$dir_arg" ]; then
# Do something with valid directory
fi
Note that this code will work in any POSIX-compliant shell, not just bash.
Trying to make a script that will take a command line argument as a pathname and then test if it is a working directory. Then I wish to run commands (tests) on the directory such as how many files in what sub directories etc.
I am unable to think of a logic to test this with. How would you determine if the argument is a directory?
This is what I have tried
if [ -d "$1" ]
then
echo "It works"
fi
I dont get "It works" when I try this. So I switched it to -a for a file because it is easier to test. And again, I do not get the echo output.
Use the -d option to the test command.
if [ -d "$1" ]
then ...
fi
The title mentions a while loop, but none of the previous commentary seems to mention that. You might try a simple script like:
#!/bin/sh
for arg; do
if test -d "$arg";
echo "$arg is a directory"
else
echo "$arg is not a directory"
fi
done
For variety, you could rewrite that as:
#!/bin/sh
for arg; do
not=$(test -d "$arg" || echo "NOT ")
echo "$arg is ${not}a directory"
done