When I run this command in terminal it displays a number.
The number changes when I mount another folder using cd command.
ls -la // List files in your current directory (hidden files are included thanks to the -a)
sort // Sort entries returned by the previous command
wc -l // Returns the number of lines returned by the previous command
«|» send the output of one command to the input of the next one
Related
As I am very new to UNIX and struggling with two commands to get the desired output.
What will be the command to print the name of the largest file in size in
/usr/include directory.
I tried but not giving the filename whose size is largest :
find /usr/include -type f -exec ls -s {} \; | sort -n | tail -n 5
command to print user ids on a Linux environment which can’t be
used as login ids. [hint: consider /etc/passwd file]
I tried: cat /etc/passwd
Please let me know if these are the correct commands.
ls -lS /usr/include | head -2 | grep -v total
Flag S is to sort files by size. Head then grabs the first 2 lines of the long output and grep removes the summary line.
awk -F: '{ if($3<100) print $1;}' /etc/passwd
UID is the third field in /etc/passwd. Generally UIDs 0-99 are reserved for predefined system accounts. So this command prints all usernames with UID less than 100.
I am new to linux. I have a folder with many files in it and i need to get the latest file depending on the file name. Example: I have 3 files RAT_20190111.txt RAT_20190212.txt RAT_20190321.txt . I need a linux command to move the latest file here RAT20190321.txt to a specific directory.
If file pattern remains the same then you can try below command :
mv $(ls RAT*|sort -r|head -1) /path/to/directory/
As pointed out by #wwn, there is no need to use sort, Since the files are lexicographically sortable ls should do the job already of sorting them so the command will become :
mv $(ls RAT*|tail -1) /path/to/directory
The following command works.
ls | grep -v '/$' |sort | tail -n 1 | xargs -d '\n' -r mv -- /path/to/directory
The command first splits output of ls with newline. Then sorts it, takes the last file and then it moves this to the required directory.
Hope it helps.
Use the below command
cp ls |tail -n 1 /data...
I have a directory with over 100,000 files. I want to know if the string "str1" exists as part of the content of any of these files.
The command:
grep -l 'str1' * takes too long as it reads all of the files.
How can I ask grep to stop reading any further files if it finds a match? Any one-liner?
Note: I have tried grep -l 'str1' * | head but the command takes just as much time as the previous one.
Naming 100,000 filenames in your command args is going to cause a problem. It probably exceeds the size of a shell command-line.
But you don't have to name all the files if you use the recursive option with just the name of the directory the files are in (which is . if you want to search files in the current directory):
grep -l -r 'str1' . | head -1
Use grep -m 1 so that grep stops after finding the first match in a file. It is extremely efficient for large text files.
grep -m 1 str1 * /dev/null | head -1
If there is a single file, then /dev/null above ensures that grep does print out the file name in the output.
If you want to stop after finding the first match in any file:
for file in *; do
if grep -q -m 1 str1 "$file"; then
echo "$file"
break
fi
done
The for loop also saves you from the too many arguments issue when you have a directory with a large number of files.
I'm trying to find a way to list all jpg images in all subdirectories and in csv format without the dir name present.
ls -R -1 -m . | grep '.jpg'
The ls command does output to csv fine, but the grep command breaks the csv format making each file appear on a new line instead of comma seperated.
I know I can use 'find' to list images but it seems to output the files in a different order than 'ls' and I don't see a output to csv parameter for 'find'
I need the images in each subdirectory on 1 comma seperated line.
I believe this does what you want. Each outputted line is a list of jpgs in a single directory separated by commas.
ls -d */ | xargs -i{} sh -c 'cd {};ls -m *jpg'
If you wanted to know which line was which directory you could run it in 2 steps like this
ls -d */ > dirs.txt
cat dirs.txt | xargs -i{} sh -c 'cd {};ls -m *txt'
and then the first line in dirs.txt would correspond to each line of output.
I want to sort the output of the "ls -al" command according to date. I am able to easily do that for one column with command:
$ ls -al | sort -k6 -M -r
But how to do it for both collumn 6 and 7 simultaneously? The command:
$ ls -al | sort -k6 -M -r | sort -k7 -r
prints out results I do not understand.
The final goal would be to see all the files from the most recently modified (or v.v.).
Here is the attached example for the data to be sorted and the command used:
With sort, if you specify -k6, the key starts at field 6 and extends to the end of the line. To truncate it and only use field 6, you should specify -k6,6. To sort on multiple keys, just specify -k multiple times. Also, you need to apply the M modifier only to the month, and the n modifier to the day. So:
ls -al | sort -k 6,6M -k 7,7n -r
Do note Charles' comment about abusing ls though. Its output cannot be reliably parsed. A good demonstration of this is that the image you've posted shows the month/date in columns 4 and 5, so it's not clear why you want to sort on columns 6 and 7.
The final goal would be to see all the files from the most recently modified
ls -t
or (for reverse, most recent at bottom):
ls -tr
The ls man page describes this in more details, and lists other options.
You could try ls -lsa -it -r
sample
enter image description here