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I have a file with a records like the below
FIRST 1: SECOND 2: THREE 4: FIVE 255: SIX 255
I want to remove values between space and :
FIRST:SECOND:THREE:FIVE:SIX
with code
awk -F '[[:space:]]*,:*' '{$1=$1}1' OFS=, file
tried on gnu awk:
awk -F' [0-9]*(: *|$)' -vOFS=':' '{print $1,$2,$3,$4,$5}' file
tried on gnu sed:
sed -E 's/\s+[0-9]+(:|$)\s*/\1/g' file
Explanation of awk,
regex , a space, followed by [0-9]+ one or more number followed by literal : followed by one or more space: *, if all such matched, then collect everything else than this matched pattern, ie. FIRST, SECOND,... so on because -F option determine it as field separator (FS) and $1, $2 .. so on is always else than FS. But the output needs nice look ie. has FS so that'd be : and it'd be awk variable definition -vOFS=':'
You can add [[:digit:]] also with a ending asterisk, and leave only a space just after OFS= :
$ awk -F '[[:space:]][[:digit:]]*' '{$1=$1}1' OFS= file
FIRST:SECOND:THREE:FIVE:SIX
To get the output we want in idiomatic awk, we make the input field separator (with -F) contain all the stuff we want to eliminate (anchored with :), and make the output field separator (OFS) what we want it replaced with. The catch is that this won't eliminate the space and numbers at the end of the line, and for this we need to do something more. GNU’s implementation of awk will allow us to use a regular expression for the input record separator (RS), but we could just do a simple sub() with POSIX complaint awk as well. Finally, force recalculation via $1=$1... the side effects for this pattern/statement are that the buffer will be recalculated doing FS/RS substitution for us, and that non-blank lines will take the default action -- which is to print.
gawk -F '[[:space:]]*[[:digit:]]*:[[:space:]]*' -v OFS=: -v RS='[[:space:]]*[[:digit:]]*\n' '$1=$1' file
Or:
awk -F '[[:space:]]*[[:digit:]]*:[[:space:]]*' -v OFS=: '{ sub(/[[:space:]]*[[:digit:]]*$/, “”) } $1=$1' file
A sed implementation is fun but probably slower (because current versions of awk have better regex implementations).
sed 's/[[:space:]]*[[:digit:]]*:[[:space:]]/:/g; s/[[:space:]]*[[:digit:]]*[[:space:]]*$//' file
Or if POSIX character classes are not available...
sed 's/[\t ]*[0-9]*:[\t ]/:/g; s/[\t ]*[0-9]*[\t ]*$//' file
Something tells me that your “FIRST, SECOND, THIRD...” might be more complicated, and might contain digits... in this case, you might want to experiment with replacing * with + for awk or with \+ for sed.
file.csv:
XA90;"standard"
XA100;"this is
the multi-line"
XA110;"other standard"
I want to grep the "XA100" entry like this:
grep XA100 file.csv
to obtain this result:
XA100;"this is
the multi-line"
but grep return only one line:
XA100;"this is
source.csv contains 3 entries.
The "XA100" entry contain a multi-line field.
And grep doesn't seem to be the right tool to "grep" CSV file including multilines fields.
Do you know the way to make the job ?
Edit: the real world file contains many columns. The researched term can be in any column (not at begin of line, nor at the begin of field). All fields are encapsulated by ". Any field can contain a multi-line, from 1 line to any, and this cannot be predicted.
Give this line a try:
awk '/^XA100;/{p=1}p;p&&/"$/{p=0}' file
I extended your example a bit:
kent$ cat f
XA90;"standard"
XA100;"this is
the
multi-
line"
XA110;"other standard"
kent$ awk '/^XA100;/{p=1}p;p&&/"$/{p=0}' f
XA100;"this is
the
multi-
line"
In the comments you mention: In the real world file, each line start with ". I assume they also end with " and present you this:
Test file:
$ cat file
"single line"
"multi-
lined"
Code and outputs:
$ awk 'BEGIN{RS=ORS="\"\n"} /single/' file
"single line"
$ awk 'BEGIN{RS=ORS="\"\n"} /m/' file
"multi-
lined"
You can also parametrize the search:
$ awk -v s="multi" 'BEGIN{RS=ORS="\"\n"} match($0,s)' file
"multi-
lined"
try:
Solution 1:
awk -v RS="XA" 'NR==3{gsub(/$\n$/,"");print RS $0}' Input_file
Making Record separator as string XA then looking for line 3rd here and then globally substituting the $\n$(which is to remove the extra line at the end of the line) with NULL. Then printing the Record Separator with the current line.
Solution 2:
awk '/XA100/{print;getline;while($0 !~ /^XA/){print;getline}}' Input_file
Looking for string XA100 then printing the current line and using getline to go to next line, using while loop then which will run and print the lines until a line is starting from XA.
If this file was exported from MS-Excel or similar then lines end with \r\n while the newlines inside quotes are just \ns so then all you need is:
$ awk -v RS='\r\n' '/XA100/' file
XA100;"this is
the multi-line"
The above uses GNU awk for multi-char RS. On some platforms, e.g. cygwin, you'll have to add -v BINMODE=3 so gawk sees the \rs rather than them getting stripped by underlying C primitives.
Otherwise, it's extremely hard to parse CSV files in general without a real CSV parser (which awk currently doesn't have but is in the works for GNU awk) but you could do this (again with GNU awk for multi-char RS):
$ cat file
XA90;"standard"
XA100;"this is
the multi-line"
XA110;"other standard"
$ awk -v RS="\"[^\"]*\"" -v ORS= '{gsub(/\n/," ",RT); print $0 RT}' file
XA90;"standard"
XA100;"this is the multi-line"
XA110;"other standard"
to replace all newlines within quotes with blank chars and then process it as regular 1-line-per-record file.
Using PS response, this works for the small example:
sed 's/^X/\n&/' file.csv | awk -v RS= '/XA100/ {print}'
For my real world CSV file, with many columns, with researched term anywhere, with unknown count of multi-lines, with characters " replaced by "", with multi-lines lines beginning with ", with all fields encapsulated by ", this works. Note the exclusion of the second character " in sed part:
sed 's/^"[^"]/\n&/' file.csv | awk -v RS= '/RESEARCH_TERM/ {print}'
Because first column of any entry cannot start with "". First column allways looks like "XXXXXXXXX", where X is any character but ".
Thank you all for so much responses, maybe others solutions are working depending the CSV file format you use.
I wish to use a string (BIRCH) as a field delimiter in awk to print second field. I am trying the following command:
cat tmp.log|awk -FBirch '{ print $2}'
Below output is getting printed:
irch2014/06/23,04:36:45,3,1401503,xml-harlan,P12345-1,temp,0a653356353635635,temp,L,Success
Desired output:
2014/06/23,04:36:45,3,1401503,xml-harlan,P12345-1,temp,0a653356353635635,temp,L,Success
Contents of tmp.log file.
-bash-3.2# cat tmp.log
Dec 05 13:49:23 [x.x.x.x.180.100] business-log-dev/int [TEST][0x80000001][business-log][info] mpgw(Test): trans(8497187)[request][10.x.x.x]:
Birch2014/06/23,04:36:45,3,1401503,xml-harlan,P12345-1,temp,0a653356353635635,temp,L,Success
Am I doing something wrong?
OS: Solaris10
Shell: Bash
Tried below command suggested in one of the ansers below. I am getting the desired output, but with an extra empty line at the top. How can this be eliminated from the output?
-bash-3.2# /usr/xpg4/bin/awk -FBirch '{print $2}' tmp.log
2014/06/23,04:36:45,3,1401503,xml-harlan,P12345-1,temp,0a653356353635635,temp,L,Success
Originally, I suggested putting quotes around "Birch" (-F'Birch') but actually, I don't think that should make any difference.
I'm not at all experienced working with Solaris but you may want to also try using nawk ("new awk") instead of awk.
nawk -FBirch '{print $2}' file
If this works, you may want to consider creating an alias so that you always use the newer version of awk with more features.
You may also want to try using the version of awk in the /usr/xpg4/bin directory, which is a POSIX compliant implementation so should support multi-character FS:
/usr/xpg4/bin/awk -FBirch '{print $2}' file
If you only want to print lines which have more than one field, you can add a condition:
/usr/xpg4/bin/awk -FBirch 'NF>1{print $2}' file
This only prints the second field when there is more than one field.
From the man page of the default awk on solaris usr/bin/awk
-Fc Uses the character c as the field separator
(FS) character. See the discussion of FS
below.
As you can see solaris awk only takes a single character as a Field separator
Also in the man page is split
split(s, a, fs)
Split the string s into array elements a[1], a[2], ...
a[n], and returns n. The separation is done with the
regular expression fs or with the field separator FS if
fs is not given.
As you can see here it takes a regular expression as a separator so we can use.
awk 'split($0,a,"Birch"){print a[2]}' file
To print the second field split by Birch
I have input file as
ab,1,3,qqq,bbc
b,445,jj,abc
abcqwe,234,23,123
abc,12,bb,88
uirabc,33,99,66
I have to select the rows which has only 'abc'. And note that abc string can appear in any of the column. Please help me how to achieve this using awk.
Output:
b,445,jj,abc
abc,12,bb,88
You could also use plain grep:
grep "(^|,)abc(,|$)" file
Or if you have to use awk
awk '/(^|,)abc(,|$)/' file
Using awk
awk 'gsub(/(^|,)abc(,|$)/,"&")' file
b,445,jj,abc
abc,12,bb,88
Based on Beny23s regex.
It does look for abc where its starting from ^ start or from a , and
ends with a , or end of line $
Another one using beny23 regex:
awk 'NF>1' FS="(^|,)abc(,|$)" infile
Not asked but if you feel the need to filter just the lines with one ocurrence:
$ cat infile
ab,1,3,qqq,bbc
b,445,jj,abc
abcqwe,234,23,123
abc,12,bb,88
abc,12,bb,abc
uirabc,33,99,66
This will be handy:
$ awk 'NF==2' FS="(^|,)abc(,|$)" infile
b,445,jj,abc
abc,12,bb,88
Also possible using Jotne solution:
$ awk 'gsub(/(^|,)abc(,|$)/,"&")==1' infile
Through awk,
$ awk -F, '{for(i=1;i<=NF;i++){if($i=="abc") print $0;}}' file | uniq
b,445,jj,abc
abc,12,bb,88
OR
$ awk -F, '{for(i=1;i<=NF;i++){if($i=="abc") {print; next}}}' file
b,445,jj,abc,abc
abc,12,bb,88
In the above awk command Field Separator variable is set to , . AWk parses the input file line by line. for function is used to traverse all the fields in a line. If a value of a particular field is abc, then it prints the whole line.
I have a file with fields separated by pipe characters and I want to print only the second field. This attempt fails:
$ cat file | awk -F| '{print $2}'
awk: syntax error near line 1
awk: bailing out near line 1
bash: {print $2}: command not found
Is there a way to do this?
Or just use one command:
cut -d '|' -f FIELDNUMBER
The key point here is that the pipe character (|) must be escaped to the shell. Use "\|" or "'|'" to protect it from shell interpertation and allow it to be passed to awk on the command line.
Reading the comments I see that the original poster presents a simplified version of the original problem which involved filtering file before selecting and printing the fields. A pass through grep was used and the result piped into awk for field selection. That accounts for the wholly unnecessary cat file that appears in the question (it replaces the grep <pattern> file).
Fine, that will work. However, awk is largely a pattern matching tool on its own, and can be trusted to find and work on the matching lines without needing to invoke grep. Use something like:
awk -F\| '/<pattern>/{print $2;}{next;}' file
The /<pattern>/ bit tells awk to perform the action that follows on lines that match <pattern>.
The lost-looking {next;} is a default action skipping to the next line in the input. It does not seem to be necessary, but I have this habit from long ago...
The pipe character needs to be escaped so that the shell doesn't interpret it. A simple solution:
$ awk -F\| '{print $2}' file
Another choice would be to quote the character:
$ awk -F'|' '{print $2}' file
Another way using awk
awk 'BEGIN { FS = "|" } ; { print $2 }'
And 'file' contains no pipe symbols, so it prints nothing. You should either use 'cat file' or simply list the file after the awk program.