I need to use ls -l and I would like to have as result just the first word of the file name for instance for a result like this
-rw-r--r-- 1 root root 9 Sep 21 23:11 best file 1.txt
I would like to have only
best
as result because I need to put this value into a variable. It is ok as well if there is another way instead of using ls -l.
...sorry to bother you again...if the file is under a sub-directory, how can I hide the folder from the result? Thanks
You don't need to use ls -l (L).
Instead, use ls -1 (number one), that just outputs the names of the files, and then filter out the first column with cut:
ls -1 | cut -d' ' -f1
^
number one, not letter L
To store the value into a variable, do:
var=$(ls -1 | cut -d' ' -f1)
Note it is not a good thing to parse ls: the number of columns may vary, etc. You can read more about the topic in Why you shouldn't parse the output of ls
Update
Note there is no even need to use -1 (one), ls alone suffices:
ls | cut -d' ' -f1
As BroSlow comments below, "because they are EOL (end of line) separated across a pipe".
If you have only one row to output, this will work fine:
var=`ls -l | awk '{ print $9 }'`
echo ${var}
Or you need to use grep to filter your output for the correct file.
set -- $(ls -l)
echo ${11} # Assumes the file is the FIRST one listed.
Should do the trick. But I'm not sure if that's really what you want. For one thing, ls -l also prints an extra header line. Why do you say that you need to use ls -l? If you could state the actual problem, maybe we can find a much better solution together...
awk can pick the first word for you;
ls | awk '{print $1}'
Try:
ls -al|awk 'NR==4{ print $9 }'
Row number 4 will have first line of files. $9 indicates column 9 which will have desired word.
Related
I need to display the filesize and the filename. Like this:
4.0K Desktop
I'm extracting these two fields using cut from the ls -l output:
ls -lhS | cut -d' ' -f5,9
Due to multiple spaces in the ls -l output, I'm getting a few erroneous outputs, like:
4.0K 19:54
4.0K 19:55
6
18:39
31
25
How should I fix this?
I need to accomplish this task using pipes only and no bash scripting ( output could be multiple pipes ) and preferably no sed, awk.
If no alternative to sed or awk is available- use of sed is OK.
You can avoid parsing ls output and use the stat command which comes as part of GNU coreutils in bash for detailed file information.
# -c --format=FORMAT
# use the specified FORMAT instead of the default; output a newline after each use of FORMAT
# %n File name
# %s Total size, in bytes
stat -c '%s %n' *
You can use translate character command before using cut.
ls -lhS | tr -s ' ' | cut -d' ' -f 5,9
Or you could just submit to awk:
$ ls -lhS | awk '$0=$5 OFS $9'
ie. replace whole record $0 with fields $5 and $9 separated by output field separator OFS.
I'm trying to combine the below outputs into one command. The issue is that the field I'm trying to grab is in reverse order. I was told that cut doesn't support a "reverse" option and to use AWK for this purpose but it didn't end up working for my purpose. I'm trying to take the output of the ls- l against the /dev/block to return the partitions and automatically build a dd if= / of= for each outputted line based on the output of the command.
I tried piping the output to awk:
cut -d' ' -f23,25 ... | awk '{print $2,$1}'
however, the result was when using sed to input the prefix and suffix, it wasn't in the appropriate order.
I built the two statements below which individually return the expected output, just looking for the "right" way to combine both of these statements in the most efficient manner using sed / awk.
ls -l /dev/block/platform/msm_sdcc.1/by-name/ | cut -d' ' -f 25 | sed "s/^/dd if=/"
ls -l /dev/block/platform/msm_sdcc.1/by-name/ | cut -d' ' -f 23 | sed "s/.*/of=\/external_sd\/&.dsk/"
Any assistance will be appreciated.
Thank you.
If you're already using awk, I don't think you'll need cut or sed. You can probably do something like the following, though I'll have to trust you on the field numbers
ls -l /dev/block/platform/msm_sdcc.1/by-name | awk '{print "dd if=/"$25 " of=/" $23 ".dsk"}'
awk will split on all whitespace, not just the space character, so it's possible the fields will shift some, though it may be more reliable too.
I'm connecting to an exadata and want to get information about "ORACLE_HOME" variable inside them. So i'm using this command:
ls -l /proc/<pid>/cwd
this is the output:
2 oracle oinstall 0 Jan 23 21:20 /proc/<pid>/cwd -> /u01/app/database/11.2.0/dbs/
i need the get the last part :
/u01/app/database/11.2.0 (i dont want the "/dbs/" there)
i will be using this command several times in different machines. So how can i get this substring from whole output?
Awk and grep are good for these types of issues.
New:
ls -l /proc/<pid>/cwd | awk '{print ($NF) }' | sed 's#/dbs/##'
Old:
ls -l /proc/<pid>/cwd | awk '{print ($NF) }' | egrep -o '^.+[.0-9]'
Awk prints the last column of the input which is your ls command and then grep grabs the beginning of that string up the last occurrence of numbers and dots. This is a situational solution and perhaps not the best.
Parsing the output of ls is generally considered sub-optimal. I would use something more like this instead:
dirname $(readlink -f /proc/<pid>/cwd)
I have a bunch of massive text files, about 100MB each.
I want to grep to find entries that have 'INDIANA JONES' in it:
$ grep -ir 'INDIANA JONES' ./
Then, I would like to find the entries where there is the word PORTUGAL within 5,000 characters of the INDIANA JONES term. How would I do this?
# in pseudocode
grep -ir 'INDIANA JONES' ./ | grep 'PORTUGAL' within 5000 char
Use grep's -o flag to output the 5000 characters surround the match, then search those characters for the second string. For example:
grep -ioE ".{5000}INDIANA JONES.{5000}" file.txt | grep "PORTUGAL"
If you need the original match, add the -n flag to the second grep and pipe into:
cut -f1 -d: > line_numbers.txt
then you could use awk to print those lines:
awk 'FNR==NR { a[$0]; next } FNR in a' line_numbers.txt file.txt
To avoid the temporary file, this could be written like:
awk 'FNR==NR { a[$0]; next } FNR in a' <(grep -ioE ".{50000}INDIANA JONES.{50000}" file.txt | grep -n "PORTUGAL" | cut -f1 -d:) file.txt
For multiple files, use find and a bash loop:
for i in $(find . -type f); do
awk 'FNR==NR { a[$0]; next } FNR in a' <(grep -ioE ".{50000}INDIANA JONES.{50000}" "$i" | grep -n "PORTUGAL" | cut -f1 -d:) "$i"
done
One way to deal with this is with gawk. You could set the record separator to either INDIANA JONES or PORTUGAL and then perform a length check on the record (after stripping newlines, assuming newlines do not count towards the limit of 5000). You may have to resort to find to run this recursively within a directory
awk -v RS='INDIANA JONES|PORTUGAL' '{a = $0;
gsub("\n", "", a)};
((RT ~ /IND/ && prevRT ~/POR/) || (RT ~ /POR/ && prevRT ~/IND/)) && length(a) < 5000{found=1};
{prevRT=RT};
END{if (found) print FILENAME}' file.txt
Consider installing ack-grep.
sudo apt-get install ack-grep
ack-grep is a more powerful version of grep.
There's no trivial solution to your question (that i can think of) outside of a full batch script, but you can use the -A and -B flags on ack-grep to specify a number of trailing or leading lines to output, resp.
This may not be a number of chars, but is a step further in that direction.
While this may not be a solution, it might give you some idea as to how to do this. Lookup filters like ack, awk, sed, etc. and see if you can find one with a flag for this kind of behaviour.
The ack-grep manual:
http://manpages.ubuntu.com/manpages/hardy/man1/ack-grep.1p.html
EDIT:
I think the sad news is, what you might think you're looking for is something like:
grep "\(INDIANA JONES\).\{1,5000\}PORTUGAL" filename
The problem is, even on a small file, querying this is going to be impossible time-wise.
I got this one to work with a different number. it's a size problem.
For such a large set of files, you'll need to do this in more than one step.
A Solution:
The only solution I know of is the leading and trailing output from ack-grep.
Step 1: how long are your lines?
If you knew how many lines out you had to go
(and you could estimate/calculate this a few ways) then you'd be able to grep the output of the first grep. Depending on what's in your file, you should be able to get a decent upper bound as to how many lines is 5000 chars (if a line has 100 chars average, 50+ lines should cover you, but if it has 10 chars, you'll need 500+).
You've got to determine the maximum number of lines that could be 5000 chars. You could guess or pick a high range if you like, but that'll be up to you. It's your data.
With that, call: (if you needed 100 lines for 5000 chars)
ack-grep -ira "PORTUGAL" -A 100 -B 100 filename
and
ack-grep -ira "INDIANA JONES" -A 100 -B 100 filename
replace the 100s with what you need.
Step 2: parse the output
you'll need to take the matches that ack-grep returns and parse them, looking for any matches again, within these sub-ranges.
Look for INDIANA JONES in the first PORTUGAL ack-grep match output, and look for PORTUGAL in the second set of matches.
This should take a bit more work, likely involving a bash script (I might see if I can get one working this week), but it solves your massive-data problem, by breaking it down into more manageable chunks.
grep 'INDIANA JONES' . -iR -l | while read filename; do head -c 5000 "$filename" | grep -n PORTUGAL -H --label="$filename" ; done
This works as follows:
grep 'INDIANA JONES' . -iR -l. Search for all files in or below the current directory. Case insensitive (-i). And only print the names of the files that match (-l), don't print any content.
| while read filename; do ...|...|...; done for each line of input, store it in variable $filename and execute the pipeline.
Now, for each file that matched 'INDIANA JONES', we do
head -c 5000 "$filename" - extract the first 5000 characters
grep ... - search for PORTUGAL. Print the filename (-H), but where we tell us the 'filename' we want to use with --label="$filename". Print line numbers too, -n.
So here is the task which I can't solve. I have a directory with .h files and a directory with .i files, which have the same names as the .h files. I want just by typing a command to have all .h files which are not found as .i files. It's not a hard problem, I can do it in some programming language, but I'm just curious how it will look like in cmd :). To be more specific here is the algo:
get file names without extensions from ls *.h
get file names without extensions from ls *.i
compare them
print all names from 1 that are not met in 2
Good luck!
diff \
<(ls dir.with.h | sed 's/\.h$//') \
<(ls dir.with.i | sed 's/\.i$//') \
| grep '$<' \
| cut -c3-
diff <(ls dir.with.h | sed 's/\.h$//') <(ls dir.with.i | sed 's/\.i$//') executes ls on the two directories, cuts off the extensions, and compares the two lists. Then grep '$<' finds the files that are only in the first listing, and cut -c3- cuts off the "< " characters that diff inserted.
ls ./dir_h/*.h | sed -r -n 's:.*dir_h/([^.]*).h$:dir_i/\1.i:p' | xargs ls 2>&1 | \
grep "No such file or directory" | awk '{print $4}' | sed -n -r 's:dir_i/([^:]*).*:dir_h/\1:p'
ls -1 dir1/*.hh dir2/*.ii | awk -F"/" '{print $NF}' |awk -F"." '{a[$1]++;b[$0]}END{for(i in a)if(a[i]==1 && b[i".hh"]) print i}'
explanation:
ls -1 dir1/*.hh dir2/*.ii
above will list all the files *.hh and *.ii files in both the directories.
awk -F"/" '{print $NF}'
above will just print the file name excluding the complete path of the file.
awk -F"." '{a[$1]++;b[$0]}END{for(i in a)if(a[i]==1 && b[i".hh"]) print i}'
above will create two associative arrays one with file name and one with excluding the extension.
if both hh and ii files exist the value in the assosciative array will 2 if there is only one file then the value will be 1.so we need array item whose value is 1 and it should be a header file (.hh).
this can be checked using the asso..array b which is done in the END block.
Assuming bash is your shell:
for file in $( ls dir_with_h/*.h ); do
name=${file%\.h}; # trim trailing ".h" file extension
name=${name#dir_with_h/}; # trim leading folder name
if [ ! -e dir_with_i/${name}.i ]; then
echo ${name};
fi
done
Undoubtedly this can be ported to virtually all other shells. I find this less cryptic than some other approaches (although this is surely my problem) but it is a little wordy. As such. a shell script might help recall it.