I'm building a script in bash for use on Linux, and I use the output of a executable to fill parameters:
version=$("${path_exec}" -version | awk '{if($1=="kernel" && $2=="release") print $3}')
patch=$("${path_exec}" -version | awk '{if($1=="patch" && $2=="number") print $3}')
This will run the executable defined in "path_exec" twice, which is time consuming. Is there a way to assign the version and path variable with a value using only one execution of "path_exec"?
An example of what I've tried to tackle this is shown below, but I don't think this will do what I want:
${path_hostexec} -version | awk '{if($1=="kernel" && $2=="release") {version_agent = $3;} else if($1=="patch" && $2=="number") {patch_agent = $3;}}'
Could you please try following. Since I didn't have output of path_exec command so couldn't test it.
myarr=($("${path_exec}" -version | awk '{if($1=="kernel" && $2=="release");val=$3} {if($1=="patch" && $2=="number") print val,$3}'))
#Now get every element in the array
for i in "${myarr[#]}"
do
echo $i
done
What I have done is:
Merged your both awk programs into one to make it run on a single time.
Now creating an array by output of awk command output(which should be val1 val2 as an example format)
Once an array created then we could get its all values by for loop or you could get its specific value by mentioning its index eg--> myarr[1] to print 2nd element.
Output both values on a single line, and let read separate the line into its two parts.
IFS=, read version patch < <($path_exec -version |
awk '/kernel release/ {v=$3}
/patch number/ {p=$3}
END {print v","p}
')
Thanks guys, I've managed to get it working thanks to your input:
version_patch_agent=$("${path_hostexec}" -version | awk '/kernel release/ {v=$3} /patch number/ {p=$3} END {print v" patch "p}')
This puts the version and patch number into a variable that I can just echo to get the info on the screen. Thanks again all!!
Related
I'm trying to use "sort -V" command (aka version-sort) in a sh file.
Specifically, I have the following line of code in a sh file:
SOME_PATH="$(ls dir_1/dir_2/v*/filename.txt | sort -V | tail -n1)"
What I'm trying to accomplish through the above command is that given a list of file paths with different version numbers, I want to get the file path with the greatest version number.
For example, let's assume that I have the following list of file paths:
dir_1/dir_2/v1/filename.txt,
dir_1/dir_2/v2/filename.txt,
dir_1/dir_2/v11/filename.txt
Then, I want the command to return dir_1/dir_2/v11/filename.txt instead of dir_1/dir_2/v2/filename.txt since the former has the greatest version value, "11".
From my understanding the above linux command precisely accomplishes this.
I confirmed it working on the Linux bash terminal.
However, when I run a sh file with the above command in it, I'm getting a
"ERROR: Unknown command line flag 'V'" error message.
Is there a way to make version-sort work in a sh file?
If not, is there a way to implement it not using -V flag?
Thank you.
Using shell's printf and awk:
SOME_PATH=$(printf %s\\0 dir_1/dir_2/v*/filename.txt |
awk 'BEGIN{FS="/";RS="\0";v=0}{match($3,/v([[:digit:]]+)/,m);if(m[1]>v){v=m[1];l=$0}}END{print l}')
Using awk only:
SOME_PATH=$(awk 'BEGIN{delete ARGV[0];v=0;for(i in ARGV){split(ARGV[i],s,"/");match(s[3],/v([[:digit:]]+)/,m);if(m[1]>v){v=m[1];l=ARGV[i]}}}END{print l}' dir_1/dir_2/v*/filename.txt)
Formatted awk script:
#!/usr/bin/env -S awk -f
BEGIN {
delete ARGV[0]
v=0
for (i in ARGV) {
split(ARGV[i], s, "/")
match(s[3], /v([[:digit:]]+)/, m)
if (m[1]>v) {
v=m[1]
l=ARGV[i]
}
}
}
END {
print l
}
Using a null delimited list stream, and not parsing the output of ls 1:
SOME_PATH=$(
printf '%s\0' dir_1/dir_2/v*/filename.txt |
sort -z -t'/' -k3V |
tail -zn1 |
tr -d '\0'
)
How it works:
printf '%s\0' dir_1/dir_2/v*/filename.txt: Expands the paths into a null delimited stream output.
sort -z -t'/' -k3V: Sorts the null delimited input stream on -k3V version number from the 3rd column, -t'/' using / as a delimiter.
tail -zn1: Outputs the least null delimited entry from the input stream.
tr -d '\0': Trim-out any remaining null to prevent the shell from complaining with error: warning: command substitution: ignored null byte in input.
StackExchange: Why not parse ls (and what to do instead)?
My input.csv file is semicolon separated, with the first line being a header for attributes. The first column contains customer numbers. The function is being called through a script that I activate from the terminal.
I want to delete all lines containing the customer numbers that are entered as arguments for the script. EDIT: And then export the file as a different file, while keeping the original intact.
bash deleteCustomers.sh 1 3 5
Currently only the last argument is filtered from the csv file. I understand that this is happening because the output file gets overwritten each time the loop runs, restoring all previously deleted arguments.
How can I match all the lines to be deleted, and then delete them (or print everything BUT those lines), and then output it to one file containing ALL edits?
delete_customers () {
echo "These customers will be deleted: "$#""
for i in "$#";
do
awk -F ";" -v customerNR=$i -v input="$inputFile" '($1 != customerNR) NR > 1 { print }' "input.csv" > output.csv
done
}
delete_customers "$#"
Here's some sample input (first piece of code is the first line in the csv file). In the output CSV file I want the same formatting, with the lines for some customers completely deleted.
Klantnummer;Nationaliteit;Geslacht;Title;Voornaam;MiddleInitial;Achternaam;Adres;Stad;Provincie;Provincie-voluit;Postcode;Land;Land-voluit;email;gebruikersnaam;wachtwoord;Collectief ;label;ingangsdatum;pakket;aanvullende verzekering;status;saldo;geboortedatum
1;Dutch;female;Ms.;Josanne;S;van der Rijst;Bliek 189;Hellevoetsluis;ZH;Zuid-Holland;3225 XC;NL;Netherlands;JosannevanderRijst#dayrep.com;Sourawaspen;Lae0phaxee;Klant;CZ;11-7-2010;best;tand1;verleden;-137;30-12-1995
2;Dutch;female;Mrs.;Inci;K;du Bois;Castorweg 173;Hengelo;OV;Overijssel;7557 KL;NL;Netherlands;InciduBois#gustr.com;Hisfireeness;jee0zeiChoh;Klant;CZ;30-8-2015;goed ;geen;verleden;188;1-8-1960
3;Dutch;female;Mrs.;Lusanne;G;Hijlkema;Plutostraat 198;Den Haag;ZH;Zuid-Holland;2516 AL;NL;Netherlands;LusanneHijlkema#dayrep.com;Digum1969;eiTeThun6th;Klant;Achmea;12-2-2010;best;mix;huidig;-335;9-3-1973
4;Dutch;female;Dr.;Husna;M;Hoegee;Tiendweg 89;Ameide;ZH;Zuid-Holland;4233 VW;NL;Netherlands;HusnaHoegee#fleckens.hu;Hatimon;goe5OhS4t;Klant;VGZ;9-8-2015;goed ;gezin;huidig;144;12-8-1962
5;Dutch;male;Mr.;Sieds;D;Verspeek;Willem Albert Scholtenstraat 38;Groningen;GR;Groningen;9711 XA;NL;Netherlands;SiedsVerspeek#armyspy.com;Thade1947;Taexiet9zo;Intern;CZ;17-2-2004;beter;geen;verleden;-49;12-10-1961
6;Dutch;female;Ms.;Nazmiye;R;van Spronsen;Noorderbreedte 180;Amsterdam;NH;Noord-Holland;1034 PK;NL;Netherlands;NazmiyevanSpronsen#jourrapide.com;Whinsed;Oz9ailei;Intern;VGZ;17-6-2003;beter;mix;huidig;178;8-3-1974
7;Dutch;female;Ms.;Livia;X;Breukers;Everlaan 182;Veenendaal;UT;Utrecht;3903
Try this in loop..
awk -v variable=$var '$1 != variable' input.csv
awk - to make decision based on columns
-v - to use a variable into a awk command
variable - store the value for awk to process
$var - to search for a specific string in run-time
!= - to check if not exist
input.csv - your input file
It's awk's behavior, when you use -v it can will work with variable on run-time and provide an output that doesn't contain the value you passed. This way, you get all the values that are not matching to your variable. Hope this is helpful. :)
Thanks
This bash script should work:
!/bin/bash
FILTER="!/(^"$(echo "$#" | sed -e "s/ /\|^/g")")/ {print}"
awk "$FILTER" input.csv > output.csv
The idea is to build an awk relevant FILTER and then use it.
Assuming the call parameters are: 1 2 3, the filter will be: !/(^1|^2|^3)/ {print}
!: to invert matching
^: Beginning of the line
The input data are in the input.csv file and output result will be in the output.csv file.
I have a directory /user/reports under which many files are there, one of them is :
report.active_user.30092018.77325.csv
I need output as number after date i.e. 77325 from above file name.
I created below command to find a value from file name:
ls /user/reports | awk -F. '/report.active_user.30092018/ {print $(NF-1)}'
Now, I want current date to be passed in above command as variable and get result:
ls /user/reports | awk -F. '/report.active_user.$(date +'%d%m%Y')/ {print $(NF-1)}'
But not getting required output.
Tried bash script:
#!/usr/bin/env bash
_date=`date +%d%m%Y`
active=$(ls /user/reports | awk -F. '/report.active_user.${_date}/ {print $(NF-1)}')
echo $active
But still output is blank.
Please help with proper syntax.
As #cyrus said you must use double quotes in your variable assignment because simple quote are use only for string and not for containing variables.
Bas use case
number=10
string='I m sentence with or wihtout var $number'
echo $string
Correct use case
number=10
string_with_number="I m sentence with var $number"
echo $string_with_number
You can use simple quote but not englobe all the string
number=10
string_with_number='I m sentence with var '$number
echo $string_with_number
Don't parse ls
You don't need awk for this: you can manage with the shell's capabilities
for file in report.active_user."$(date "+%d%m%Y")"*; do
tmp=${file%.*} # remove the extension
number=${tmp##*.} # remove the prefix up to and including the last dot
echo "$number"
done
See https://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion
I have a log that returns thousands of lines of data, I want to extract a few values from that.
In the log there is only one line containing the unquie unit reference so I can grep for that using:
grep "unit=Central-C152" logfile.txt
That produces a line of output similar to the following:
a3cd23e,85d58f5,53f534abef7e7,unit=Central-C152,locale=32325687-8595-9856-1236-12546975,11="School",1="Mr Green",2="Qual",3="SWE",8="report",5="channel",7="reset",6="velum"
The format of the line may change in that the order of the values won't always be in the same position.
I'm trying to work out how to get the value of 2 and 7 in to separate variables.
I had thought about cut on , or = but as the values aren't in a set order I couldn't work out that best way to do it.
I' trying to get:
var state=value of 2 without quotes
var mode=value of 7 without quotes
Can anyone advise on the best way to do this ?
Thanks
Could you please try following to create variable's values.
state=$(awk '/unit=Central-C152/ && match($0,/2=\"[^"]*/){print substr($0,RSTART+3,RLENGTH-3)}' Input_file)
mode=$(awk '/unit=Central-C152/ && match($0,/7=\"[^"]*/){print substr($0,RSTART+3,RLENGTH-3)}' Input_file)
You could print them too by doing following.
echo "$state"
echo "$mode"
Explanation: Adding explanation of command too now.
awk ' ##Starting awk program here.
/unit=Central-C152/ && match($0,/2=\"[^"]*/){ ##Checking condition if a line has string (unit=Central-C152) and using match using REGEX to check from 2 to till "
print substr($0,RSTART+3,RLENGTH-3) ##Printing substring starting from RSTART+3 till RLENGTH-3 characters.
}
' Input_file ##Mentioning Input_file name here.
You are probably better off doing all of the processing in Awk.
awk -F, '/unit=Central-C152/ {
for(i=1;i<=NF;++i)
if($i ~ /^[27]="/) {
b[++k] = $i
sub(/^[27]="/, "", b[k])
sub(/"$/, "", b[k])
gsub(/\\/, "", b[k])
}
print "state " b[1] ", mode " b[2]
}' logfile.txt
This presupposes that the fields always occur in the same order (2 before 7). Maybe you need to change or disable the gsub to remove backslashes in the values.
If you want to do more than print the values, refactoring whatever Bash code you have into Awk is often a better approach than doing this processing in Bash.
Assuming you already have the line in a variable such as with:
line="$(grep 'unit=Central-C152' logfile.txt | head -1)"
You can then simply use the built-in parameter substitution features of bash:
f2=${line#*2=\"} ; f2=${f2%%\"*} ; echo ${f2}
f7=${line#*7=\"} ; f7=${f7%%\"*} ; echo ${f7}
The first command on each line strips off the first part of the line up to and including the <field-number>=". The second command then strips everything off that beyond (and including) the first quote. The third, of course, simply echos the value.
When I run those commands against your input line, I see:
Qual
reset
which is, from what I can see, what you were after.
I have one command to cut string.
I wonder detail of control index of command in Linux "awk"
I have two different case.
I want to get word "Test" in below example string.
1. "Test-01-02-03"
2. "01-02-03-Test-Ref1-Ref2
First one I can get like
substr('Test-01-02-03',0,index('Test-01-02-03',"-"))
-> Then it will bring result only "test"
How about Second case I am not sure how can I get Test in that case using index function.
Do you have any idea about this using awk?
Thanks!
This is how to use index() to find/print a substring:
$ cat file
Test-01-02-03
01-02-03-Test-Ref1-Ref2
$ awk -v tgt="Test" 's=index($0,tgt){print substr($0,s,length(tgt))}' file
Test
Test
but that may not be the best solution for whatever your actual problem is.
For comparison here's how to do the equivalent with match() for an RE:
$ awk -v tgt="Test" 'match($0,tgt){print substr($0,RSTART,RLENGTH)}' file
Test
Test
and if you like the match() synopsis, here's how to write your own function to do it for strings:
awk -v tgt="Test" '
function strmatch(source,target) {
SSTART = index(source,target)
SLENGTH = length(target)
return SSTART
}
strmatch($0,tgt){print substr($0,SSTART,SLENGTH)}
' file
If these lines are the direct input to awk then the following work:
echo 'Test-01-02-03' | awk -F- '{print $1}' # First field
echo '01-02-03-Test-Ref1-Ref2' | awk -F- '{print $NF-2}' # Third field from the end.
If these lines are pulled out of a larger line in an awk script and need to be split again then the following snippets will do that:
str="Test-01-02-03"; split(str, a, /-/); print a[1]
str="01-02-03-Test-Ref1-Ref2"; numfields=split(str, a, /-/); print a[numfields-2]