how can i search for files and give only a list of mimes or types out.
example:
dir
-file1.pdf
-file2.pdf
-dir2
--file3.png
--file4.pdf
wished output:
pdf
png
Edit
Found also a solution, but does not make a difference between upper and lowercase and also not .peng and .png
find . -type f -printf '%f\n' | sed 's/^.*\.//' | sort -u
Use find's -exec option together with any solution that extracts the extension. Then pipe through sort -u to remove duplicates.
find dir -type f -exec bash -c 'printf %s\\n "${###*.}"' x {} + | sort -u
Files without extensions will be listed too. To filter them out add the option -name '*\.*' before -exec.
Related
I'm trying to create a little script using bash in linux. That allows me to find if there is any tag 103=16 inside a log
I have multiple folders named for example l51prdsrv-api1.nebex.local, l51prdsrv-oe1.nebex.local, etc... inside those folders are .log files like TRADX_gsoe3.log, TRADX_gseuoe2.log, etc... .
I need to find if inside those logs there is the tag 103=16
I'm trying this command
find . /opt/FIXLOGS/l51prdsrv* -iname "TRADX_" -type f | grep -e 103=16
But what it does is that is showing just the logs names and not the content to see if there is a tag 103=16
First of all, you are not searching files of the form TRADX_something.log, but only files which are just named TRADX_ (case-insensitively, so TradX_ would also be found).
Then you are feeding to grep the names of the files, but never look into the content of those files. From the grep man page, you see that the file content can be supplied either via stdin, or by specifying the file name on the command line. In your case, the latter is the way to go. Therefore you can either do a
find . /opt/FIXLOGS/l51prdsrv* -iname "TRADX_*.log" -type f -exec grep -F 103=16 {} \;
if you are only interested in the matchin lines, or a
find . /opt/FIXLOGS/l51prdsrv* -iname "TRADX_*.log" -type f -exec grep -F 103=16 {} /dev/null \;
if you also want to see the file names where the pattern matches. The reason is that grep is printing the filename only if it sees more than 1 filename on the command line and the /dev/null provides a second dummy file. find replaces the {} by the filename.
BTW, I used -f for grep instead of your -e, because you don't seem to use any specific regular expression pattern anyway.
But you don't need find for this task. An alternative would be an explicit loop:
shopt -s nocasematch # make globbing case-insensitive
shopt -s globstar # turn on ** globbing
for f in {.,/opt/FIXLOGS/l51prdsrv*}/**/tradx_*.log
do
[[ -f $f ]] && grep -F 103=16 "$f" /dev/null
done
While the loop looks more complicated at first glance, it is easier to extend the logic in case you want to do more with the files instead of just grepping the lines, for instance taking specific actions on those files which contain the pattern.
You are doing:
find . /opt/FIXLOGS/l51prdsrv* -iname "TRADX_" -type f | grep -e 103=16
I propose you do:
find . /opt/FIXLOGS/l51prdsrv* -iname "TRADX_" -type f -exec grep -e "103=16" {} /dev/null \;
What's the difference?
find ... -type f
=> gives you a list of files.
When you add | grep -e 103=16, then you perform that on the filenames.
When you add -exec grep ..., then you perform that on the files itselfs.
I'm trying to find all the files from a folder and then print them but sorted.
I have this so far
find . -type f -exec cat {} \;
and it print's all files but I need to sort them too but when I do
find . -type f -exec sort cat {};
I get the next error
sort:cannot read:cat:No such file or directory
and if I switch sort and cat like this
find . -type f -exec cat sort {} \;
I get the same error the it print's the file(I have only one file to print)
It's not clear to me if you want to display the contents of the files unchanged sorting the files by name, or if you want to sort the contents of each file. If the latter:
find . -type f -exec sort {} \;
If the former, use bsd find's -s option:
find -s . -type f -exec cat {} \;
If you don't have bsd find, use:
find . -type f -print0 | sort -z | xargs -0 cat
Composing commands using pipes is often the simplest solution.
find . -print0 -type f | sort | xargs -0 cat
Explanation: you can sort filenames after the fact using ... | sort, then pass the output (the list of files) to cat using xargs, i.e. ... | xargs cat.
As #arkaduisz points out, when using pipes, should carefully handle filenames containing whitespaces (thus using -print0 and -0).
I have a case where multiple .bz2 files are situated in subdirectories. And I want to search for a text, from all files, using bzcat and grep command linux commands.
I am able to search one-one file by using the following command:
bzcat <filename.bz2> | grep -ia 'text string' | less
But I now I need to do the above for all files in subdirectories.
You can use bzgrep instead of bzcat and grep. This is faster.
To grep recursively in a directory tree use find:
find -type f -name '*.bz2' -execdir bzgrep "pattern" {} \;
find is searching recursively for all files with the *.bz2 extension and applies the command specified with -execdir to them.
There are several methods:
bzgrep regexp $(find -name \*.bz2)
This method will work if number of the found files is not very big (and they have no special characters in the pathes). Otherwise you better use this one:
find -name \*.bz2 -exec bzgrep regexp {} /dev/null \;
Please note /dev/null in the second method. You use it to make bzgrep print the filename,
where the regexp was found.
Just try to use:
bzgrep --help
grep through bzip2 files
Usage: bzgrep [grep_options] pattern [files]
For example, I need grep information from list of files by number 1941974:
'billing_log_1.bz'
'billing_log_2.bz'
'billing_log_3.bz'
'billing_log_4.bz'
'billing_log_5.bz'
What can I do?
bzgrep '1941974' billing_log_1
Continuous your code with fixes by bzcat:
find . -type f -name "*.bz2" |while read file
do
bzcat $file | grep -ia 'text string' | less
done
I have a list of file names as output of certain command.
I need to find each of these files in a given directory.
I tried following command:
ls -R /home/ABC/testDir/ | grep "\.java" | xargs find /home/ABC/someAnotherDir -iname
But it is giving me following error:
find: paths must precede expression: XYZ.java
What would be the right way to do it?
ls -R /home/ABC/testDir/ | grep -F .java |
while read f; do find . -iname "$(basename $f)"; done
You can also use ${f##*/} instead of basename. Or;
find /home/ABC/testDir -iname '*.java*' |
while read f; do find . -iname "${f##*/}"; done
Note that, undoubtedly, many people will object to parsing the output of ls or find without using a null byte as filename separater, claiming that whitespace in filenames will cause problems. Those people usually ignore newlines in filenames, and their objections can be safely ignored. (As long as you don't allow whitespace in your filenames, that is!)
A better option is:
find /home/ABC/testDir -iname '*.java' -exec find . -iname {}
The reason xargs doesn't work is that is that you cannot pass 2 arguments to -iname within find.
find /home/ABC/testDir -name "\.java"
I have a Textfile with one Filename per row:
Interpret 1 - Song 1.mp3
Interpret 2 - Song 2.mp3
...
(About 200 Filenames)
Now I want to search a Folder recursivly for this Filenames to get the full path for each Filename in Filenames.txt.
How to do this? :)
(Purpose: Copied files to my MP3-Player but some of them are broken and i want to recopy them all without spending hours of researching them out of my music folder)
The easiest way may be the following:
cat orig_filenames.txt | while read file ; do find /dest/directory -name "$file" ; done > output_file_with_paths
Much faster way is run the find command only once and use fgrep.
find . -type f -print0 | fgrep -zFf ./file_with_filenames.txt | xargs -0 -J % cp % /path/to/destdir
You can use a while read loop along with find:
filecopy.sh
#!/bin/bash
while read line
do
find . -iname "$line" -exec cp '{}' /where/to/put/your/files \;
done < list_of_files.txt
Where list_of_files.txt is the list of files line by line, and /where/to/put/your/files is the location you want to copy to. You can just run it like so in the directory:
$ bash filecopy.sh
+1 for #jm666 answer, but the -J option doesn't work for my flavor of xargs, so i chaned it to:
find . -type f -print0 | fgrep -zFf ./file_with_filenames.txt | xargs -0 -I{} cp "{}" /path/to/destdir/