I have lot of files starting with processConfig-. I wanted to rename it to processCfg-. What's the easy way to change the first part of file name to processCfg- in linux?
But I don't want to rename this file processConfig.json since it doesn't match with my prefix.
> ls -lrth
total 467
-rw-r--r-- 1 david staff 9.8K May 26 15:14 processConfig-data-1234.json
-rw-r--r-- 1 david staff 11K May 26 15:14 processConfig-data-8762.json
-rw-r--r-- 1 david staff 4.9K May 26 15:14 processConfig-dataHold-1.json
-rw-r--r-- 1 david staff 6.6K May 26 15:14 processConfig-letter.json
-rw-r--r-- 1 david staff 5.6K May 26 16:44 processConfig-data-90987.json
-rw-r--r-- 1 david staff 284K May 28 18:44 processConfig.json
Like this :
rename -n 's/^processConfig-/processCfg-/' processConfig-*.json
Remove -n switch when the output looks good to rename for real.
man rename
There are other tools with the same name which may or may not be able to do this, so be careful.
The rename command that is part of the util-linux package, won't.
If you run the following command (GNU)
$ file "$(readlink -f "$(type -p rename)")"
and you have a result that contains Perl script, ASCII text executable and not containing ELF, then this seems to be the right tool =)
If not, to make it the default (usually already the case) on Debian and derivative like Ubuntu :
$ sudo apt install rename
$ sudo update-alternatives --set rename /usr/bin/file-rename
If you don't have this command with another distro, search your package manager to install it or do it manually (no deps...)
This tool was originally written by Larry Wall, the Perl's dad.
Related
I'm unsure about some small obscure details that I worry will have large effects. On my Raspbian Debian 11 running on a Raspberry Pi, sudo apt update && sudo apt full-upgrade only updates my libxml2 library to version 2.9.10, no further. However, I need version 2.9.14 for the security patches contained within. With help from this question (thank you Esther!), I decided to compile version 2.9.14 from source. Everything went well, and the library was placed into /usr/local/lib. I then updated ldconfig by following this answer. However, although that should have made Debian use the new 2.9.14 version, apt-cache policy libxml2 still shows:
libxml2:
Installed: 2.9.10+dfsg-6.7+deb11u2
Candidate: 2.9.10+dfsg-6.7+deb11u2
Version table:
*** 2.9.10+dfsg-6.7+deb11u2 500
500 http://raspbian.raspberrypi.org/raspbian stable/main armhf Packages
100 /var/lib/dpkg/status
I think I know why this is. If I was installing a never-before-seen library, everything might have worked properly. However, since I now have a second libxml2 library without removing the 1st, any time the system needs to use libxml2, the search first reaches /usr/lib/arm-linux-gnueabihf where the old libxml2 is, so the system finds the old version, is satisfied, and so stops searching before finding the new version.
For context before I continue:
(link to below but in color: https://i.stack.imgur.com/OJLJW.png)
pi#fuelightcontrol:~ $ cd /usr/lib/arm-linux-gnueabihf/
pi#fuelightcontrol:/usr/lib/arm-linux-gnueabihf $ ls -l | grep libxml2
lrwxrwxrwx 1 root root 17 May 15 14:58 libxml2.so.2 -> libxml2.so.2.9.10
-rw-r--r-- 1 root root 1510312 May 15 14:58 libxml2.so.2.9.10
pi#fuelightcontrol:/usr/lib/arm-linux-gnueabihf $ cd /usr/local/lib
pi#fuelightcontrol:/usr/local/lib $ ls -l
total 12120
drwxr-xr-x 3 root root 4096 Jun 14 18:17 cmake
-rw-r--r-- 1 root root 7145994 Jun 14 18:17 libxml2.a
-rwxr-xr-x 1 root root 944 Jun 14 18:17 libxml2.la
lrwxrwxrwx 1 root root 17 Jun 14 18:17 libxml2.so -> libxml2.so.2.9.14
lrwxrwxrwx 1 root root 17 Jun 14 18:17 libxml2.so.2 -> libxml2.so.2.9.14
-rwxr-xr-x 1 root root 5242072 Jun 14 18:17 libxml2.so.2.9.14
drwxr-xr-x 2 root root 4096 Jun 14 18:17 pkgconfig
drwxr-xr-x 3 root root 4096 Jun 13 21:43 python3.9
-rw-r--r-- 1 root root 205 Jun 14 18:17 xml2Conf.sh
pi#fuelightcontrol:/usr/local/lib $
The question is, what would be the best way to go about fixing the problem of the old version still being used by apt-cache policy libxml2 and other programs? I could:
Just delete /usr/lib/arm-linux-gnueabihf/libxml2.so.2.9.10 (the old one) and its symbolic link, so the system keeps searching past that point and eventually finds /usr/local/lib/libxml2.so.2.9.14 (the new one). However, something feels... off about having my libraries scattered around in different directories. My gut tells me to keep them in one place. Also, see paragraph below the next list item.
I could delete /usr/lib/arm-linux-gnueabihf/libxml2.so.2.9.10 (the old one) and its symbolic link, then move the new version into /usr/lib/arm-linux-gnueabihf to replace the old version. However, there's more libxml2 related files and 1 more symbolic link in /usr/local/lib that are not present in /usr/lib/arm-linux-gnueabihf. Do I need to move those too, or should I just move libxml2.so.2.9.14 and one (both?) of the symbolic links? If only 1 link, which?
Should I delete the files left behind after I move the required ones over? Also, see paragraph below.
What concerns me about deleting anything is if some other script comes looking for libxml2.2.9.10, can't find it, and fails. I don't know how to tell the rest of the programs that libxml2's filename is different now. I suppose both options 1 and 2 might work, but is one option a cleaner, smarter idea? I'm trying to save myself some work in the future.
Sorry this is such a small silly question. Thank you for your help!
Edit: After making backups of both directories, I tried option 1 first, then option 2. Neither changed the output of apt-cache policy libxml2 - it still says I have libxml2 2.9.10 installed, even though I deleted /usr/lib/arm-linux-gnueabihf/libxml2.so.2.9.10 and its symbolic link, rebooted, and ran sudo apt update
Here's how I updated ldconfig (same as the second link), to clear up loose ends. The link to /usr/local/lib was done for me already, which was nice.
Link to screenshot of below but in color: https://i.stack.imgur.com/7w6XR.png
pi#fuelightcontrol:/etc $ ls -l ld.so.conf
ld.so.conf ld.so.conf.d/
pi#fuelightcontrol:/etc $ cat ld.so.conf
include /etc/ld.so.conf.d/*.conf
pi#fuelightcontrol:/etc $ ls -l ld.so.conf.d
total 16
-rw-r--r-- 1 root root 12 Dec 1 2021 00-vmcs.conf
-rw-r--r-- 1 root root 109 May 14 2019 arm-linux-gnueabihf.conf
-rw-r--r-- 1 root root 41 Jun 25 2018 fakeroot-arm-linux-gnueabihf.conf
-rw-r--r-- 1 root root 44 Jun 14 19:08 libc.conf
pi#fuelightcontrol:/etc $ cat ld.so.conf.d/libc.conf
# libc default configuration
/usr/local/lib
pi#fuelightcontrol:/etc $ sudo ldconfig
pi#fuelightcontrol:/etc $ sudo ldconfig /usr/local/lib
pi#fuelightcontrol:/etc $ sudo ldconfig -n /usr/local/lib
pi#fuelightcontrol:/etc $ cat ld.so.conf.d/00-vmcs.conf
/opt/vc/lib
pi#fuelightcontrol:/etc $ cat ld.so.conf.d/arm-linux-gnueabihf.conf clear
# Multiarch support
/usr/local/lib/arm-linux-gnueabihf
/lib/arm-linux-gnueabihf
/usr/lib/arm-linux-gnueabihf
cat: clear: No such file or directory
pi#fuelightcontrol:/etc $ cat ld.so.conf.d/fakeroot-arm-linux-gnueabihf.conf
/usr/lib/arm-linux-gnueabihf/libfakeroot
pi#fuelightcontrol:/etc $
There are two directories that contains these files:
First one /usr/local/nagios/etc/hosts
[root#localhost hosts]$ ll
total 12
-rw-rw-r-- 1 apache nagios 1236 Feb 7 10:10 10.80.12.53.cfg
-rw-rw-r-- 1 apache nagios 1064 Feb 27 22:47 10.80.12.62.cfg
-rw-rw-r-- 1 apache nagios 1063 Feb 22 12:02 localhost.cfg
And the second one /usr/local/nagios/etc/services
[root#localhost services]$ ll
total 20
-rw-rw-r-- 1 apache nagios 2183 Feb 27 22:48 10.80.12.62.cfg
-rw-rw-r-- 1 apache nagios 1339 Feb 13 10:47 Check usage _etc.cfg
-rw-rw-r-- 1 apache nagios 7874 Feb 22 11:59 localhost.cfg
And I have a script that goes through file in Hosts directory and paste some lines from that file in the file in the Services directory.
The script is ran like this:
./nagios-contacts.sh /usr/local/nagios/etc/hosts/10.80.12.62.cfg /usr/local/nagios/etc/services/10.80.12.62.cfg
How can I achieve that another script calls my script and goes through every file in the Hosts directory and does its job for the files with the same name in the Service directory?
In my script I´m pulling out contacts from the 10.80.12.62.cfg in the Hosts directory and appending them to the file with the same name in the Service directory.
Don't use ls output as an input to for loop instead use the built-in wild-cards. See why it's not a good idea.
for f in /usr/local/nagios/etc/hosts/*.cfg
do
basef=$(basename "$f")
./nagios-contacts.sh "$f" "/usr/local/nagios/etc/services/${basef}"
done
It sounds like you just need to do some iteration.
echo $(pwd)
for file in $(ls); do ./nagious-contacts.sh $file; done;
So it will loop over all files in the current directory.
You can also modify it as well by doing something more absolute.
abspath=$1
for file in $(ls $abspath); do ./nagious-contacts.sh $abspath/$file; done
which would loop over all files in a set directory, and then pass the abspath/filename into your script.
Aim is to run the file sendData.c with passing argument. But it shows file not found while it is already there. Let me know where it is going wrong.
root#OpenWrt:/tmp/sendData# ls -l
-rw-r--r-- 1 root root 75 Dec 19 07:02 Makefile
-rw-r--r-- 1 root root 5627 Dec 18 07:33 sendData.c
-rw-r--r-- 1 root root 13162 Dec 18 07:33 send_Data
-rw-r--r-- 1 root root 10744 Dec 20 07:46 send_Data_loop
root#OpenWrt:/tmp/sendData# ./sendData wlan0 E8:DE:27:C3:E6:07
-ash: ./sendData: not found
The file you want to use is called differently
# ./send_Data wlan0 E8:DE:27:C3:E6:07
I think you have a confusing naming scheme, because your .c file is named without and underscore
SOLVED
Did not execute the "make" for which the sendData file was not generated in the first place.
Firstly, make your file executable:
chmod +x <filename> and run command again.
I have some problem with the Linux grep command, it don't work !!!
I am trying the following test on my Ubuntu system:
I have create the following folder: /home/andrea/Scrivania/prova
Inside this folder I have created a txt file named prova.txt and inside this file I have write the string test and I have save it
In the shell I have first access the folder /home/andrea/Scrivania/prova and so I have launched the grep command in the following way:
~/Scrivania/prova$ grep test
The problem is that the cursor continues to blink endlessly and cannot find NOTHING! Why? What is the problem?
You've not provided files for the grep command to scan
grep "test" *
or for recursive
grep -r "test" *
Because grep searches standard input if no files are given. Try this.
grep test *
You are not running the command you were looking for.
grep test * will look for test in all files in your current directory.
grep test prova.txt will look for test specifically in prova.txt
(grep test will grep the test string in stdin, and will not return until EOF.)
You need to pipe in something to grep - you cant just call grep test without any other arguments as it is actually doing nothing. try grep test *
Another use for grep is to pipe in a command
e.g. This is my home directory:
drwx------+ 3 oliver staff 102 12 Nov 21:57 Desktop
drwx------+ 10 oliver staff 340 17 Nov 18:34 Documents
drwx------+ 17 oliver staff 578 20 Nov 18:57 Downloads
drwx------# 12 oliver staff 408 13 Nov 20:53 Dropbox
drwx------# 52 oliver staff 1768 11 Nov 12:05 Library
drwx------+ 3 oliver staff 102 12 Nov 21:57 Movies
drwx------+ 5 oliver staff 170 17 Nov 10:40 Music
drwx------+ 3 oliver staff 102 20 Nov 19:17 Pictures
drwxr-xr-x+ 4 oliver staff 136 12 Nov 21:57 Public
If i run
l | grep Do
I get the result
drwx------+ 10 oliver staff 340 17 Nov 18:34 Documents
drwx------+ 17 oliver staff 578 20 Nov 18:57 Downloads
remember to pipe the grep command
From grep man page:
Grep searches the named input FILEs (or standard input if no files
are
named, or the file name - is given) for lines containing a match to the
given PATTERN.
If you don't provide file name(s) for it to use, it will try to read from stdin.
Try grep test *
As per GNU Grep 3.0
A file named - stands for standard input. If no input is specified,
grep searches the working directory . if given a command-line
option specifying recursion; otherwise, grep searches standard input.
So for OP's command, without any additional specification, grep tries to search in standard input, which is not actually provided there.
A simple approach is grep -r [pattern], as per the above, to specify recursion with -r and search in current directory and sub-directories.
Also note that wildcard * only includes files, not directories. If used, a prompt might be shown for hint:
grep: [directory_name]: Is a directory
I know this is really basic, but I cannot find this information
in the ls man page, and need a refresher:
$ ls -ld my.dir
drwxr-xr-x 1 smith users 4096 Oct 29 2011 my.dir
What is the meaning of the number 1 after drwxr-xr-x ?
Does it represent the number of hard links to the direcory my.dir?
I cannot remember. Where can I find this information?
Thanks,
John Goche
I found it on Wikipedia:
duuugggooo (hard link count) owner group size modification_date name
The number is the hard link count.
If you want a more UNIXy solution, type info ls. This gives more detailed information including:
`-l'
`--format=long'
`--format=verbose'
In addition to the name of each file, print the file type, file
mode bits, number of hard links, owner name, group name, size, and
timestamp (*note Formatting file timestamps::), normally the
modification time. Print question marks for information that
cannot be determined.
That is the number of named (hard links) of the file. And I suppose, there is an error here. That must be at least 2 here for a directory.
$ touch file
$ ls -l
total 0
-rw-r--r-- 1 igor igor 0 Jul 15 10:24 file
$ ln file file-link
$ ls -l
total 0
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file-link
$ mkdir a
$ ls -l
total 0
drwxr-xr-x 2 igor igor 40 Jul 15 10:24 a
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file
-rw-r--r-- 2 igor igor 0 Jul 15 10:24 file-link
As you can see, as soon as you make a directory, you get 2 at the column.
When you make subdirectories in a directory, the number increases:
$ mkdir a/b
$ ls -ld a
drwxr-xr-x 3 igor igor 60 Jul 15 10:41 a
As you can see the directory has now three names ('a', '.' in it, and '..' in its subdirectory):
$ ls -id a ; cd a; ls -id .; ls -id b/..
39754633 a
39754633 .
39754633 b/..
All these three names point to the same directory (inode 39754633).
Trying to explain why for directory the initial link count value =2.
Pl. see if this helps.
Any file/directory is indentified by an inode.
Number of Hard Links = Number of references to the inode.
When a directory/file is created, one directory entry (of the
form - {myname, myinodenumber}) is created in the parent directory.
This makes the reference count of the inode for that file/directory =1.
Now when a directory is created apart from this the space for directory is also created which by default should be having two directory entries
one for the directory which is created and another for the
parent directory that is two entries of the form {., myinodenumber}
and {.., myparent'sinodenumber}.
Current directory is referred by "." and the parent is referred by ".." .
So when we create a directory the initial number of Links' value = 1+1=2,
since there are two references to myinodenumber. And the parent's number
of link value is increased by 1.