What command should I use to sort all the files in a project directory according to the date of modification?
Tried ls -t but it does not find the file and used find to find all the files of the specified type, but could not sort them.
You can use this to find all the files of the type you are looking for and sorted by the last modified date.
find $directory -type f -name "*.type" -print0 | xargs -0 stat -c "%y %n" | sort
The above looks for all the files of type "type" and then sorts it according to the last modified timestamp.
Related
I want to use the GNU find command to find files based on a pattern, and then have them displayed in order of the most recently modified file to the least recently modified.
I understand this:
find / -type f -name '*.md'
but then what would be added to sort the files from the most recently modified to the least?
find can't sort files, so you can instead output the modification time plus filename, sort on modification time, then remove the modification time again:
find . -type f -name '*.md' -printf '%T# %p\0' | # Print time+name
sort -rnz | # Sort numerically, descending
cut -z -d ' ' -f 2- | # Remove time
tr '\0' '\n' # Optional: make human readable
This uses \0-separated entries to avoid problems with any kind of filenames. You can pass this directly and safely to a number of tools, but here it instead pipes to tr to show the file list as individual lines.
find <dir> -name "*.mz" -printf "%Ts - %h/%f\n" | sort -rn
Print the modified time in epoch format (%Ts) as well as the directories (%h) and file name (%f). Pipe this through to sort -rn to sort in reversed number order.
Pipe the output of find to xargs and ls:
find / -type f -name '*.md' | xargs ls -1t
I'm looking for a specific directory file count that returns a number. I would type it into the terminal and it can give me the specified directory's file count.
I've already tried echo find "'directory' | wc -l" but that didn't work, any ideas?
You seem to have the right idea. I'd use -type f to find only files:
$ find some_directory -type f | wc -l
If you only want files directly under this directory and not to search recursively through subdirectories, you could add the -maxdepth flag:
$ find some_directory -maxdepth 1 -type f | wc -l
Open the terminal and switch to the location of the directory.
Type in:
find . -type f | wc -l
This searches inside the current directory (that's what the . stands for) for all files, and counts them.
The fastest way to obtain the number of files within a directory is by obtaining the value of that directory's kMDItemFSNodeCount metadata attribute.
mdls -name kMDItemFSNodeCount directory_name -raw|xargs
The above command has a major advantage over find . -type f | wc -l in that it returns the count almost instantly, even for directories which contain millions of files.
Please note that the command obtains the number of files, not just regular files.
I don't understand why folks are using 'find' because for me it's a lot easier to just pipe in 'ls' like so:
ls *.png | wc -l
to find the number of png images in the current directory.
I'm using tree, this is the way :
tree ph
I'd like to construct a Linux command to list all files (with their full paths) within a specific directory (and subdirectories) ordered by access time.
ls can order by access time, but doesn't give the full path. find gives the full path, but the only control you have over the access time is to specify a range with -atime N (accessed at least 24*N hours ago), which isn't what I want.
Is there a way to order by access time and get the full path at once? I could just write a script, but it seems there should be a way to do this with the standard Linux programs.
find . -type f -exec ls -l {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7
This will find all files, and sort them by date and then time. You can then use awk or cut to extract the dates and files name from the ls -l output
you could try:
ls -l $(find /foo/bar -type f )
you can add other options (e.g. -t for sorting) to ls command to achieve your goal.
also you could add your searching criteria to find cmd
find . -type f | xargs ls -ldt should do the trick as long as there's not so many files that you hit the command like argument limit and spawn 2 instances of ls.
pwd | xargs -I % find % -type f
find . -type f -exec ls -l --full-time {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7
Alex's answer did not work for me since I had files older than one year and the sorting got messed up. The above adds the --full-time parameter which nuetralizes the date/time values and makes them sortable regardless of how old they are.
This should be pretty simple, but I am not figuring it out. I have a large code base more than 4GB under Linux. A few header files and xml files are generated during build (using gnu make). If it matters the header files are generated based on xml files.
I want to search for a keyword in header file that was last modified after a time instance ( Its my start compile time), and similarly xml files, but separate grep queries.
If I run it on all possible header or xml files, it take a lot of time. Only those that were auto generated. Further the search has to be recursive, since there are a lot of directories and sub-directories.
You could use the find command:
find . -mtime 0 -type f
prints a list of all files (-type f) in and below the current directory (.) that were modified in the last 24 hours (-mtime 0, 1 would be 48h, 2 would be 72h, ...). Try
grep "pattern" $(find . -mtime 0 -type f)
To find 'pattern' in all files newer than some_file in the current directory and its sub-directories recursively:
find -newer some_file -type f -exec grep 'pattern' {} +
You could specify the timestamp directly in date -d format and use other find tests e.g., -name, -mmin.
The file list could also be generate by your build system if find is too slow.
More specific tools such as ack, etags, GCCSense might be used instead of grep.
Use this. Because if find doesn't return a file, then grep will keep waiting for an input halting the script.
find . -mtime 0 -type f | xargs grep "pattern"
I need to search a file in unix which starts with "catalina"
find ... what to be used effectively -name, -exec ? Whats the expression
Also I need to show few files at a time, then show some more. There are huge set of log files in there. I know there is some expression, but forgot...
find /path/to/search/in -name 'catalina*'
Use iname to match case-insensitively.
To not be overwhelmed with a long list of files, filter through less (append |less). You can also use more instead of less.
If catalina is the file name, then use
find -name 'catalina*'
If catalina is the first word contained in the file, then use
find -type f | xargs head -v -n 1 | grep -B 1 -A 1 -e '^catalina'