I'm trying to create a really simple bash script, which will list the first few lines of every file in a specific directory. The directory should be specified by the argument.
I think that the Grep command should be used, but I have really no idea how.
My existing script does not seem to work at all, so it's no use putting it in here.
Use head command:
head -3 /path/to/dir/*
For any answer using head and *, redirect stderr to /dev/null unless you want to see errors like:
head: error reading ‘tmp’: Is a directory
for file in dir/*; do
echo "-- $file --"
head "$file"
echo
done
If you want the first few lines of all files ending in .txt, try
head *.txt
or
head --lines=3 *.txt
Because bash does filename expansion (globbing) by default, you can just let your shell expand input and let head do the rest:
head *
The * wildcard expands to all the filenames in the working directory. On zsh you can see this nicely, when it autocompletes your commandline when you press tab.
You can change the amount of lines with the -n argument to head.
If you want to do this recursively:
find . \! -type d -exec head '{}' +
Related
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$f" "$(echo "$f" | sed s/IMG/VACATION/)"; done
In this example, I am assuming that all your image files contain the string IMG and you want to replace IMG with VACATION.
The shell automatically evaluates *.jpg to all the matching files.
The second argument of mv (the new name of the file) is the output of the sed command that replaces IMG with VACATION.
If your filenames include whitespace pay careful attention to the "$f" notation. You need the double-quotes to preserve the whitespace.
You can use rename utility to rename multiple files by a pattern. For example following command will prepend string MyVacation2011_ to all the files with jpg extension.
rename 's/^/MyVacation2011_/g' *.jpg
or
rename <pattern> <replacement> <file-list>
this example, I am assuming that all your image files begin with "IMG" and you want to replace "IMG" with "VACATION"
solution : first identified all jpg files and then replace keyword
find . -name '*jpg' -exec bash -c 'echo mv $0 ${0/IMG/VACATION}' {} \;
for file in *.jpg ; do mv $file ${file//IMG/myVacation} ; done
Again assuming that all your image files have the string "IMG" and you want to replace "IMG" with "myVacation".
With bash you can directly convert the string with parameter expansion.
Example: if the file is IMG_327.jpg, the mv command will be executed as if you do mv IMG_327.jpg myVacation_327.jpg. And this will be done for each file found in the directory matching *.jpg.
IMG_001.jpg -> myVacation_001.jpg
IMG_002.jpg -> myVacation_002.jpg
IMG_1023.jpg -> myVacation_1023.jpg
etcetera...
find . -type f |
sed -n "s/\(.*\)factory\.py$/& \1service\.py/p" |
xargs -p -n 2 mv
eg will rename all files in the cwd with names ending in "factory.py" to be replaced with names ending in "service.py"
explanation:
In the sed cmd, the -n flag will suppress normal behavior of echoing input to output after the s/// command is applied, and the p option on s/// will force writing to output if a substitution is made. Since a sub will only be made on match, sed will only have output for files ending in "factory.py"
In the s/// replacement string, we use "& " to interpolate the entire matching string, followed by a space character, into the replacement. Because of this, it's vital that our RE matches the entire filename. after the space char, we use "\1service.py" to interpolate the string we gulped before "factory.py", followed by "service.py", replacing it. So for more complex transformations youll have to change the args to s/// (with an re still matching the entire filename)
Example output:
foo_factory.py foo_service.py
bar_factory.py bar_service.py
We use xargs with -n 2 to consume the output of sed 2 delimited strings at a time, passing these to mv (i also put the -p option in there so you can feel safe when running this). voila.
NOTE: If you are facing more complicated file and folder scenarios, this post explains find (and some alternatives) in greater detail.
Another option is:
for i in *001.jpg
do
echo "mv $i yourstring${i#*001.jpg}"
done
remove echo after you have it right.
Parameter substitution with # will keep only the last part, so you can change its name.
Can't comment on Susam Pal's answer but if you're dealing with spaces, I'd surround with quotes:
for f in *.jpg; do mv "$f" "`echo $f | sed s/\ /\-/g`"; done;
You can try this:
for file in *.jpg;
do
mv $file $somestring_${file:((-7))}
done
You can see "parameter expansion" in man bash to understand the above better.
I have a folder with lots of files which name has the following structure:
01.artist_name - song_name.mp3
I want to go through all of them and rename them using the regexp:
/^d+\./
so i get only :
artist_name - song_name.mp3
How can i do this in bash?
You can do this in BASH:
for f in [0-9]*.mp3; do
mv "$f" "${f#*.}"
done
Use the Perl rename utility utility. It might be installed on your version of Linux or easy to find.
rename 's/^\d+\.//' -n *.mp3
With the -n flag, it will be a dry run, printing what would be renamed, without actually renaming. If the output looks good, drop the -n flag.
Use 'sed' bash command to do so:
for f in *.mp3;
do
new_name="$(echo $f | sed 's/[^.]*.//')"
mv $f $new_name
done
...in this case, regular expression [^.].* matches everything before first period of a string.
This might be a very simple thing for a shell scripting programmer but am pretty new to it. I was trying to execute the below command in a shell script and save the output into a variable
inputfile=$(ls -ltr *.{PDF,pdf} | head -1 | awk '{print $9}')
The command works fine when I fire it from terminal but fails when executed through a shell script (sh). Why is that the command fails, does it mean that shell script doesn't support the command or am I doing it wrong? Also how do I know if a command will work in shell or not?
Just to give you a glimpse of my requirement, I was trying to get the oldest file from a particular directory (I also want to make sure upper case and lower case extensions are handled). Is there any other way to do this ?
The above command will work correctly only if BOTH *.pdf and *.PDF files are in the directory you are currently.
If you would like to execute it in a directory with only one of those you should consider using e.g.:
inputfiles=$(find . -maxdepth 1 -type f \( -name "*.pdf" -or -name "*.PDF" \) | xargs ls -1tr | head -1 )
NOTE: The above command doesn't work with files with new lines, or with long list of found files.
Parsing ls is always a bad idea. You need another strategy.
How about you make a function that gives you the oldest file among the ones given as argument? the following works in Bash (adapt to your needs):
get_oldest_file() {
# get oldest file among files given as parameters
# return is in variable get_oldest_file_ret
local oldest f
for f do
[[ -e $f ]] && [[ ! $oldest || $f -ot $oldest ]] && oldest=$f
done
get_oldest_file_ret=$oldest
}
Then just call as:
get_oldest_file *.{PDF,pdf}
echo "oldest file is: $get_oldest_file_ret"
Now, you probably don't want to use brace expansions like this at all. In fact, you very likely want to use the shell options nocaseglob and nullglob:
shopt -s nocaseglob nullglob
get_oldest_file *.pdf
echo "oldest file is: $get_oldest_file_ret"
If you're using a POSIX shell, it's going to be a bit trickier to have the equivalent of nullglob and nocaseglob.
Is perl an option? It's ubiquitous on Unix.
I would suggest:
perl -e 'print ((sort { -M $b <=> -M $a } glob ( "*.{pdf,PDF}" ))[0]);';
Which:
uses glob to fetch all files matching the pattern.
sort, using -M which is relative modification time. (in days).
fetches the first element ([0]) off the sort.
Prints that.
As #gniourf_gniourf says, parsing ls is a bad idea. Such as leaving unquoted globs, and generally not counting for funny characters in file names.
find is your friend:
#!/bin/sh
get_oldest_pdf() {
#
# echo path of oldest *.pdf (case-insensitive) file in current directory
#
find . -maxdepth 1 -mindepth 1 -iname "*.pdf" -printf '%T# %p\n' \
| sort -n \
| tail -1 \
| cut -d\ -f1-
}
whatever=$(get_oldest_pdf)
Notes:
find has numerous ways of formatting the output, including
things like access time and/or write time. I used '%T# %p\n',
where %T# is last write time in UNIX time format incl.fractal part.
This will never containt space so it's safe to use as separator.
Numeric sort and tail get the last item, sorting by the time,
cut removes the time from the output.
I used IMO much easier to read/maintain pipe notation, with help of \.
the code should run on any POSIX shell,
You could easily adjust the function to parametrize the pattern,
time used (access/write), control the search depth or starting dir.
I'm pretty sure I've seen this done before but I can't remember the exact syntax.
Suppose you have a couple of files with different file extensions:
foo.txt
bar.rtf
index.html
and instead of doing something with all of them (cat *), you only want to run a command on 2 of the 3 file extensions.
Can't you do something like this?
cat ${*.txt|*.rtf}
I'm sure there's some find trickery to identify the files first and pipe them to a command, but I think bash supports what I'm talking about without having to do that.
The syntax you want is cat *.{txt,rft}. A comma is used instead of a pipe.
$ echo foo > foo.txt
$ echo bar > bar.rft
$ echo "bar txt" > bar.txt
$ echo "test" > index.html
$ cat *.{txt,rft}
bar txt
foo
bar
$ ls *.{txt,rft}
bar.rft bar.txt foo.txt
But as Anthony Geoghegan said in their answer there's a simpler approach you can use.
Shell globbing is much more basic than regular expressions. If you want to cat all the files which have a .txt or .rtf suffix, you'd simply use:
cat *.txt *.rtf
The glob patterns will be expanded to list all the filenames that match the pattern. In your case, the above command would call the cat command with foo.txt and bar.rtf as its arguments.
Here's a simple way i use to do it using command substitution.
cat $(find . -type f \( -name "*.txt" -o -name "*.rtf" \))
But Anthony Geoghegan's answer is much simpler. I learned from it too.
I just downloaded about 600 files from my server and need to remove the last 11 characters from the filename (not including the extension). I use Ubuntu and I am searching for a command to achieve this.
Some examples are as follows:
aarondyne_kh2_13thstruggle_or_1250556383.mus should be renamed to aarondyne_kh2_13thstruggle_or.mus
aarondyne_kh2_darknessofunknow_1250556659.mp3 should be renamed to aarondyne_kh2_darknessofunknow.mp3
It seems that some duplicates might exist after I do this, but if the command fails to complete and tells me what the duplicates would be, I can always remove those manually.
Try using the rename command. It allows you to rename files based on a regular expression:
The following line should work out for you:
rename 's/_\d+(\.[a-z0-9A-Z]+)$/$1/' *
The following changes will occur:
aarondyne_kh2_13thstruggle_or_1250556383.mus renamed as aarondyne_kh2_13thstruggle_or.mus
aarondyne_kh2_darknessofunknow_1250556659.mp3 renamed as aarondyne_kh2_darknessofunknow.mp3
You can check the actions rename will do via specifying the -n flag, like this:
rename -n 's/_\d+(\.[a-z0-9A-Z]+)$/$1/' *
For more information on how to use rename simply open the manpage via: man rename
Not the prettiest, but very simple:
echo "$filename" | sed -e 's!\(.*\)...........\(\.[^.]*\)!\1\2!'
You'll still need to write the rest of the script, but it's pretty simple.
find . -type f -exec sh -c 'mv {} `echo -n {} | sed -E -e "s/[^/]{10}(\\.[^\\.]+)?$/\\1/"`' ";"
one way to go:
you get a list of your files, one per line (by ls maybe) then:
ls....|awk '{o=$0;sub(/_[^_.]*\./,".",$0);print "mv "o" "$0}'
this will print the mv a b command
e.g.
kent$ echo "aarondyne_kh2_13thstruggle_or_1250556383.mus"|awk '{o=$0;sub(/_[^_.]*\./,".",$0);print "mv "o" "$0}'
mv aarondyne_kh2_13thstruggle_or_1250556383.mus aarondyne_kh2_13thstruggle_or.mus
to execute, just pipe it to |sh
I assume there is no space in your filename.
This script assumes each file has just one extension. It would, for instance, rename "foo.something.mus" to "foo.mus". To keep all extensions, remove one hash mark (#) from the first line of the loop body. It also assumes that the base of each filename has at least 12 character, so that removing 11 doesn't leave you with an empty name.
for f in *; do
ext=${f##*.}
new_f=${base%???????????.$ext}
if [ -f "$new_f" ]; then
echo "Will not rename $f, $new_f already exists" >&2
else
mv "$f" "$new_f"
fi
done