I've been trying to make a bash script that starts at a folder, namely my home folder, then gets each file recursively and checks for some properties. Say I want to check to see if my files have a certain size and have text (not binary data) in them. It should take care of the special cases where the files are hidden or starting with a hyphen. This is what I came up with:
for i in $(cd "/home/user" && ls -aR);
do
if [[ $(file ./"$i") == "./\"$i\": ASCII text" ]] && [[ $(du -b ./"$i" | grep -oE "[0-9]+") == "1015" ]]; then
echo ./"$i"
fi
done
I don't know how many subfolders there are, and I need it to echo the path of the files that meet the criteria. It works ok for files in /home/user/ but it doesn't seem to find (and thus check) the files in any subfolder. How may I fix this?
I'm going to assume you are using bash 4 unless otherwise stated.
shopt -s globstar
for f in /home/user/**/*:
if [[ $(file -- "$f") != *": ASCII text" ]]; then
continue
fi
# This is the syntax for GNU stat; consult your manual for
# other implementations
size=$(stat -c %s -- "$f")
if (( size != 1015 )); then
continue
fi
echo "$f"
done
I would separate traversing the file tree from checking the individual files.
Start by writing a script which examines a single file and prints the file name to stdout if the file matches your criteria. Let's call this script check_file. Now use, for instance,
find /home/user -type f -exec check_file {} \;
Related
I would want to iterate over contents of a directory and list only ordinary files.
The path of the directory is given as an user input. The script works if the input is current directory but not with others.
I am aware that this can be done using ls.. but i need to use a for .. in control structure.
#!/bin/bash
echo "Enter the path:"
read path
contents=$(ls $path)
for content in $contents
do
if [ -f $content ];
then
echo $content
fi
done
ls is only returning the file names, not including the path. You need to either:
Change your working directory to the path in question, or
Combine the path with the names for your -f test
Option #2 would just change:
if [ -f $content ];
to:
if [ -f "$path/$content" ];
Note that there are other issues here; ls may make changes to the output that break this, depending on wrapping. If you insist on using ls, you can at least make it (somewhat) safer with:
contents="$(command ls -1F "$path")"
You have two ways of doing this properly:
Either loop through the * pattern and test file type:
#!/usr/bin/env bash
echo "Enter the path:"
read -r path
for file in "$path/"*; do
if [ -f "$file" ]; then
echo "$file"
fi
done
Or using find to iterate a null delimited list of file-names:
#!/usr/bin/env bash
echo "Enter the path:"
read -r path
while IFS= read -r -d '' file; do
echo "$file"
done < <(
find "$path" -maxdepth 1 -type f -print0
)
The second way is preferred since it will properly handle files with special characters and offload the file-type check to the find command.
Use file, set to search for files (-type f) from $path directory:
find "$path" -type f
Here is what you could write:
#!/usr/bin/env bash
path=
while [[ ! $path ]]; do
read -p "Enter path: " path
done
for file in "$path"/*; do
[[ -f $file ]] && printf '%s\n' "$file"
done
If you want to traverse all the subdirectories recursively looking for files, you can use globstar:
shopt -s globstar
for file in "$path"/**; do
printf '%s\n' "$file"
done
In case you are looking for specific files based on one or more patterns or some other condition, you could use the find command to pick those files. See this post:
How to loop through file names returned by find?
Related
When to wrap quotes around a shell variable?
Why you shouldn't parse the output of ls
Is double square brackets [[ ]] preferable over single square brackets [ ] in Bash?
#!/bin/bash
for filenames in $( ls $1 )
do
echo $filenames | grep "\.old$"
if [ ! $filenames = 0 ]
then
$( mv "$1/$filenames" "$1/$filenames.old" )
fi
done
So I think most of the script works. It is intended to take the output of ls for a directory inputed in the first parameter, and search for any files with .old at the end. Any files that do not contain .old will then be renamed.
The script successfully renames the files, but it will add .old to a file already containing the extension. I am assuming that the if variable is wrong, but I cannot figure out which variable to use in this case.
Answer is in the key but if anyone needs to do this here is an even easier way:
#!/bin/bash
for filenames in $( ls $1 | grep -v "\.old$" )
do
$( mv "$1/$filenames" "$1/$filenames.old" )
done
Use `find for this
find /directory/here -type f ! -iname "*.old" -exec mv {} {}.old \;
Problems the original approach
for filenames in $( ls $1 ) Never parse ls output. Check [ this ]
Variables are not double quoted, say in if [ ! $filenames = 0 ]. This results in word-splitting. Use "$filenames" unless you expect word splitting.
So the final script would be
#!/bin/bash
if [ -d "$1" ]
then
find "$1" -type f ! -iname "*.old" -exec mv {} {}.old \;
# use -maxdepth 1 with find if you don't wish to recursively check subdirectories
else
echo "Directory : $1 doesn't exist !"
fi
Usage
./script '/path/to/directory'
Don't use ls in scripts.
#!/bin/bash
for filename in "$1"/*
do
case $filename in *.old) continue;; esac
mv "$filename" "$filename.old"
done
I prefer case over if because it supports wildcard matching naturally and portably. (You could run this with /bin/sh just as well.) If you wanted to use if instead, that'd be
if echo "$filename" | grep -q '\.old$'; then
or more idiomatically, but recent shells only,
if [[ "$filename" == *.old ]]; then
You want to avoid calling additional utility functions if simple shell builtins will do. Why? Each additional utility you call grep, etc. spawns and runs in a separate subshell of its own. (if you are spawning a subshell for every iteration in your loop -- things will really slow down) If the shell doesn't provide a feature, then sure... calling a utility is the right thing to do.
As mentioned above, shell globbing along with parameter expansion with substring removal provides a simple test for determining if a file has an .old extension. All you need is:
for i in "$1"/*; do
[ "${i##*.}" = "old" ] || mv "$i" "${i}.old"
done
(note: this will skip add the .old extension to single file named 'old', but that can be handled separately if needed -- unlikely. Additionally, the solution with find is a fine approach as well)
I solved the problem, as I was misled by my instructor!
$? is the variable which represents the pipeline output which is currently in the forground (which would be grep). The new code is unedited except for
if [ ! $? = 0 ]
Currently learning some bash scripting and having an issue with a question involving listing all files in a given directory and stating if they are a file or directory. The issue I am having is that I only get either my current directory or if a specify a directory it will just say that it is a directory eg. /home/user/shell_scripts will return shell_scipts is a directory rather than the files contained within it.
This is what I have so far:
dir=$dir
for file in $dir; do
if [[ -d $file ]]; then
echo "$file is a directory"
if [[ -f $file ]]; then
echo "$file is a regular file"
fi
done
Your line:
for file in $dir; do
will expand $dir just to a single directory string. What you need to do is expand that to a list of files in the directory. You could do this using the following:
for file in "${dir}/"* ; do
This will expand the "${dir}/"* section into a name-only list of the current directory. As Biffen points out, this should guarantee that the file list wont end up with split partial file names in file if any of them contain whitespace.
If you want to recurse into the directories in dir then using find might be a better approach. Simply use:
for file in $( find ${dir} ); do
Note that while simple, this will not handle files or directories with spaces in them. Because of this, I would be tempted to drop the loop and generate the output in one go. This might be slightly different than what you want, but is likely to be easier to read and a lot more efficient, especially with large numbers of files. For example, To list all the directories:
find ${dir} -maxdepth 1 -type d
and to list the files:
find ${dir} -maxdepth 1 -type f
if you want to iterate into directories below, then remove the -maxdepth 1
This is a good use for globbing:
for file in "$dir/"*
do
[[ -d "$file" ]] && echo "$file is a directory"
[[ -f "$file" ]] && echo "$file is a regular file"
done
This will work even if files in $dir have special characters in their names, such as spaces, asterisks and even newlines.
Also note that variables should be quoted ("$file"). But * must not be quoted. And I removed dir=$dir since it doesn't do anything (except break when $dir contains special characters).
ls -F ~ | \
sed 's#.*/$#/& is a Directory#;t quit;s#.*#/& is a File#;:quit;s/[*/=>#|] / /'
The -F "classify" switch appends a "/" if a file is a directory. The sed code prints the desired message, then removes the suffix.
for file in $(ls $dir)
do
[ -f $file ] && echo "$file is File"
[ -d $file ] && echo "$file is Directory"
done
or replace the
$(ls $dir)
with
`ls $`
If you want to list files that also start with . use:
for file in "${dir}/"* "${dir}/"/.[!.]* "${dir}/"/..?* ; do
Could someone please help with this script. I need to use grep to loop to through the filenames that need to be changed.
#!/bin/bash
file=
for file in $(ls $1)
do
grep "^.old" | mv "$1/$file" "$1/$file.old"
done
bash can handle regular expressions without using grep.
for f in "$1"/*; do
[[ $f =~ \.old ]] && continue
# Or a pattern instead
# [[ $f == *.old* ]] && continue
mv "$f" "$f.old"
done
You can also move the name checking into the pattern itself:
shopt -s extglob
for f in "$1/"!(*.old*); do
mv "$f" "$f.old"
done
If I understand your question correctly, you want to make rename a file (i.e. dir/file.txt ==> dir/file.old) only if the file has not been renamed before. The solution is as follow.
#!/bin/bash
for file in "$1/"*
do
backup_file="${file%.*}.old"
if [ ! -e "$backup_file" ]
then
echo mv "$file" "$backup_file"
fi
done
Discussion
The script currently does not actual make back up, it only displays the action. Run the script once and examine the output. If this is what you want, then remove the echo from the script and run it again.
Update
Here is the no if solution:
ls "$1/"* | grep -v ".old" | while read file
do
echo mv "$file" "${file}.old"
done
Discussion
The ls command displays all files.
The grep command filter out those files that has the .old extension so they won't be displayed.
The while loop reads the file names that do not have the .old extension, one by one and rename them.
I'm trying to do a script which lists files on a directory and then searchs one by one every file in other directory. For dealing with spaces and special characters like "[" or "]" I'm using $(printf %q "$FILENAME") as input for the find command: find /directory/to/search -type f -name $(printf %q "$FILENAME").
It works like a charm for every filename except in one case: when there's multibyte characters (UTF-8). In that case the output of printf is an external quoted string, i.e.: $'file name with blank spaces and quoted characters in the form of \NNN\NNN', and that string is not being expanded without the $'' quoting, so find searchs for a file with a name including that quote: «$'filename'».
Is there an alternative solution in order to be able to pass to find any kind of filename?
My script is like follows (I know some lines can be deleted, like the "RESNAME="):
#!/bin/bash
if [ -d $1 ] && [ -d $2 ]; then
IFSS=$IFS
IFS=$'\n'
FILES=$(find $1 -type f )
for FILE in $FILES; do
BASEFILE=$(printf '%q' "$(basename "$FILE")")
RES=$(find $2 -type f -name "$BASEFILE" -print )
if [ ${#RES} -gt 1 ]; then
RESNAME=$(printf '%q' "$(basename "$RES")")
else
RESNAME=
fi
if [ "$RESNAME" != "$BASEFILE" ]; then
echo "FILE NOT FOUND: $FILE"
fi
done
else
echo "Directories do not exist"
fi
IFS=$IFSS
As an answer said, I've used associative arrays, but with no luck, maybe I'm not using correctly the arrays, but echoing it (array[#]) returns nothing. This is the script I've written:
#!/bin/bash
if [ -d "$1" ] && [ -d "$2" ]; then
declare -A files
find "$2" -type f -print0 | while read -r -d $'\0' FILE;
do
BN2="$(basename "$FILE")"
files["$BN2"]="$BN2"
done
echo "${files[#]}"
find "$1" -type f -print0 | while read -r -d $'\0' FILE;
do
BN1="$(basename "$FILE")"
if [ "${files["$BN1"]}" != "$BN1" ]; then
echo "File not found: "$BN1""
fi
done
fi
Don't use for loops. First, it is slower. Your find has to complete before the rest of your program can run. Second, it is possible to overload the command line. The enter for command must fit in the command line buffer.
Most importantly of all, for sucks at handling funky file names. You're running conniptions trying to get around this. However:
find $1 -type f -print0 | while read -r -d $'\0' FILE
will work much better. It handles file names -- even file names that contain \n characters. The -print0 tells find to separate file names with the NUL character. The while read -r -d $'\0 FILE will read each file name (separate by the NUL character) into $FILE.
If you put quotes around the file name in the find command, you don't have to worry about special characters in the file names.
Your script is running find once for each file found. If you have 100 files in your first directory, you're running find 100 times.
Do you know about associative (hash) arrays in BASH? You are probably better off using associative arrays. Run find on the first directory, and store those files names in an associative array.
Then, run find (again using the find | while read syntax) for your second directory. For each file you find in the second directory, see if you have a matching entry in your associative array. If you do, you know that file is in both arrays.
Addendum
I've been looking at the find command. It appears there's no real way to prevent it from using pattern matching except through a lot of work (like you were doing with printf. I've tried using the -regex matching and using \Q and \E to remove the special meaning of pattern characters. I haven't been successful.
There comes a time that you need something a bit more powerful and flexible than shell to implement your script, and I believe this is the time.
Perl, Python, and Ruby are three fairly ubiquitous scripting languages found on almost all Unix systems and are available on other non-POSIX platforms (cough! ...Windows!... cough!).
Below is a Perl script that takes two directories, and searches them for matching files. It uses the find command once and uses associative arrays (called hashes in Perl). I key the hash to the name of my file. In the value portion of the hash, I store an array of the directories where I found this file.
I only need to run the find command once per directory. Once that is done, I can print out all the entries in the hash that contain more than one directory.
I know it's not shell, but this is one of the cases where you can spend a lot more time trying to figure out how to get shell to do what you want than its worth.
#! /usr/bin/env perl
use strict;
use warnings;
use feature qw(say);
use File::Find;
use constant DIRECTORIES => qw( dir1 dir2 );
my %files;
#
# Perl version of the find command. You give it a list of
# directories and a subroutine for filtering what you find.
# I am basically rejecting all non-file entires, then pushing
# them into my %files hash as an array.
#
find (
sub {
return unless -f;
$files{$_} = [] if not exists $files{$_};
push #{ $files{$_} }, $File::Find::dir;
}, DIRECTORIES
);
#
# All files are found and in %files hash. I can then go
# through all the entries in my hash, and look for ones
# with more than one directory in the array reference.
# IF there is more than one, the file is located in multiple
# directories, and I print them.
#
for my $file ( sort keys %files ) {
if ( #{ $files{$file} } > 1 ) {
say "File: $file: " . join ", ", #{ $files{$file} };
}
}
Try something like this:
find "$DIR1" -printf "%f\0" | xargs -0 -i find "$DIR2" -name \{\}
How about this one-liner?
find dir1 -type f -exec bash -c 'read < <(find dir2 -name "${1##*/}" -type f)' _ {} \; -printf "File %f is in dir2\n" -o -printf "File %f is not in dir2\n"
Absolutely 100% safe regarding files with funny symbols, newlines and spaces in their name.
How does it work?
find (the main one) will scan through directory dir1 and for each file (-type f) will execute
read < <(find dir2 -name "${1##*/} -type f")
with argument the name of the current file given by the main find. This argument is at position $1. The ${1##*/} removes everything before the last / so that if $1 is path/to/found/file the find statement is:
find dir2 -name "file" -type f
This outputs something if file is found, otherwise has no output. That's what is read by the read bash command. read's exit status is true if it was able to read something, and false if there wasn't anything read (i.e., in case nothing is found). This exit status becomes bash's exit status which becomes -exec's status. If true, the next -printf statement is executed, and if false, the -o -printf part will be executed.
If your dirs are given in variables $dir1 and $dir2 do this, so as to be safe regarding spaces and funny symbols that could occur in $dir2:
find "$dir1" -type f -exec bash -c 'read < <(find "$0" -name "${1##*/}" -type f)' "$dir2" {} \; -printf "File %f is in $dir2\n" -o -printf "File %f is not in $dir2\n"
Regarding efficiency: this is of course not an efficient method at all! the inner find will be executed as many times as there are found files in dir1. This is terrible, especially if the directory tree under dir2 is deep and has many branches (you can rely a little bit on caching, but there are limits!).
Regarding usability: you have fine-grained control on how both find's work and on the output, and it's very easy to add many more tests.
So, hey, tell me how to compare files from two directories? Well, if you agree on loosing a little bit of control, this will be the shortest and most efficient answer:
diff dir1 dir2
Try it, you'll be amazed!
Since you are only using find for its recursive directory following, it will be easier to simply use the globstar option in bash. (You're using associative arrays, so your bash is new enough).
#!/bin/bash
shopt -s globstar
declare -A files
if [[ -d $1 && -d $2 ]]; then
for f in "$2"/**/*; do
[[ -f "$f" ]] || continue
BN2=$(basename "$f")
files["$BN2"]=$BN2
done
echo "${files[#]}"
for f in "$1"/**/*; do
[[ -f "$f" ]] || continue
BN1=$(basename $f)
if [[ ${files[$BN1]} != $BN1 ]]; then
echo "File not found: $BN1"
fi
done
fi
** will match zero or more directories, so $1/**/* will match all the files and directories in $1, all the files and directories in those directories, and so forth all the way down the tree.
If you want to use associative arrays, here's one possibility that will work well with files with all sorts of funny symbols in their names (this script has too much to just show the point, but it is usable as is – just remove the parts you don't want and adapt to your needs):
#!/bin/bash
die() {
printf "%s\n" "$#"
exit 1
}
[[ -n $1 ]] || die "Must give two arguments (none found)"
[[ -n $2 ]] || die "Must give two arguments (only one given)"
dir1=$1
dir2=$2
[[ -d $dir1 ]] || die "$dir1 is not a directory"
[[ -d $dir2 ]] || die "$dir2 is not a directory"
declare -A dir1files
declare -A dir2files
while IFS=$'\0' read -r -d '' file; do
dir1files[${file##*/}]=1
done < <(find "$dir1" -type f -print0)
while IFS=$'\0' read -r -d '' file; do
dir2files[${file##*/}]=1
done < <(find "$dir2" -type f -print0)
# Which files in dir1 are in dir2?
for i in "${!dir1files[#]}"; do
if [[ -n ${dir2files[$i]} ]]; then
printf "File %s is both in %s and in %s\n" "$i" "$dir1" "$dir2"
# Remove it from dir2 has
unset dir2files["$i"]
else
printf "File %s is in %s but not in %s\n" "$i" "$dir1" "$dir2"
fi
done
# Which files in dir2 are not in dir1?
# Since I unset them from dir2files hash table, the only keys remaining
# correspond to files in dir2 but not in dir1
if [[ -n "${!dir2files[#]}" ]]; then
printf "File %s is in %s but not in %s\n" "$dir2" "$dir1" "${!dir2files[#]}"
fi
Remark. The identification of files is only based on their filenames, not their contents.