This takes a directory as a parameter:
#!/bin/bash
ls -l $1 |awk '$3!=$4{print $9}'
Now what I need is to be able to do ANOTHER ls -l on the just the files that are found from the awk statement.
Yeah, it sounds dumb, and I know of like 3 other ways to do this, but not with awk.
Use awk system command:
ls -l $1 |awk '$3!=$4{system("ls -l " $9)}'
The command to use is xargs.
man xargs
should give some clues.
#!/bin/bash
ls -l $1 | awk '$3!=$4{ system( "ls -l '$1'/" $9}'
If it is allowed to adjust the first ls -l you could try to do:
ls -ld "$1"/* |awk '$3!=$4{print $9}' | xargs ls -l
In this case ls will prefix the directory. But I don't know if this is portable behavior.
Related
I want total size of file in a folder using certain file starting with name like abc_1_* in sun solaris os because here i cannot use du -ch, current i am using find command i am getting required output but i want round up output after decimal
Current code :-
echo `find $DUMPDIR -name "${DUMPFILE}*" -exec ls -ltr {} \; | awk ' {s+=$5} END {print s/1024/1024/1024}'`
output:-
1.768932
Desired output:-
1.7G
Kindly help me with this i am new to solaris
find $DUMPDIR -name "${DUMPFILE}*" -exec ls -lh {} \; | awk '{print $5}'
You can use GNU extension -h option for ls to print sizes in human readable format. Note: as suggested by #Andrew Henle, -h is not guaranteed to be supported in ls.
Or simply use,
ls -lh "$DUMPDIR/${DUMPFILE}*" | cut -d' ' -f 5
You can round up a float in awk with
awk 'BEGIN {fl=1.768932; printf("%.1f G\n", fl)}'
I'm trying to use the /etc/passwd file to list home directories of users in the system, sorted and without repetitions, such that nonexisting directories would not be printed..
This is my command:
cut -f 6 -d ':' /etc/passwd | sort -su | ls -ld
It acts as if I just ran ls -ld with no arguments from the command pipe at all.
You can't pipe stuff into ls.. You could do something like:
ls -ld $(cut -f 6 -d ':' /etc/passwd | sort -su)
By spawning a new bash to execute the cut | sort and passing it as a ls argument
ls does not take piped output. You could, however, use forward quotes to execute it on a list of directories:
ls `cut -f 6 -d ':' /etc/passwd | sort -su `
You could use xargs
cut -f 6 -d ':' /etc/passwd | sort -su | xargs ls
You were not far away, it was enough to add a xargs before ls :
cut -f 6 -d ':' /etc/passwd | sort -u | xargs ls -ld
I have to do below operation in one script.
/etc/passwd file to display user home directory ownership:
ls -lLd /<usershomedirectory>
I can grep the home path as following, please help me with running the ls on the home path based on the user home path that is identified using single line script.
grep "" /etc/passwd | cut -f 6 -d :
Updated
How to do it for some more ls
# ls -al /<usershomedirectory>/.login
# ls -al /<usershomedirectory>/.cschrc
# ls -al /<usershomedirectory>/.logout
# ls -al /<usershomedirectory>/.profile
# ls -al /<usershomedirectory>/.bash_profile
# ls -al /<usershomedirectory>/.bashrc
# ls -al /<usershomedirectory>/.bash_logout
# ls -al /<usershomedirectory>/.env
# ls -al /<usershomedirectory>/.dtprofile
# ls -al /<usershomedirectory>/.dispatch
# ls -al /<usershomedirectory>/.emacs
# ls -al /<usershomedirectory>/.exrc
You can use backticks/command output substitution:
ls -lLd "$(grep "^$username:" /etc/passwd | cut -f6 -d: )"
or piping to xargs:
grep "^$username:" /etc/passwd | cut -f 6 -d: | xargs -r ls -lLd
(I also added some more grep context to make sure the user matches exactly and supports blanks in file names)
The ~ expansion could be used with eval, not sure if thats a good idea:
eval "ls -lLd ~$username"
As #EtanReisner commented, the usual way to do this would be:
ls -lLd ~user
However, if you must use parsing of /etc/passwd and substitution, I would recommend this particular recipe:
ls -lLd $(awk -v acct="user" -F: '$1 == acct { print $NF-1 }' /etc/passwd)
But there's quite a few ways, as the other answer(s) will demonstrate as well...
I need help doing this I tried:
#!/bin/bash
ls -l [0-9][0-9]*
I am trying to get all the entries with 2 integers in them.
says ls cannot access no such file or directory
ls -l | awk '{print $9}' | grep '[0-9][0-9]'
Or if you simply want to count them:
ls -l | awk '{print $9}' | grep '[0-9][0-9]' | wc -l
I have search string in one variable ($AUD_DATE) and replace string in another variable ($YEST_DATE). I need to search file name in a folder using $AUD_DATE and then replace it with $YEST_DATE.
I tried using this link to do it but its not working with variables.
Find and replace filename recursively in a directory
shrivn1 $ AUD_DATE=140101
shrivn1 $ YEST_DATE=140124
shrivn1 $ ls *$AUD_DATE*
NULRL.PREM.DATA.CLRSFIFG.140101.dat NULRL.PREM.DATA.CLRTVEH.140101.dat
shrivn1 $ ls *$AUD_DATE*.dat | awk '{a=$1; gsub("$AUD_DATE","$YEST_DATE");printf "mv \"%s\" \"%s\"\n", a, $1}'
mv "NULRL.PREM.DATA.CLRSFIFG.140101.dat" "NULRL.PREM.DATA.CLRSFIFG.140101.dat"
mv "NULRL.PREM.DATA.CLRTVEH.140101.dat" "NULRL.PREM.DATA.CLRTVEH.140101.dat"
Actual output I need is
mv "NULRL.PREM.DATA.CLRSFIFG.140101.dat" "NULRL.PREM.DATA.CLRSFIFG.140124.dat"
mv "NULRL.PREM.DATA.CLRTVEH.140101.dat" "NULRL.PREM.DATA.CLRTVEH.140124.dat"
Thanks in advance
Approach 1
I generally create mv commands using sed and then pipe the output to sh. This approach allows me to see the commands that will be executed beforehand.
For example:
$ AUD_DATE=140101
$ YEST_DATE=140124
$ ls -1tr | grep "${AUD_DATE}" | sed "s/\(.*\)/mv \1 \1_${YEST_DATE}"
Once you are happpy with the output of the previous command;repeat it and pipe it's output to sh, like so:
$ ls -1tr | grep "${AUD_DATE}" | sed "s/\(.*\)/mv \1 \1_${YEST_DATE}" | sh
Approach 2
You could use xargs command.
ls -1tr | grep ${AUD_DATE}" | xargs -I target_file mv target_file target_file${YEST_DATE}