I am trying to understand how
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
worked and what the pieces mean. The man page I read just confused me more and I tried the info sai Id but had no idea how to work it! I'm pretty new to Linux. Debian is my first distro but seemed like a rather logical place to start as it is a root of many others and has been around a while so probably is doing stuff well and fairly standardized. I am running Wheezy 64 bit as fyi if needed.
The sed command is a stream editor, reading its file (or STDIN) for input, applying commands to the input, and presenting the results (if any) to the output (STDOUT).
The general syntax for sed is
sed [OPTIONS] COMMAND FILE
In the shell command you gave:
sed 's/\^\[/\o33/g;s/\[1G\[/\[27G\[/' /var/log/boot
the sed command is s/\^\[/\o33/g;s/\[1G\[/\[27G\[/' and /var/log/boot is the file.
The given sed command is actually two separate commands:
s/\^\[/\o33/g
s/\[1G\[/\[27G\[/
The intent of #1, the s (substitute) command, is to replace all occurrences of '^[' with an octal value of 033 (the ESC character). However, there is a mistake in this sed command. The proper bash syntax for an escaped octal code is \nnn, so the proper way for this sed command to have been written is:
s/\^\[/\033/g
Notice the trailing g after the replacement string? It means to perform a global replacement; without it, only the first occurrence would be changed.
The purpose of #2 is to replace all occurrences of the string \[1G\[ with \[27G\[. However, this command also has a mistake: a trailing g is needed to cause a global replacement. So, this second command needs to be written like this:
s/\[1G\[/\[27G\[/g
Finally, putting all this together, the two sed commands are applied across the contents of the /var/log/boot file, where the output has had all occurrences of ^[ converted into \033, and the strings \[1G\[ have been converted to \[27G\[.
Related
Write a wildcard to match all files (does not matter the files are in which directory, just ask for the wildcard) named in the following rule: starts with a string “image”, immediately followed by a one-digit number (in the range of 0-9), then a non-digit char plus anything else, and ends with either “.jpg” or “.png”. For example, image7.jpg and image0abc.png should be matched by your wildcard while image2.txt or image11.png should not.
My folder contained these files imag2gh.jpeg image11.png image1agb.jpg image1.png image2gh.jpg image2.txt image5.png image70.jpg image7bn.jpg Screenshot .png
If my command work it should only display image1agb.jpg image1.png image2gh.jpg image5.png image70.jpg image7bn.jpg
This is the command I used (ls -ad image[0-9][^0-9]*{.jpg,.png}) but I'm only getting this image1agb.jpg image2gh.jpg image7bn.jpg so I'm missing (image1.png image5.png)Kali Terminal and what I did
ls -ad image[0-9][!0-9]*{.jpg,.png}
Info
Character ranges like [0-9] are usually seen in RegEx statements and such. They won't work as shell globs (wildcards) like that.
Possible solution
Pipe output of command ls -a1
to standard input of the grep command (which does support RegEx).
Use a RegEx statement to make grep filter filenames.
ls -a1|grep "image"'[[:digit:]]\+[[:alpha:]]*\.\(png\|jpg\)'
What I want:
There is a file /scripts/backup/config.cfg which contains variables. In my specific case the important ones are:
BACKUPLOCATION=""
ROOTLOCATION="/backup"
Then there is a script /scripts/backup/performBackup.sh
For a specific reason I want a part of the script do the following operations:
read the value of the variable ROOTLOCATION
add a ("/" and) timestamp (Date&Time)
safe the new created value to BACKUPLOCATION (by replacing its current value)
Example
If this is the previous state of the config.cfg:
BACKUPLOCATION="dummy"
ROOTLOCATION="/backup"
After the script ran it should be:
BACKUPLOCATION="/backup/2020-05-02-23-00"
ROOTLOCATION="/backup/"
What I tried
First of all the config file gets "loaded" using
source /scripts/backup/config.cfg
I then tried to use the sed command but the quotes are messing with me. Here is one try (which didn't work):
sed -i 's/BACKUPLOCATION\=.*/BACKUPLOCATION="'$ROOTLOCATION/$(date +%Y-%m-%d-%H-%M)'"/' /scripts/backup/config.cfg
Try this:
source /scripts/backup/config.cfg
sed -i 's|BACKUPLOCATION=.*|BACKUPLOCATION="'"$ROOTLOCATION/$(date +%Y-%m-%d-%H-%M)"'"|' /scripts/backup/config.cfg
The problem with your sed is that you use / as delimiter, which is present in $ROOTLOCATION after expansion, therefore sed fails. I used |, which is usually is not present in filenames. If you ever create a file with |, that sed will fail too! So, "know your data" :)
I'm using a shell & TCL script to login to a switch and get the output of certain commands and in some places I can see ^D coming up. I tried to use the dos2unix utility but still it didn't go away.
Eth1/37 NOM: xcvrAbsen routed auto auto --
^DEth1/38 NOM: xcvrAbsen routed auto auto --
Eth1/39 NOM: xcvrAbsen routed auto auto --
Eth101/1/45 eth 1000 NOM:NO_PATCHING CABLE
^DEth101/1/46 eth 1000 NOM:NO_PATCHING CABLE
Eth101/1/47 eth 1000 NOM:NO_PATCHING CABLE
How can this be eliminated, are there any standard tools like dos2unix which can get rid of such data?
What I'm trying to do is to compare two files which are from the same switch and the same command and the same output, but due to these ^D, Vimdiff shows it as different lines.
How to get this eliminated?
Command I'm using is something like this:
$cdir/ciscocmd -Y -u $operator -p $password -s $password -t $switch -r rfc_sa_commands | sed 's/^^D//' > $switch.$NOW
dos2unix removes carriage returns, no other control characters.
The tool to remove all occurrences of an arbitrary character is called tr.
tr -d '\004' <inputfile >outputfile
This assumes you have literal ctrl-D characters, not sequences of caret ^ and D. The tr utility cannot remove a specific sequence; it just processes individual characters. To remove a sequence, you'd need
sed 's/\^D//g' inputfile >outputfile
where the backslash is required because the caret alone has a special meaning in regular expressions (it matches beginning of line). Doubling it does not escape it; ^^ probably still just matches beginning of line, though it's not really well-defined, and could introduce apparently random behavior.
Even if the special character is visible as '^D', it may be NOT catchable like this.
Interesting readings, are:
https://en.wikipedia.org/wiki/ASCII#Character_groups
https://en.wikipedia.org/wiki/End-of-Transmission_character
I think a way to do it would be:
<your command>|sed -e 's/\x04//g'
Does it solve your issue?
I'm currently working on a Unix box and came across this post which I found helpful, in order to learn about cat command in Unix. At the bottom of the page found this line saying: -A = Equivalent to -vET
As I'm new into Unix, I'm unaware of what does this mean actually? For example lets say I've created a file called new using cat and then apply this command to the file:
cat -A new, I tried this command but an error message comes up saying it's and illegal option.
To cut short, wanted to know what does cat -A really mean and how does it effect when I apply it to a file. Any help would be appreciated.
It means show ALL.
Basically its a combination of -vET
E : It will display '$' at the end of every line.
T : It will display tab character as ^I
v : It will use ^ and M-notation
^ and M-notation:
(Display control characters except for LFD(LineFeed or NewLine) and TAB using '^' notation and precede characters that have the high bit set with
'M-') M- notation is a way to display high-bit characters as low bit ones by preceding them with M-
You should read about little-endian and big-endian if you like to know more about M notation.
For example:
!http://i.imgur.com/0DGET5k.png?1
Check your manual page as below and it will list all options avaialable with your command and check is there -A present, if it is not present it is an illegal option.
man cat
It displays non-printing characters
In Mac OS you need to use -e flag and
-e Display non-printing characters (see the -v option), and display a dollar sign (`$') at the end of each line.
I am new to Linux.
I was debugging some code. I encountered the following command:
PROGRAM_ID=$(echo $PROGRAM_ID|sed 's/-/,/g')
Can anybody explain what the g represents here?
I understand hyphen is being replaced with comma.
The /g flag means, perform the substitution globally on a line. Without that flag, only the first hyphen on every line would get substituted.
A better way with Bash would be
PROGRAM_ID=${PROGRAM_ID//-/,}
but if you have to be portable to Bourne shell in general, this replacement facility is not available.
(In which case you should take care to keep "$PROGRAM_ID" in double quotes in the echo.)
Its easy to see how g (global) works with these two example:
echo "test-one-two-three" | sed 's/-/,/g'
test,one,two,three
echo "test-one-two-three" | sed 's/-/,/'
test,one-two-three
Without the g it only replace the first hit.