Terminal Command to write mp3 info to text file - unix

I'm hoping to run a command that will get the info from mp3's in a directory and write each MP3's info to a text file with the same name.
Using id3lib to get the MP3 info. This is what I have so far but not working as expected.
This gives me a text file named "{}.txt" with the info
find . -name \*.mp3 -exec info=$(id3info {}); name=({}); echo $info >> {}.txt
Working toward something like this
find . -name \*.mp3 -exec info=$(id3info {}) : filename=$({}); echo $info >> $filename.text : restart ;\
Thanks for any help.

You'd need to execute a shell command as an argument to -exec option. The following might work for you:
find . -name "*.mp3" -exec sh -c "id3info {} > {}.txt" \;
If you also need the mp3 filename to be added to the output, say:
find . -name "*.mp3" -exec sh -c "{ echo {} ; id3info {} ; } > {}.txt" \;

Related

bash find with two commands in an exec ~ How to find a specific Java class within a set of JARs

My use case is I want to search a collection of JARs for a specific class file. More specifically, I want to search recursively within a directory for all *.jar files, then list their contents, looking for a specific class file.
So this is what I have so far:
find . -name *.jar -type f -exec echo {} \; -exec jar tf {} \;
This will list the contents of all JAR files found recursively. I want to put a grep within the seconed exec because I want the second exec to only print the contents of the JAR that grep matches.
If I just put a pipe and pipe it all to grep afterward, like:
find . -name *.jar -type f -exec echo {} \; -exec jar tf {} \; | grep $CLASSNAME
Then I lose the output of the first exec, which tells me where the class file is (the name of JAR file is likely to not match the class file name).
So if there was a way for the exec to run two commands, like:
-exec "jar tf {} | grep $CLASSNAME" \;
Then this would work. Using a grep $(...) in the exec command wouldn't work because I need the {} from the find to take the place of the file that was found.
Is this possible?
(Also I am open to other ways of doing this, but the command line is preferred.)
i find it difficult to execute multiple commands within find-exec, so i usually only grab the results with find and loop around the results.
maybe something like this might help?
find . -type f -name *.jar | while read jarfile; do echo $jarfile; jar tf $jarfile; done
I figured it out - still using "one" command. What I was looking for was actually answered in the question How to use pipe within -exec in find. What I have to do is use a shell command with my exec. This ends up making the command look like:
find . -name *.jar -type f -exec echo {} \; -exec sh -c "jar tf {} | grep --color $CLASSNAME" \;
The --color will help the final result to stick out while the command is recursively listing all JAR files.
A couple points:
This assumes I have a $CLASSNAME set. The class name has to appear as it would in a JAR, not within a Java package. So com.ibm.json.java.JSONObject would become com/ibm/json/java/JSONObject.class.
This requires a JDK - that is where we get the jar command. The JDK must be accessible on the system path. If you have a JDK that is not on the system path, you can set an environment variable, such as JAR to point to the jar executable. I am running this from cygwin, so it turns out my jar installation is within the "Program Files" directory. The presence of a space breaks this, so I have to add these two commands:
export JAR=/cygdrive/c/Program\ Files/Java/jdk1.8.0_65/bin/jar
find . -name *.jar -type f -exec echo {} \; -exec sh -c "\"$JAR\" tf {} | grep --color $CLASSNAME" \;
The $JAR in the shell command must be escaped otherwise the terminal will not know what to do with the space in "Program Files".

Recursively remove filename suffix from files in shell

When we develop locally, we append ".dev" or ".prod" to files that should be made available only to the development/production server respectively.
What I would like to do is; after deploying the site to the server, recursively find all files with the ".dev" suffix (for example) and remove it (renaming the file). How would I go about doing this, preferably entirely in the shell (without scripts) so I can add it to our deployment script?
Our servers run Ubuntu 10.04.
Try this (not entirely shell-only, requires the find and mv utilities):
find . '(' -name '*.dev' -o -name '*.prod' ')' -type f -execdir sh -c 'mv -- "$0" "${0%.*}"' '{}' ';'
If you have the rename and xargs utilities, you can speed this up a lot:
find . '(' -name '*.dev' -o -name '*.prod' ')' -type f -print0 | xargs -0 rename 's/\.(dev|prod)$//'
Both versions should work with any file name, including file names containing newlines.
It's totally untested, but this should work in the POSIX-like shell of your choice:
remove-suffix () {
local filename
while read filename; do
mv "$filename" "$(printf %s "$filename" | sed "s/\\.$1\$//")"
done
}
find -name '*.dev' | remove-suffix .dev
Note: In the very unusual case that one or more of your filenames contains a newline character, this won't work.
for file in `ls *.dev`; do echo "Old Name $file"; new_name=`echo $file | sed -e 's/dev//'` ; echo "New Name $new_name"; mv $file $new_name; done
In an example of something I used recently this code looks for any file that ends with new.xml changes a date in the filename (filenames were of the form xmlEventLog_2010-03-23T11:16:16_PFM_1_1.xml), removes the _new from the name and renames the filename to the new name :
for file in `ls *new.xml`; do echo "Old Name $file"; new_name=`echo $file | sed -e 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}/2010-03-23/g' | sed 's/_new//g'` ; echo "New Name $new_name"; mv $file $new_name; done
Is this the type of thing you wanted?
find /fullpath -type f -name "*.dev"|sed 's|\(.*\)\(\.pdf\)|mv & \1.sometag|' | sh

Unix - how to source multiple shell scripts in a directory?

when I want to execute some shell script in Unix (and let's say that I am in the directory where the script is), I just type:
./someShellScript.sh
and when I want to "source" it (e.g. run it in the current shell, NOT in a new shell), I just type the same command just with the "." (or with the "source" command equivalent) before it:
. ./someShellScript.sh
And now the tricky part. When I want to execute "multiple" shell scripts (let's say all the files with .sh suffix) in the current directory, I type:
find . -type f -name *.sh -exec {} \;
but "what command should I use to "SOURCE" multiple shell scripts in a directory"?
I tried this so far but it DIDN'T work:
find . -type f -name *.sh -exec . {} \;
and it only threw this error:
find: `.': Permission denied
Thanks.
for file in *.sh
do . $file
done
Try the following version of Jonathan's code:
export IFSbak = $IFS;export IFS="\0"
for file in `find . -type f -name '*.sh' -print0`
do source "$file"
end
export IFS=$IFSbak
The problem lies in the way shell's work, and that '.' itself is not a command (neither is source in this). When you run find, the shell will fork itself and then turn into the find process, meaning that any environment variables or other environmental stuff goes into the find process (or, more likely, find forks itself for new processes for each exec).
Also, note that your command (and Jonathan's) will not work if there are spaces in the file names.
You can use find and xargs to do this:
find . -type f -name "*.sh" | xargs -I sh {}

How do you recursively unzip archives in a directory and its subdirectories from the Unix command-line?

The unzip command doesn't have an option for recursively unzipping archives.
If I have the following directory structure and archives:
/Mother/Loving.zip
/Scurvy/Sea Dogs.zip
/Scurvy/Cures/Limes.zip
And I want to unzip all of the archives into directories with the same name as each archive:
/Mother/Loving/1.txt
/Mother/Loving.zip
/Scurvy/Sea Dogs/2.txt
/Scurvy/Sea Dogs.zip
/Scurvy/Cures/Limes/3.txt
/Scurvy/Cures/Limes.zip
What command or commands would I issue?
It's important that this doesn't choke on filenames that have spaces in them.
If you want to extract the files to the respective folder you can try this
find . -name "*.zip" | while read filename; do unzip -o -d "`dirname "$filename"`" "$filename"; done;
A multi-processed version for systems that can handle high I/O:
find . -name "*.zip" | xargs -P 5 -I fileName sh -c 'unzip -o -d "$(dirname "fileName")/$(basename -s .zip "fileName")" "fileName"'
A solution that correctly handles all file names (including newlines) and extracts into a directory that is at the same location as the file, just with the extension removed:
find . -iname '*.zip' -exec sh -c 'unzip -o -d "${0%.*}" "$0"' '{}' ';'
Note that you can easily make it handle more file types (such as .jar) by adding them using -o, e.g.:
find . '(' -iname '*.zip' -o -iname '*.jar' ')' -exec ...
Here's one solution that extracts all zip files to the working directory and involves the find command and a while loop:
find . -name "*.zip" | while read filename; do unzip -o -d "`basename -s .zip "$filename"`" "$filename"; done;
You could use find along with the -exec flag in a single command line to do the job
find . -name "*.zip" -exec unzip {} \;
This works perfectly as we want:
Unzip files:
find . -name "*.zip" | xargs -P 5 -I FILENAME sh -c 'unzip -o -d "$(dirname "FILENAME")" "FILENAME"'
Above command does not create duplicate directories.
Remove all zip files:
find . -depth -name '*.zip' -exec rm {} \;
Something like gunzip using the -r flag?....
Travel the directory structure recursively. If any of the file names specified on the command line are directories, gzip will descend into the directory and compress all the files it finds there (or decompress them in the case of gunzip ).
http://www.computerhope.com/unix/gzip.htm
If you're using cygwin, the syntax is slightly different for the basename command.
find . -name "*.zip" | while read filename; do unzip -o -d "`basename "$filename" .zip`" "$filename"; done;
I realise this is very old, but it was among the first hits on Google when I was looking for a solution to something similar, so I'll post what I did here. My scenario is slightly different as I basically just wanted to fully explode a jar, along with all jars contained within it, so I wrote the following bash functions:
function explode {
local target="$1"
echo "Exploding $target."
if [ -f "$target" ] ; then
explodeFile "$target"
elif [ -d "$target" ] ; then
while [ "$(find "$target" -type f -regextype posix-egrep -iregex ".*\.(zip|jar|ear|war|sar)")" != "" ] ; do
find "$target" -type f -regextype posix-egrep -iregex ".*\.(zip|jar|ear|war|sar)" -exec bash -c 'source "<file-where-this-function-is-stored>" ; explode "{}"' \;
done
else
echo "Could not find $target."
fi
}
function explodeFile {
local target="$1"
echo "Exploding file $target."
mv "$target" "$target.tmp"
unzip -q "$target.tmp" -d "$target"
rm "$target.tmp"
}
Note the <file-where-this-function-is-stored> which is needed if you're storing this in a file that is not read for a non-interactive shell as I happened to be. If you're storing the functions in a file loaded on non-interactive shells (e.g., .bashrc I believe) you can drop the whole source statement. Hopefully this will help someone.
A little warning - explodeFile also deletes the ziped file, you can of course change that by commenting out the last line.
Another interesting solution would be:
DESTINY=[Give the output that you intend]
# Don't forget to change from .ZIP to .zip.
# In my case the files were in .ZIP.
# The echo were for debug purpose.
find . -name "*.ZIP" | while read filename; do
ADDRESS=$filename
#echo "Address: $ADDRESS"
BASENAME=`basename $filename .ZIP`
#echo "Basename: $BASENAME"
unzip -d "$DESTINY$BASENAME" "$ADDRESS";
done;
You can also loop through each zip file creating each folder and unzip the zip file.
for zipfile in *.zip; do
mkdir "${zipfile%.*}"
unzip "$zipfile" -d "${zipfile%.*}"
done
this works for me
def unzip(zip_file, path_to_extract):
"""
Decompress zip archives recursively
Args:
zip_file: name of zip archive
path_to_extract: folder where the files will be extracted
"""
try:
if is_zipfile(zip_file):
parent_file = ZipFile(zip_file)
parent_file.extractall(path_to_extract)
for file_inside in parent_file.namelist():
if is_zipfile(os.path.join(os.getcwd(),file_inside)):
unzip(file_inside,path_to_extract)
os.remove(f"{zip_file}")
except Exception as e:
print(e)

How do I get the find command to print out the file size with the file name?

If I issue the find command as follows:
find . -name *.ear
It prints out:
./dir1/dir2/earFile1.ear
./dir1/dir2/earFile2.ear
./dir1/dir3/earFile1.ear
I want to 'print' the name and the size to the command line:
./dir1/dir2/earFile1.ear 5000 KB
./dir1/dir2/earFile2.ear 5400 KB
./dir1/dir3/earFile1.ear 5400 KB
find . -name '*.ear' -exec ls -lh {} \;
just the h extra from jer.drab.org's reply. saves time converting to MB mentally ;)
You need to use -exec or -printf. Printf works like this:
find . -name *.ear -printf "%p %k KB\n"
-exec is more powerful and lets you execute arbitrary commands - so you could use a version of 'ls' or 'wc' to print out the filename along with other information. 'man find' will show you the available arguments to printf, which can do a lot more than just filesize.
[edit] -printf is not in the official POSIX standard, so check if it is supported on your version. However, most modern systems will use GNU find or a similarly extended version, so there is a good chance it will be implemented.
A simple solution is to use the -ls option in find:
find . -name \*.ear -ls
That gives you each entry in the normal "ls -l" format. Or, to get the specific output you seem to be looking for, this:
find . -name \*.ear -printf "%p\t%k KB\n"
Which will give you the filename followed by the size in KB.
Using GNU find, I think this is what you want. It finds all real files and not directories (-type f), and for each one prints the filename (%p), a tab (\t), the size in kilobytes (%k), the suffix " KB", and then a newline (\n).
find . -type f -printf '%p\t%k KB\n'
If the printf command doesn't format things the way you want, you can use exec, followed by the command you want to execute on each file. Use {} for the filename, and terminate the command with a semicolon (;). On most shells, all three of those characters should be escaped with a backslash.
Here's a simple solution that finds and prints them out using "ls -lh", which will show you the size in human-readable form (k for kilobytes and M for megabytes):
find . -type f -exec ls -lh \{\} \;
As yet another alternative, "wc -c" will print the number of characters (bytes) in the file:
find . -type f -exec wc -c \{\} \;
find . -name '*.ear' -exec du -h {} \;
This gives you the filesize only, instead of all the unnecessary stuff.
Awk can fix up the output to give just what the questioner asked for. On my Solaris 10 system, find -ls prints size in KB as the second field, so:
% find . -name '*.ear' -ls | awk '{print $2, $11}'
5400 ./dir1/dir2/earFile2.ear
5400 ./dir1/dir2/earFile3.ear
5400 ./dir1/dir2/earFile1.ear
Otherwise, use -exec ls -lh and pick out the size field from the output.
Again on Solaris 10:
% find . -name '*.ear' -exec ls -lh {} \; | awk '{print $5, $9}'
5.3M ./dir1/dir2/earFile2.ear
5.3M ./dir1/dir2/earFile3.ear
5.3M ./dir1/dir2/earFile1.ear
Try the following commands:
GNU stat:
find . -type f -name *.ear -exec stat -c "%n %s" {} ';'
BSD stat:
find . -type f -name *.ear -exec stat -f "%N %z" {} ';'
however stat isn't standard, so du or wc could be a better approach:
find . -type f -name *.ear -exec sh -c 'echo "{} $(wc -c < {})"' ';'
Just list the files (-type f) that match the pattern (-name '*.ear) using the disk-usage command (du -h) and sort the files by the human-readable file size (sort -h):
find . -type f -name '*.ear' -exec du -h {} \; | sort -h
Output
5.0k ./dir1/dir2/earFile1.ear
5.4k ./dir1/dir2/earFile2.ear
5.4k ./dir1/dir3/earFile1.ear
I struggled with this on Mac OS X where the find command doesn't support -printf.
A solution that I found, that admittedly relies on the 'group' for all files being 'staff' was...
ls -l -R | sed 's/\(.*\)staff *\([0-9]*\)..............\(.*\)/\2 \3/'
This splits the ls long output into three tokens
the stuff before the text 'staff'
the file size
the file name
And then outputs tokens 2 and 3, i.e. output is number of bytes and then filename
8071 sections.php
54681 services.php
37961 style.css
13260 thumb.php
70951 workshops.php
Why not use du -a ? E.g.
find . -name "*.ear" -exec du -a {} \;
Works on a Mac
This should get you what you're looking for, formatting included (i.e. file name first and size afterward):
find . -type f -iname "*.ear" -exec du -ah {} \; | awk '{print $2"\t", $1}'
sample output (where I used -iname "*.php" to get some result):
./plugins/bat/class.bat.inc.php 20K
./plugins/quotas/class.quotas.inc.php 8.0K
./plugins/dmraid/class.dmraid.inc.php 8.0K
./plugins/updatenotifier/class.updatenotifier.inc.php 4.0K
./index.php 4.0K
./config.php 12K
./includes/mb/class.hwsensors.inc.php 8.0K
You could try this:
find. -name *.ear -exec du {} \;
This will give you the size in bytes. But the du command also accepts the parameters -k for KB and -m for MB. It will give you an output like
5000 ./dir1/dir2/earFile1.ear
5400 ./dir1/dir2/earFile2.ear
5400 ./dir1/dir3/earFile1.ear
find . -name "*.ear" | xargs ls -sh
$ find . -name "test*" -exec du -sh {} \;
4.0K ./test1
0 ./test2
0 ./test3
0 ./test4
$
Scripter World reference
find . -name "*.ear" -exec ls -l {} \;
If you need to get total size, here is a script proposal
#!/bin/bash
totalSize=0
allSizes=`find . -type f -name *.ear -exec stat -c "%s" {} \;`
for fileSize in $allSizes; do
totalSize=`echo "$(($totalSize+$fileSize))"`
done
echo "Total size is $totalSize bytes"
You could try for loop:
for i in `find . -iname "*.ear"`; do ls -lh $i; done

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