I need to print only the 900 in this line: auth required pam_faillock.so preauth silent deny=3 unlock_time=604800 fail_interval=900
However, this line will not always be in this order.
I need to find out how to print the value after the =.
I will need to do this for unlock_time and fail_interval
I have been searching all night for something that will work exactly for me and cannot find it. I have been toying around with sed and awk and have not nailed this down yet.
Let's define your string:
s='auth required pam_faillock.so preauth silent deny=3 unlock_time=604800 fail_interval=900'
Using awk:
$ printf %s "$s" | awk -F= '$1=="fail_interval"{print $2}' RS=' '
900
Or:
$ printf %s "$s" | awk -F= '$1=="unlock_time"{print $2}' RS=' '
604800
How it works
Awk divides its input into records. We tell it to use a space as the record separator. Each record is divided into fields. We tell awk to use = as the field separator. In more detail:
printf %s "$s"
This prints the string. printf is safer than echo in cases where the string might begin with -.
-F=
This tells awk to use = as the field separator.
$1=="fail_interval" {print $2}
If the first field is fail_interval, then we tell awk to print the second field.
RS=' '
This tells awk to use a space as the record separator.
You may use sed for this
Command
echo "...stuff.... unlock_time=604800 fail_interval=900" | sed -E '
s/^.*unlock_time=([[:digit:]]*).*fail_interval=([[:digit:]]*).*$/\1 \2/'
Output
604800 900
Notes
The (..) in sed is used for selections.
[[:digit:]]* or 0-9 is used to match any number of digits
The \1 and \2 is used to replace the matched stuff, in order.
Given an input variable:
input="auth required pam_faillock.so preauth silent deny=3 unlock_time=604800 fail_interval=900"
With GNU grep:
$ grep -oP 'fail_interval=\K([0-9]*)' <<< "$input"
900
$ grep -oP 'unlock_time=\K([0-9]*)' <<< "$input"
604800
Try using this.
unlock_time=$(echo "auth required pam_faillock.so preauth silent deny=3 unlock_time=604800 fail_interval=900" | awk -F'unlock_time=' '{print $2}' | awk '{print $1}')
echo "$unlock_time"
fail_interval=$(echo "auth required pam_faillock.so preauth silent deny=3 unlock_time=604800 fail_interval=900" | awk -F'fail_interval=' '{print $2}' | awk '{print $1}')
echo "$fail_interval"
Related
I have a line from which I need to cut the branch name to the first comma:
commit 2bea9e0351dae65f18d2de11621049b465b1e868 (HEAD, origin/MGB-322, refs/pipelines/36877)
I need to cut out MGB-322.
The number of characters in a line is always different.
awk -F "origin/" '{print $2}' - this is how I cut out
MGB-322, refs/pipelines/36877)
But how to tell it to trim to the first comma?
I tried doing it via substr,
awk -F "origin/" '{print substr ($2,1, index $2 ,)}'
But it is not clear how to correctly specify the comma in index
With any awk. Use / and , as field separator:
awk '{print $3}' FS='[/,]' file
Output:
MGB-322
See: 8 Powerful Awk Built-in Variables – FS, OFS, RS, ORS, NR, NF, FILENAME, FNR
With OP's code fix: considered that you have only occurrence of origin in case you have more than occurrence then change $NF to $2 in following code. Written and tested in https://ideone.com/xjv2we
awk -F"origin/" '{print $NF}' Input_file
sed could be also helpful here, generic solution it's based on first occurrence of comma and / as per OP's thread title. I have written this on mobile so couldn't test it as of now should with though and will test it after sometime.
sed 's/\([^,]*\),\([^/]*\)\/\(.*\)/\3/' Input_file
"I need to cut out MGB-322."
You can use cut in two steps:
echo "${line}" | cut -d"/" -f2 | cut -d"," -f1
I would prefer one step with awk (already anwered by others) or sed
echo "${line}" | sed -r 's/.*origin.(.*), refs.*/\1/'
Why spawn procs? bash's built-in parameter parsing will handle this.
If
$: line="commit 2bea9e0351dae65f18d2de11621049b465b1e868 (HEAD, origin/MGB-322, refs/pipelines/36877)"
then
$: [[ "$line" =~ .*origin.(.*), ]] && echo "${BASH_REMATCH[1]}"
MGB-322
or maybe
$: tmp=${line#*, origin/}; echo ${tmp%,*}
MGB-322
or even
$: IFS=",/" read _ _ x _ <<< "$line" && echo $x
MGB-322
c.f. https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
/Home/in/test_file.txt
echo /Home/in/test_file.txt | awk -F'/' '{ print $2,$3 }'
Gives the result as:
Home in
But I need /Home/in/ as the result .I have to get all except test_file.txt
How to achieve this?
$ echo '/Home/in/test_file.txt' | awk '{sub("/[^/]+$","")} 1'
/Home/in
$ echo '/Home/in/test_file.txt' | awk '{sub("[^/]+$","")} 1'
/Home/in/
$ echo '/Home/in/test_file.txt' | sed 's:/[^/]*$::'
/Home/in
$ echo '/Home/in/test_file.txt' | sed 's:[^/]*$::'
/Home/in/
$ dirname '/Home/in/test_file.txt'
/Home/in
Your attempt awk -F'/' '{ print $2,$3 }' didn't do what you wanted as -F'/' is telling awk to split the input into fields at every / and then print $2,$3 is telling awk to print the 2nd and 3rd fields separated by a blank char (the default value for OFS). You could do:
$ echo '/Home/in/test_file.txt' | awk 'BEGIN{FS=OFS="/"} { print "",$2,$3,"" }'
/Home/in/
to get the expected output but it'd be the wrong approach since it's removing the field you don't want AND removing the input separators AND then adding new output separators which happen to the have the same value as the input separators rather than simply removing the field you don't want like the other solutions above do.
echo /Home/in/test_file.txt | awk -F'/[^/]*$' '{ print $1 }'
..will print the everything but the trailing slash
There are several ways to achieve this:
Using dirname:
$ dirname /home/in/test_file.txt
/home/in
Using Shell substitution:
$ var="/home/in/test_file.txt"
$ echo "${var%/*}"
/home/in
Using sed: (See Ed Morton)
Using AWK:
$ echo "/home/in/test_file.txt" | awk -F'/' '{OFS=FS;$NF=""}1'
/home/in/
Remark: all these work since you can't have a filename with a forward slash (Is it possible to use "/" in a filename?)
Note: all but dirname will fail if you just have a single file_name without a path. While dirname foo will return ./ all others will return foo
awk behaves as it should.
When you define slash / as a separator, the fields in your expression become the content between the separators.
If you need the separator to be printed as well, you need to do it explicitly, like:
echo /Home/in/test_file.txt | awk -F'/' '{ printf "%s/%s/",$2,$3 }'
replace your last field with an empty string and
put the slash back in as the (builtin) Output Field Separator (OFS)
echo /Home/in/test_file.txt | awk -F'/' -vOFS='/' '{$NF="";print}
I would like to use awk for a variable
that has the form abc,def
I also don't know how to use awk for a variable instead of a file
I tried the following but it doesn't works
awk -F, '{$1" "$2}' $varand
awk -F, '{$1" "$2}' "$var"
Use a herestring
awk 'commands' <<< "$string"
Also if you want to print the first two fields of a comma separated string, change the command to
awk -F, '{print $1, $2}' <<< "$string"
You can do something like this:
echo "$variable" | awk -F, '{print $1 " " $2}'
Given a hostname in format of aaa0.bbb.ccc, I want to extract the first substring before ., that is, aaa0 in this case. I use following awk script to do so,
echo aaa0.bbb.ccc | awk '{if (match($0, /\./)) {print substr($0, 0, RSTART - 1)}}'
While the script running on one machine A produces aaa0, running on machine B produces only aaa, without 0 in the end. Both machine runs Ubuntu/Linaro, but A runs newer version of awk(gawk with version 3.1.8 while B with older awk (mawk with version 1.2)
I am asking in general, how to write a compatible awk script that performs the same functionality ...
You just want to set the field separator as . using the -F option and print the first field:
$ echo aaa0.bbb.ccc | awk -F'.' '{print $1}'
aaa0
Same thing but using cut:
$ echo aaa0.bbb.ccc | cut -d'.' -f1
aaa0
Or with sed:
$ echo aaa0.bbb.ccc | sed 's/[.].*//'
aaa0
Even grep:
$ echo aaa0.bbb.ccc | grep -o '^[^.]*'
aaa0
Or just use cut:
echo aaa0.bbb.ccc | cut -d'.' -f1
I am asking in general, how to write a compatible awk script that
performs the same functionality ...
To solve the problem in your quesiton is easy. (check others' answer).
If you want to write an awk script, which portable to any awk implementations and versions (gawk/nawk/mawk...) it is really hard, even if with --posix (gawk)
for example:
some awk works on string in terms of characters, some with bytes
some supports \x escape, some not
FS interpreter works differently
keywords/reserved words abbreviation restriction
some operator restriction e.g. **
even same awk impl. (gawk for example), the version 4.0 and 3.x have difference too.
the implementation of certain functions are also different. (your problem is one example, see below)
well all the points above are just spoken in general. Back to your problem, you problem is only related to fundamental feature of awk. awk '{print $x}' the line like that will work all awks.
There are two reasons why your awk line behaves differently on gawk and mawk:
your used substr() function wrongly. this is the main cause. you have substr($0, 0, RSTART - 1) the 0 should be 1, no matter which awk do you use. awk array, string idx etc are 1-based.
gawk and mawk implemented substr() differently.
You don't need awk for this...
echo aaa0.bbb.ccc | cut -d. -f1
cut -d. -f1 <<< aaa0.bbb.ccc
echo aaa0.bbb.ccc | { IFS=. read a _ ; echo $a ; }
{ IFS=. read a _ ; echo $a ; } <<< aaa0.bbb.ccc
x=aaa0.bbb.ccc; echo ${x/.*/}
Heavier options:
sed:
echo aaa0.bbb.ccc | sed 's/\..*//'
sed 's/\..*//' <<< aaa0.bbb.ccc
awk:
echo aaa0.bbb.ccc | awk -F. '{print $1}'
awk -F. '{print $1}' <<< aaa0.bbb.ccc
You do not need any external command at all, just use Parameter Expansion in bash:
hostname=aaa0.bbb.ccc
echo ${hostname%%.*}
if you don't want to change the input field separator, then it's possible to use split function:
echo "some aaa0.bbb.ccc text" | awk '{split($2, a, "."); print a[1]}'
documentation:
split(string, array [, fieldsep [, seps ] ])
Divide string into pieces separated by fieldsep
and store the pieces in array and the separator
strings in the seps array.
awk is still the cleanest approach :
mawk NF=1 FS='[.]' <<< aaa0.bbb.ccc
aaa0
If there's stuff before or after :
mawk ++NF FS='[.].+$|^[^ ]* ' OFS= <<< 'some aaa0.bbb.ccc text'
mawk '$!NF=$2' FS='[ .]' <<< 'some aaa0.bbb.ccc text'
aaa0
assume I have a string
"1,2,3,4"
Now I want to replace, e.g. the 3rd field of the string by some different value.
"1,2,NEW,4"
I managed to do this with the following command:
echo "1,2,3,4" | awk -F, -v OFS=, '{$3="NEW"; print }'
Now the index for the column to be replaced should be passed as a variable. So in this case
index=3
How can I pass this to awk? Because this won't work:
echo "1,2,3,4" | awk -F, -v OFS=, '{$index="NEW"; print }'
echo "1,2,3,4" | awk -F, -v OFS=, '{$($index)="NEW"; print }'
echo "1,2,3,4" | awk -F, -v OFS=, '{\$$index="NEW"; print }'
Thanks for your help!
This might work for you:
index=3
echo "1,2,3,4" | awk -F, -v OFS=, -v INDEX=$index '{$INDEX="NEW"; print }'
or:
index=3
echo "1,2,3,4" | sed 's/[^,]*/NEW/'$index
Have the shell interpolate the index in the awk program:
echo "1,2,3,4" | awk -F, -v OFS=, '{$'$index'="NEW"; print }'
Note how the originally single quoted awk program is split in three parts, a single quoted beginning '{$', the interpolated index value, followed by the single quoted remainder of the program.
Here's a seductive way to break the awkwardness:
$ echo "1,2,3,4" | sed 's/,/\n/g' | sed -e $index's/.*/NEW/'
This is easily extendable to multiple indexes just by adding another -e $newindex's/.*/NEWNEW/'
# This should be faster than awk or sed.
str="1,2,3,4"
IFS=','
read -a f <<< "$str"
f[2]='NEW'
printf "${f[*]}"
With plain awk (I.E. Not gawk etc) I believe you'll have to use split( string, array, [fieldsep] ); change the array entry of choice and then join them back together with sprintf or similar in a loop.
gawk allows you to have a variable as a field name, $index in your example. See here.
gawk is usually the default awk on Linux, so change your invocation to gawk "script" and see if it works.