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I have created a soft link as follows:
/bip/etl>ln -s /bip/etl bipet
And now can see the soft link being created as well..
/bip/etl>ls -lrt |tail
-rw-rw-rw- 1 cdtbipx cduserg 24988174 Jun 19 19:17 227015716_WLR3PSTN_Filtered_06202016_5of6.csv.gz.gpg
-rw-rw-rw- 1 cdtbipx cduserg 23857587 Jun 19 19:17 227015716_WLR3PSTN_Filtered_06202016_6of6.csv.gz.gpg
drwxrwxrwx 1082 prod release 61440 Jul 3 02:51 WSC
drwxrwxrwx 5 oracle oinstall 4096 Jul 4 01:22 dsl
lrwxrwxrwx 1 cdtbipx cduserg 8 Jul 4 08:43 bipet -> /bip/etl
However, I cannot refer to the soft link bipet while I try to search a specific file in the concerned folder.
ls -lrt /bipetl/227015716_WLR3PSTN_Filtered_06202016_6of6.csv.gz.gpg
ls: /bipetl/227015716_WLR3PSTN_Filtered_06202016_6of6.csv.gz.gpg: No such file or directory
What am I doing wrong here?
You created a link bipet in directory /bip/etl (current working directory when you did ln).
You you should do:
ls -lrt /bip/etl/bipetl/227015716_WLR3PSTN_Filtered_06202016_6of6.csv.gz.gpg
Or create the link using (assuming you have privileges to write to the /):
ln -s /bip/etl /bipet
And then you can do:
ls -lrt /bipetl/227015716_WLR3PSTN_Filtered_06202016_6of6.csv.gz.gpg
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I have the bzcat command with the following permissions:
-rwxr-xr-x 3 root root 39144 Sep 5 2019 /bin/bzcat
I have the bzmore command with the following permissions:
-rwxr-xr-x 1 root root 1297 Sep 5 2019 /bin/bzmore
And I have a bz2 zip file:
-r-------- 1 root root 61 Feb 17 14:37 flag.bz2
When I use bzmore flag.bz2 it works fine. But when I use bzcat flag.bz2 I get Can't open input file flag.bz2: Permission denied.. Why is this? If I put sudo in front of bzcat it works fine too but why doesn't bzmore need this when they both have the same permissions.
I tried running all the commands to test out what happened.
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It seems that both /bin/ls and /usr/bin/ls have the same inode (and the same sha-256 hash) but the number of hard links displayed by ls -li /usr/bin/ls /bin/ls is 1 instead of 2 :
user#debian:~/Documents/Unix$ ls -lai /usr/bin/ls /bin/ls
8258848 -rwxr-xr-x 1 root root 147176 24 sept. 2020 /bin/ls
8258848 -rwxr-xr-x 1 root root 147176 24 sept. 2020 /usr/bin/ls
Could somebody explain me what I have misunderstood?
I was puzzled for a while by this too, until I discovered this:
$ ls -ld /bin
lrwxrwxrwx 1 root root 7 May 31 02:39 /bin -> usr/bin
So /bin is just a symlink to /usr/bin, and there is really only one link to the file.
There are no differences between ls utility. You should know the difference is only between /bin and /usr/bin directories. /bin directory contains all programs that are used by system admin and all others users. /bin directory we can access whenever we want, but /usr/bin is accessible only for users that are locally logged.
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Ubuntu 20 LTS, Installed laradock,
in Ubuntu
$ pwd
/root/Docker
$ ls
blog laradock
$ rsync -a /media/sf_code/blog . && chmod -R 755 blog
$ cd laracock
$ docker-compose exec --user=root workspace bash
in docker
> ll
total 20
drwxr-xr-x 4 laradock laradock 4096 Nov 12 06:52 ./
drwxr-xr-x 1 root root 4096 Nov 12 02:30 ../
drwxr-xr-x 12 root 998 4096 Nov 12 03:09 blog/
drwxr-xr-x 74 laradock laradock 4096 Nov 12 06:35 laradock/
what does 998 mean?
The 4th column is the group id. It there is an entry in /etc/group with this id, then the group name will be printed otherwise the id.
The your example the group id of folder blog is 998 but no group exist inside the container with this id. Mapping a folder to a docker container does not change owner or group.
Some explanation can be found here
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I have a problem on my server.
When i try to start my server, it says that "there no left space on device"
If i execute the command "df", I see that on directory if full.
/dev/mapper/owegdc_vg-owegdc_logs_lv
10321208 9797004 0 100% /opt/application/owegdc/logs
When i get to the logs directory here what i see
ls -lrta
total 368
drwxr-x--- 2 oweadm grpowe 16384 Jan 15 2014 lost+found
drwxr-x--- 7 oweadm grpowe 4096 Jun 18 11:55 .
drwxr-xr-x 2 oweadm grpowe 12288 Aug 4 10:20 apache
drwxr-xr-x 2 oweadm grpowe 4096 Aug 5 00:56 batches
drwxr-xr-x 2 oweadm grpowe 4096 Sep 10 13:43 expl
drwxr-xr-x 2 oweadm grpowe 327680 Sep 10 13:50 jonas
drwxr-xr-x 11 oweadm grpowe 4096 Sep 10 13:50 ..
du -sk
9642792 .
I tried things like 'lsof' but it didn't work...
Do you have an idea ?
Thx
You could just try something like
du | sort -h -r
That would list the directories on your disk, ordered by their size descending. The first directory in the output list is the biggest one.
Better, if you're looking for large single files instead of a directory, this answer on Unix & Linux gives useful information, especially this:
find . -type f | xargs du -h | sort -rn
The output is the same, but it lists files instead of dirs.
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I'm looking at the output of find . -ls. For example, here is a small excerpt for /lib64 on a CentOS system:
163542 28 -rwxr-xr-x 1 root root 28448 Aug 4 2010 ./libvolume_id.so.0.66.0
163423 0 lrwxrwxrwx 1 root root 16 Mar 3 2010 ./libwrap.so.0 -> libwrap.so.0.7.6
163601 0 lrwxrwxrwx 1 root root 11 Nov 9 2010 ./libc.so.6 -> libc-2.5.so
The find(1) man page says "list current file in ls -dils format on standard output". I then tried to figure it out from ls(1) man page, but I'm stumped on the second column. Any idea?
For reference: the columns (with ref. for the first line) are:
inode 163542
??? 28 what is this? stat that file doesn't mention any field equals to '28'
permissions -rwxr-xr-x
hard-links 1
owner root
group root
size(bytes) 28448
modified Aug 4 2010
name ./libvolume_id.so.0.66.0
(for logical links: -> softlink)
Doh, a casual regression against size reveals that it's roughly the number of 1024-byte blocks...