I am doing a find command to get items sorted by access time:
$ find . -printf "%A+\t%p\n" | sort -r
The output looks like this:
2020-05-05+06:00:55.5569719990 ./form.py
2020-05-05+06:00:55.5569719990 ./amazon.js
2020-05-04+12:48:24.8209719990 ./historical.py
However, I would like to only show the filepath and remove the timestamp after it's been sorted like that, so getting:
./form.py
./amazon.js
./historical.py
What would be the best way to do this?
Like this:
find . -printf "%A+\t%p\n" | sort -r | cut -d$'\t' -f2-
Or if you want to use awk, you should use TAB as delimiter (if not, it breaks on files with spaces on filenames):
find . -printf "%A+\t%p\n" | sort -r | awk -F$'\t' '{print $2}'
You can use awk here if you want which will figure out the separator for you:
find . -printf "%A+\t%p\n" | sort -r | awk '{print $2}'
Note that it needs single quotes and fields are one-indexed.
Related
I'm trying to extract an address from a file.
grep keyword /path/to/file
is how I'm finding the line of code I want. The output is something like
var=http://address
Is there a way I can get only the part directly after the = i.e. http://address , considering the keyword I'm greping for is both in the var and http://address parts
grep keyword /path/to/file | cut -d= -f2-
Just pipe to cut:
grep keyword /path/to/file | cut -d '=' -f 2
You can avoid the needless pipes:
awk -F= '/keyword/{print $2}' /path/to/file
Is there a oneliner for for sort and uniq given a filename in unix?
I googled and found the following but its not sorting,also not sure what is the below command doing..any better ways using awk or anyother unix tool?
cut -d, -f1 file | uniq | xargs -I{} grep -m 1 "{}" file
On a side note,is there one that can be used in both windows and unix?this is not important but just checking..
C:\Users\Chola>sort -t "#" -k2,2 email-list.txt
Input text file:-
436485
422636
429228
427041
433414
425810
422636
431526
428808
If your file consists only of numbers, one per line:
sort -n FILENAME | uniq
or
sort -u -n FILENAME
(You can add -u to the sort command instead of piping through uniq in all of the following.).
If you want to extract just one column of a file, and then sort that column numerically removing duplicates, you could do this:
cut -f7 FILENAME | sort -n | uniq
Cut assumes that there is a single tab between columns. If your file is CSV, you might be able to do this:
cut -f7 -d, FILENAME | sort -n | uniq
but that won't work if there is a , in a text field in the file (where CSV will protect it with "'s).
If you want to sort by the column but remove only completely duplicate lines, then you can do this:
sort -k7,7n FILENAME | uniq
sort assumes that columns are separated by whitespace. Again, if you want to separate with ,, you can use:
sort -k7,7n -t, FILENAME | uniq
pdftk file.pdf dump_data output | grep NumberOfPages:
gives me:
NumberOfPages: 5
I don't want it to output NumberOfPages. I want to get in this case just 5. Is there a flag I can say in grep to get just that? I did a man grep and nothing seemed to do the trick.
I think grep doesn't know about how to parse strings in different formats. But other utilities like awk will help you:
pdftk file.pdf dump_data output | grep NumberOfPages: | awk '{print $2}'
pdftk file.pdf dump_data output | grep NumberOfPages: | sed 's\NumberOfPages:\\'
Yes, in GNU Grep you can use the -o operator to get "only" the matching portion of your expression. So something like;
pdftk file.pdf dump_data output | grep -o ' .*'
Could work for you. As other answers have pointed out, if you want only the number you'd be better off using something in addition to grep.
For example:
$ echo 'NumberOfPages: 5' | grep -o ' .*'
5
Notice the space before the 5 being included.
Is there a simple way to recursively find all files in a directory hierarchy, that do not have a matching file with a different extension?
For example the directory has a bunch of files ending in .dat
I want to find the .dat files that do not have an accompanying .out file.
I have a while loop that checks each entry, but that is slow for long lists...
I am using GNU find.
Perhaps something like this?
find . -name "*.dat" -print | sort > column1.txt
find . -name "*.out" -print | sort > column2.txt
diff column1.txt column2.txt
I haven't tested it, but I think it's probably close to what you're asking for.
find . -name '*.dat' -printf "[ -f %p ] || echo %p\n" | sed 's/\.dat/.out/' | sh
I had to add a bunch of bells and whistles to the 1st solution, but that was a good start, thanks...
find . -print | grep -Fi '.dat' | grep -vFi '.dat.' | sort | sed -e 's/.dat//g' > column1.txt
find . -print | grep -Fi '.out' | grep -vFi '.out.' | sort | sed -e 's/.out//g' > column2.txt
sdiff -s column1.txt column2.txt | grep -F '<' | cut -f1 -d"<" > c12diff.txt
I want to sort the owners in alphabetical order from a call to ls -l and cannot figure out a way to do it. I know something like ls-l | sort would sort the file name but how do i sort the owners in order?
The owner is the third field, so use -k 3:
ls -l | sort -k 3
You can extend this idea to sorting based on other fields, and you can have multiple -k options. For instance, maybe you want to sort by owner, and then size in descending order:
ls -l | sort -k 3,3 -k 5rn
I am not sure if you want only the owners or the whole information sorted by owner. In the former case superfo's solution is almost correct.
Additionally you need to remove repeating white spaces from ls's output with tr because otherwise cut that uses them as a delimiter won't work in all directories.*
So in the end you get this:
ls -l | tr -s ' ' | cut -d ' ' -f 3 | sort | uniq
*Some directories have a two digit value in the second field and all other lines with a single digit get an additional whitespace to preserve the layout.
How about ...
ls -l | cut -d ' ' -f 3 | sort | uniq
Try this:
ls -l | awk '{print $3, $4, $8}' | sort
It will print the user name, the group name and the file name. (File name cannot contain spaces)
ls -l | awk '{print $3, $4, $0}' | sort
This will print the user name, group name and the full ls -l output, sorted by the user name first, then the group name, then what ls -l prints first