I want to print the fields in specific format ,
Input :
col1|col2|col3|col4
I used cat file | cut -d '|' -f 3,1,4
output :
col1|col3|col4
But my expected output is:
col3|col1|col4
Can anyone help me with this?
From man cut:
Selected input is written in the same order that it is read, and is written exactly once
You should do:
$ awk -F'|' -vOFS='|' '{print $3,$1,$4}' <<< "col1|col2|col3|col4"
col3|col1|col4
even though awk is good,here is a perl solution:
perl -F"\|" -ane 'print join "|",#F[2,0,3]'
tested:
> echo "col1|col2|col3|col4" | perl -F"\|" -ane 'print join "|",#F[2,0,3]'
col3|col1|col4
Related
Below are the full file names.
qwertyuiop.abcdefgh.1234567890.txt
qwertyuiop.1234567890.txt
trying to use
awk -F'.' '{print $1}'
How can i use awk command to extract below output.
qwertyuiop.abcdefgh
qwertyuiop
Edit
i have a list of files in a directory
i am trying to extract time,size,owner,filename into seperate variables.
for filenames.
NAME=$(ls -lrt /tmp/qwertyuiop.1234567890.txt | awk -F'/' '{print $3}' | awk -F'.' '{print $1}')
$ echo $NAME
qwertyuiop
$
NAME=$(ls -lrt /tmp/qwertyuiop.abcdefgh.1234567890.txt | awk -F'/' '{print $3}' | awk -F'.' '{print $1}')
$ echo $NAME
qwertyuiop
$
expected
qwertyuiop.abcdefgh
With GNU awk and other versions that allow manipulation of NF
$ awk -F. -v OFS=. '{NF-=2} 1' ip.txt
qwertyuiop.abcdefgh
qwertyuiop
NF-=2 will effectively delete last two fields
1 is an awk idiom to print contents of $0
Note that this assumes there are at least two fields in every line, otherwise you'd get an error
Similar concept with perl, prints empty line if number of fields in the line is less than 3
$ perl -F'\.' -lane 'print join ".", #F[0..$#F-2]' ip.txt
qwertyuiop.abcdefgh
qwertyuiop
With sed, you can preserve lines if number of fields is less than 3
$ sed 's/\.[^.]*\.[^.]*$//' ip.txt
qwertyuiop.abcdefgh
qwertyuiop
EDIT: Taking inspiration from Sundeep sir's solution and adding this following too in this mix.
awk 'BEGIN{FS=OFS="."} {$(NF-1)=$NF="";sub(/\.+$/,"")} 1' Input_file
Could you please try following.
awk -F'.' '{for(i=(NF-1);i<=NF;i++){$i=""};sub(/\.+$/,"")} 1' OFS="." Input_file
OR
awk 'BEGIN{FS=OFS="."} {for(i=(NF-1);i<=NF;i++){$i=""};sub(/\.+$/,"")} 1' Input_file
Explanation: Adding explanation for above code too here.
awk '
BEGIN{ ##Mentioning BEGIN section of awk program here.
FS=OFS="." ##Setting FS and OFS variables for awk to DOT here as per OPs sample Input_file.
} ##Closing BEGIN section here.
{
for(i=(NF-1);i<=NF;i++){ ##Starting for loop from i value from (NF-1) to NF for all lines.
$i="" ##Setting value if respective field to NULL.
} ##Closing for loop block here.
sub(/\.+$/,"") ##Substituting all DOTs till end of line with NULL in current line.
}
1 ##Mentioning 1 here to print edited/non-edited current line here.
' Input_file ##Mentioning Input_file name here.
I have the following list in a text file:
10.1.2.200
10.1.2.201
10.1.2.202
10.1.2.203
I want to encase in "double quotes", comma separate and join the values as one string.
Can this be done in sed or awk?
Expected output:
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203","10.1.2.204"
The easiest is something like this (in pseudo code):
Read a line;
Put the line in quotes;
Keep that quoted line in a stack or string;
At the end (or while constructing the string), join the lines together with a comma.
Depending on the language, that is fairly straightforward to do:
With awk:
$ awk 'BEGIN{OFS=","}{s=s ? s OFS "\"" $1 "\"" : "\"" $1 "\""} END{print s}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
Or, less 'wall of quotes' to define a quote character:
$ awk 'BEGIN{OFS=",";q="\""}{s=s ? s OFS q$1q : q$1q} END{print s}' file
With sed:
$ sed -E 's/^(.*)$/"\1"/' file | sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/,/g'
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
(With Perl and Ruby, with a join function, it is easiest to push the elements onto a stack and then join that.)
Perl:
$ perl -lne 'push #a, "\"$_\""; END{print join(",", #a)}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
Ruby:
$ ruby -ne 'BEGIN{#arr=[]}; #arr.push "\"#{$_.chomp}\""; END{puts #arr.join(",")}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
here is another alternative
sed 's/.*/"&"/' file | paste -sd,
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
awk -F'\n' -v RS="\0" -v OFS='","' -v q='"' '{NF--}$0=q$0q' file
should work for given example.
Tested with gawk:
kent$ cat f
10.1.2.200
10.1.2.201
10.1.2.202
10.1.2.203
kent$ awk -F'\n' -v RS="\0" -v OFS='","' -v q='"' '{NF--}$0=q$0q' f
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
$ awk '{o=o (NR>1?",":"") "\""$0"\""} END{print o}' file
"10.1.2.200","10.1.2.201","10.1.2.202","10.1.2.203"
I have one file:
file.txt
101|aaa {rating=1, dept=10, date=10/02/2013, com=11}
106|bbb {rating=2, dept=11, date=10/03/2013, com=11}
103|vvv {rating=3, dept=12, date=10/03/2013, com=11}
102|aaa {rating=1, dept=10, date=10/04/2013, com=11}
109|bbb {rating=2, dept=11, date=10/05/2013, com=11}
104|bbb {rating=2, dept=11, date=10/07/2013, com=11}
I am greping it based on:
for i in `cat file.txt | grep -i "|aaa "`
do
echo `echo $i|cut -d' ' -f1`"|" `sed -n '/date=/,/, com/p' $i` >> output.txt
done
This error occurs
"/sysdate=/,/systime/p: No such file or directory"
Please help me?
The output should be:
output.txt
101|aaa|10/02/2013
102|aaa|10/04/2013
awk is way better for these cases:
$ awk -F"[ =,]" -v OFS="|" '/aaa/{print $1, $9}' a
101|aaa|10/02/2013
102|aaa|10/04/2013
This sets field separators to either space, = or , and fetches the first and 9th fields, whenever the text aaa is found in the line.
I am trying to use awk to get the name of a file given the absolute path to the file.
For example, when given the input path /home/parent/child/filename I would like to get filename
I have tried:
awk -F "/" '{print $5}' input
which works perfectly.
However, I am hard coding $5 which would be incorrect if my input has the following structure:
/home/parent/child1/child2/filename
So a generic solution requires always taking the last field (which will be the filename).
Is there a simple way to do this with the awk substr function?
Use the fact that awk splits the lines in fields based on a field separator, that you can define. Hence, defining the field separator to / you can say:
awk -F "/" '{print $NF}' input
as NF refers to the number of fields of the current record, printing $NF means printing the last one.
So given a file like this:
/home/parent/child1/child2/child3/filename
/home/parent/child1/child2/filename
/home/parent/child1/filename
This would be the output:
$ awk -F"/" '{print $NF}' file
filename
filename
filename
In this case it is better to use basename instead of awk:
$ basename /home/parent/child1/child2/filename
filename
If you're open to a Perl solution, here one similar to fedorqui's awk solution:
perl -F/ -lane 'print $F[-1]' input
-F/ specifies / as the field separator
$F[-1] is the last element in the #F autosplit array
Another option is to use bash parameter substitution.
$ foo="/home/parent/child/filename"
$ echo ${foo##*/}
filename
$ foo="/home/parent/child/child2/filename"
$ echo ${foo##*/}
filename
Like 5 years late, I know, thanks for all the proposals, I used to do this the following way:
$ echo /home/parent/child1/child2/filename | rev | cut -d '/' -f1 | rev
filename
Glad to notice there are better manners
It should be a comment to the basename answer but I haven't enough point.
If you do not use double quotes, basename will not work with path where there is space character:
$ basename /home/foo/bar foo/bar.png
bar
ok with quotes " "
$ basename "/home/foo/bar foo/bar.png"
bar.png
file example
$ cat a
/home/parent/child 1/child 2/child 3/filename1
/home/parent/child 1/child2/filename2
/home/parent/child1/filename3
$ while read b ; do basename "$b" ; done < a
filename1
filename2
filename3
I know I'm like 3 years late on this but....
you should consider parameter expansion, it's built-in and faster.
if your input is in a var, let's say, $var1, just do ${var1##*/}. Look below
$ var1='/home/parent/child1/filename'
$ echo ${var1##*/}
filename
$ var1='/home/parent/child1/child2/filename'
$ echo ${var1##*/}
filename
$ var1='/home/parent/child1/child2/child3/filename'
$ echo ${var1##*/}
filename
you can skip all of that complex regex :
echo '/home/parent/child1/child2/filename' |
mawk '$!_=$-_=$NF' FS='[/]'
filename
2nd to last :
mawk '$!--NF=$NF' FS='/'
child2
3rd last field :
echo '/home/parent/child1/child2/filename' |
mawk '$!--NF=$--NF' FS='[/]'
child1
4th-last :
mawk '$!--NF=$(--NF-!-FS)' FS='/'
echo '/home/parent/child000/child00/child0/child1/child2/filename' |
child0
echo '/home/parent/child1/child2/filename'
parent
major caveat :
- `gawk/nawk` has a slight discrepancy with `mawk` regarding
- how it tracks multiple,
- and potentially conflicting, decrements to `NF`,
- so other than the 1st solution regarding last field,
- the rest for now, are only applicable to `mawk-1/2`
just realized it's much much cleaner this way in mawk/gawk/nawk :
echo '/home/parent/child1/child2/filename' | …
'
awk ++NF FS='.+/' OFS= # updated such that
# root "/" still gets printed
'
filename
You can also use:
sed -n 's/.*\/\([^\/]\{1,\}\)$/\1/p'
or
sed -n 's/.*\/\([^\/]*\)$/\1/p'
Here is a query:
grep bar 'foo.txt' | awk '{print $3}'
The field name emitted by the 'awk' query are mangled C++ symbol names. I want to pass each to dem and finally output the output of 'dem'- i.e the demangled symbols.
Assume that the field separator is a ' ' (space).
awk is a pattern matching language. The grep is totally unnecessary.
awk '/bar/{print $3}' foot.txt
does what your example does.
Edit Fixed up a bit after reading the comments on the precedeing answer (I don't know a thing about dem...):
You can make use of the system call in awk with something like:
awk '/bar/{cline="dem " $3; system(cline)}' foot.txt
but this would spawn an instance of dem for each symbol processed. Very inefficient.
So lets get more clever:
awk '/bar/{list = list " " $3;}END{cline="dem " list; system(cline)}' foot.txt
BTW-- Untested as I don't have dem or your input.
Another thought: if you're going to use the xargs formulation offered by other posters, cut might well be more efficient than awk. At that point, however, you would need grep again.
How about
grep bar 'foo.txt' | awk '{ print $3 }' | xargs dem | awk '{ print $3 }'
This will print the demangled symbols, complete with argument lists in the case of methods:
awk '/bar/ { print $3 }' foo.txt | xargs dem | sed -e 's:.* == ::'
This will print the demangled symbols, without argument lists in the case of methods:
awk '/bar/ { print $3 }' foo.txt | xargs dem | sed -e 's:.* == \([^(]*\).*:\1:'
Cheers,
V.