I'm running an awk command the prints out an output with a ":" in the result. How can I remove that? Is there a way to do the whole awk command in one?
The command I'm running is:
fdisk -l | '/Disk/{print $2
Which gives:
/dev/sda:
Thanks
This should do the trick:
fdisk -l | awk -F'[ :]+' '/^Disk \// {print $2}'
/dev/sda
Explanation:
-F'[ :]+' sets the field Separator to a space or colon, as long as there are more than one.
And I match /^Disk \/, to prevent some false positives (the forward slash needs to be escaped by a backslash).
For a list of all /dev/{disks} you can try using lsblk with the -o {flags}, and you don't need to be SU either... N.B. 'PATH' is a column header
lsblk -o PATH
That'll give you all disks and partitions (including loop partitions) as you'll see from "fdisk -l".
There's a lot more information that 'lsblk -o {flags}' will give you... Try this one for fun (and google/man for more)...
lsblk -o NAME,LABEL,PATH,MOUNTPOINT
Related
I have got the following URL:
https://xcg5847#git.rz.bankenit.de/scm/smat/sma-mes-test.git
I need to pull out smat-mes-test and smat:
git config --local remote.origin.url|sed -n 's#.*/\([^.]*\)\.git#\1#p'
sma-mes-test
This works. But I also need the project name, which is smat
I am not really familiar to complex regex and sed, I was able to find the other command in another post here. Does anyone know how I am able to extract the smat value here?
With your shown samples please try following awk code. Simple explanation would be, setting field separator(s) as / and .git for all the lines and in main program printing 3rd last and 3nd last elements from the line.
your_git_command | awk -F'/|\\.git' '{print $(NF-2),$(NF-1)}'
Your sed is pretty close. You can just extend it to capture 2 values and print them:
git config --local remote.origin.url |
sed -E 's~.*/([^/]+)/([^.]+)\.git$~\1 \2~'
smat sma-mes-test
If you want to populate shell variable using these 2 values then use this read command in bash:
read v1 v2 < <(git config --local remote.origin.url |
sed -E 's~.*/([^/]+)/([^.]+)\.git$~\1 \2~')
# check variable values
declare -p v1 v2
declare -- v1="smat"
declare -- v2="sma-mes-test"
Using sed
$ sed -E 's#.*/([^/]*)/#\1 #' input_file
smat sma-mes-test.git
I would harness GNU AWK for this task following way, let file.txt content be
https://xcg5847#git.rz.bankenit.de/scm/smat/sma-mes-test.git
then
awk 'BEGIN{FS="/"}{sub(/\.git$/,"",$NF);print $(NF-1),$NF}' file.txt
gives output
smat sma-mes-test
Explanation: I instruct GNU AWK that field separator is slash character, then I replace .git (observe that . is escaped to mean literal dot) adjacent to end ($) in last field ($NF), then I print 2nd from end field ($(NF-1)) and last field ($NF), which are sheared by space, which is default output field separator, if you wish to use other character for that purpose set OFS (output field separator) in BEGIN. If you want to know more about NF then read 8 Powerful Awk Built-in Variables – FS, OFS, RS, ORS, NR, NF, FILENAME, FNR
(tested in gawk 4.2.1)
Why not sed 's!.*/\(.*/.*\)!\1!'?
string=$(config --local remote.origin.url | tail -c -21)
var1=$(echo "${string}" | cut -d'/' -f1)
var2=$(echo "${string}" | cut -d'/' -f2 | sed s'#\.git##')
If you have multiple urls with variable lengths, this will not work, but if you only have the one, it will.
var1=smat
var2=sma-mes-test.git
If I did have something variable, personally I would replace all of the forward slashes with carriage returns, throw them into a file, and then export the last and second last lines with ed, which would give me the two last segments of the url.
Regular expressions literally give me a migraine headache, but as long as I can get everything on its' own line, I can quite easily bypass the need for them entirely.
I am trying to print all files in /usr/bin/ where the filename starts with a v. This works,
ls -lA /usr/bin/ | awk '{print $9}' | grep ^v
Surprisingly, this returns no output,
ls -lA /usr/bin/ | awk '/^v/ {print $9}'.
I don't understand the difference. I am running Ubuntu 21.10 with awk -W version saying that it is on 1.3.4 20200120.
Edit: I understand that awk may not be the best way to accomplish what I am wanting to do here. But, this is an exercise in learning awk by testing my understanding via comparing it to the real output.
The difference between the two pipelines is that the first outputs the 9th column and then check to see if that starts with a v the second checks to see if the line starts with a v, change the second to:
$ ls -lA /usr/bin/ | awk '$9 ~ /^v/ {print $9}'
When writing:
/pattern/ { ... }
it's the same as writing
$0 ~ /pattern/ { ... }
but in your case you want to compare the 9th column, so write that instead.
But you really don't want to create a pipeline for this, and what would happen if your files contain a space?
You can consider using find or globs instead:
$ printf '%s\n' /usr/bin/v*
/usr/bin/vi
/usr/bin/view
...
or
$ find /usr/bin -name 'v*' -print
/usr/bin/vi
/usr/bin/view
...
cat file1.txt | awk -F '{print $1 "|~|" $2 "|~|" $3}' > file2.txt
I am using above command to filter first three columns from file1 and put into file.
But only getting the column names and not the column data.
How to do that?
|~| - is the delimiter.
file1.txt has values as :
a|~|b|~|c|~|d|~|e
1|~|2|~|3|~|4|~|5
11|~|22|~|33|~|44|~|55
111|~|222|~|333|~|444|~|555
my expedted output is :
a|~|b|~|c
1|~|2|~|3
11|~|22|~|33
111|~|222|~|333
With your shown samples, please try following awk code. You need to set field separator to |~| and remove starting space from lines, then print the lines.
awk -F'\\|~\\|' -v OFS='|~|' '{sub(/^[[:blank:]]+/,"");print $1,$2,$3}' Input_file
In case you want to keep spaces(which was in initial post before edit) then try following:
awk -F'\\|~\\|' -v OFS='|~|' '{print $1,$2,$3}' Input_file
NOTE: Had a chat with user in room and got to know why this code was not working for user because of gunzip -c file was being used wrongly, its output was being saved into a variable on which user was running awk program, so correcting that command generated right file and awk program ran fine on it. Adding this as a reference for future readers.
One approach would be:
awk -v FS="," -v OFS="|~|" '{gsub(/[|][~][|]/,","); sub(/^\s*/,""); print $1,$2,$3}' file1.txt
The approach simply replaces all "|~|" with a "," setting the output file separator to "|~|". All leading whitespace is trimmed with sub().
Example Use/Output
With your data in file1.txt, you would have:
$ awk -v FS="," -v OFS="|~|" '{gsub(/[|][~][|]/,","); sub(/^\s*/,""); print $1,$2,$3}' file1.txt
a|~|b|~|c
1|~|2|~|3
11|~|22|~|33
111|~|222|~|333
Let me know if this is what you intended. You can simply redirect, e.g. > file2.txt to write to the second file.
For such cases, my bash+awk script rcut comes in handy:
rcut -Fd'|~|' -f-3 ip.txt
The -F option enables fixed string input delimiter (which is given using the -d option). And by default, the output field separator will also be same as -d when -F is active. -f-3 is similar to cut syntax to specify first three fields.
For better speed, use hck command:
hck -Ld'|~|' -D'|~|' -f-3 ip.txt
Here, -L enables literal field separator and -D specifies output field separator.
Another benefit is that hck supports -z option to automatically handle common compressed formats based on filename extension (adding this since OP had an issue with compressed input).
Another way:
sed 's/|~|/\t/g' file1.txt | awk '{print $1"|~|"$2"|~|"$3}' > file2.txt
First replace the |~| delimiter, and use the default awk separator, then print columns what you need.
..And I know why:
I have a xml document with lots of information inside. I need to extract what I need and eventually print them on a new file.
The xml (well, part of it.. rows just keeps repeating)
<module classname="org.openas2.processor.receiver.AS2DirectoryPollingModule"
outboxdir="%home%/../../../home/samba/user/Outbound/toMartha/"
errordir="%home%/../../../home/samba/user/Outbound/toMartha/error"
sentdir="%home%/../../../home/samba/user/data/Sent/Martha"
interval="600"
defaults="sender.name=me_myself, receiver.name=Martha"
sendfilename="true"
mimetype="application/standard"/>
<module classname="org.openas2.processor.receiver.AS2DirectoryPollingModule"
outboxdir="%home%/../../../home/samba/user/Outbound/toJosh/"
errordir="%home%/../../../home/samba/user/Outbound/toJosh/error"
sentdir="%home%/../../../home/samba/user/data/Sent/Josh"
interval="600"
defaults="sender.name=me_myself, receiver.name=Josh"
sendfilename="true"
mimetype="application/standard"/>
<module classname="org.openas2.processor.receiver.AS2DirectoryPollingModule"
outboxdir="%home%/../../../home/samba/user/Outbound/toPamela/"
errordir="%home%/../../../home/samba/user/Outbound/toPamela/error"
interval="600"
defaults="sender.name=me_myself, receiver.name=Pamela"
sendfilename="true"
mimetype="application/standard"/>
I need to extract the folder after "Outbound" and clean it from quotes or slashes.
Also, I need to exclude the "/error" so I get only 1 result for each of them.
My command is:
grep -o -v "/error" "Outbound/" config.xml | awk -F"Outbound/" '{print $2}' | sed -e "s/\/\"//g" > /tmp/sync_users
The error is: grep: Outbound/: No such file or directory which of course means that I'm giving to grep too many arguments (?) - If i remove the -v "/error" it would work but would print also the names with "/error".
Can someone help me?
EDIT:
As some pointed out in their example (thanks for the time you put in), I'd need to extract these words based on the sample above:
toMartha
toJosh
toPamela
could be intersting to use sed in this case
sed -e '\#/Outbound/#!d' -e '\#/error"$#d' -e 's#.*/Outbound/##;s#/\{0,1\}"$##' Config.xml
awk version, assuming (for last print) that your line is always 1 folder below Outbound as shown
awk -F '/' '$0 !~ /\/Outbound\// || /\/error"$/ {next} {print $(NF-1)}' Config.xml
Loose the grep altogether:
$ awk '/outboxdir/{gsub(/^.+Outbound\/|\/" *\r?$/,""); print}' file
toMartha
toJosh
toPamela
/^outboxdir/ /outboxdir/only process records that have start with outboxdir on them
gsub remove unwanted parts of the record
added space removal at the end of record and CRLF fix for Windows originated files
To give grep multiples patterns they have to be separated by newlines or specified by multiples pattern option (-e, F,.. ). However -v invert the match as a whole, you can't invert only one.
For what you're after you can use PCRE (-P argument) for the lookaround ability:
grep -o -P '(?<=Outbound\/)[^\/]+(?!.*\/error)' config.xml
Regex demo here
The regex try to
match something not a slash at least once, the [^\/]+
preceded by Outbound/ the positive lookbehind (?<=Outbound\/)
and not followed by something ending with /error, the negative lookahead (?!.*\/error)
With your first sample input:
$ grep -o -P '(?<=Outbound\/)[^\/]+(?!.*\/error)' test.txt
toMartha
toJosh
toPamela
How about:
grep -i "outbound" your_file | awk -F"Outbound/" '{print $2}' | sed -e 's/error//' -e 's/\/\"//' | uniq
Should work :)
You can use match in gawkand capturing group in regex
awk 'match($0, /^.*\/Outbound\/([^\/]+)\/([^\/]*)\/?"$/, a){
if(a[2]!="error"){print a[1]}
}' config.xml
you get,
toMartha
toJosh
toPamela
grep can accept multiple patterns with the -e option (aka --regexp, even though it can be used with --fixed-strings too, go figure). However, -v (--invert-match) applies to all of the patterns as a group.
Another solution would be to chain two calls to grep:
grep -v "/error" config.xml | grep "Outbound/" | awk -F"Outbound/" '{print $2}' | sed -e "s/\/\"//g"
Why doesn't this work?
x=5
$ ls -l | awk '{print $(($x))}'
should print field 5 of ls -l command, right?
The only ways you should pass in the value of shell variable to awk are the following
$ x=5
$ ls -l | awk -v x="$x" '{print $x}'
$ ls -l | awk '{print $x}' x="$x"
The main difference between these two methods is that by using -v the value of x is set in the BEGIN block whilst the second method the value would not be set. All other methods with quoting tricks or escaping should not be used unless you like headaches.
However you don't want to being parsing ls at all, the command you really want is:
stat --printf="%s\n" *
Assuming the fifth column of your ls is the same as mine, this will display all the file sizes in the current directory.
You could access the shell variable something similar to these;
The first way is not suggested!
x=5
ls -l | awk '{print $'$x'}'
or assigning the value x to the variable shellVar, before execution of the program begins
x=5
ls -l | awk -v shellVar="$x" '{print $shellVar}'
or using an array containing the values of the current environment
export x=5
ls -l | awk '{print $ENVIRON["x"]}'
That's a shell variable, which is not expanded by the shell in single quotes. The reason we put awk scripts in single quotes is precisely to prevent the shell from interpreting things meant for awk's benefit and screwing things up, but sometimes you want the shell to interpret part of it.
For something like this, I prefer to pass the value in as an awk variable:
ls -l | awk -v "x=$x" '{print $x}'
but you could do any number of other ways. For instance, this:
ls -l | awk '{print $'$x'}'
which should really be this:
ls -l | awk '{print $'"$x"'}'
alternatively, this:
ls -l | awk "{print \$$x}"
Try this :
ls -l | awk '{print $'$x'}'