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.
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 a file with multiple lines. I'm trying to find lines that match a certain pattern and then get them appended to an output file, all on the same.
Ex:
Input file:
ABCD
other text
EFGH
other text
IJKLM
I'm trying to get the output to be :
ABCD EFGH IJKLM
An easy way to make grep output matches separated by spaces instead of newlines is to wrap it in a sub-shell with $(...) like this:
echo $(grep -o '^[A-Z]*$' input.txt) >> output.txt
Or you could use tr:
grep -o '^[A-Z]*$' input.txt | tr '\n' ' ' >> output.txt
Or perl:
grep -o '^[A-Z]*$' input.txt | perl -pe 'chomp; s/$/ /'
You can use tr to translate the newlines to spaces:
grep $EXPRESSION $INPUT_FILE | tr '\n' ' ' >> $OUTPUT_FILE
If you like perl, you can also
perl -nl40e 'print if /PATTERN/' files....
like
perl -nl40e 'print if /[A-Z]/' file
for your input produces
ABCD EFGH IJKLM
Here is an short awk
awk 'NR%2==1' ORS=" " file
ABCD EFGH IJKLM
It will print every second line into one line.
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
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.