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 $2 }'
And 'file' contains no pipe symbols, so it prints nothing. You should either use 'cat file' or simply list the file after the awk program.
Related
I currently have the following line of code:
ls /some/dir/prefix* | sed -e 's/prefix.//' | tr '\n' ' '
Which does achieve what I want it to do:
Get list of files starting with prefix
Remove path and prefix from each string
Remove newlines and replace with spaces for later processing.
For example:
/some/dir/prefix.hello
/some/dir/prefix.world
Should become
hello world
But I feel like there's a nicer way of doing this. Is there a better way to do this in one line?
Here is a two-liner using just built-ins that does it:
fnames=(some/dir/prefix*)
echo "${fnames[#]##*.}"
And here's how this works:
fnames=(some/dir/prefix*) creates an array with all the files starting with prefix and avoids all the problems that come with parsing ls
echo "${fnames[#]##*.}" is a combination of two parameter expansions: ${fnames[#]} prints all array elements, and the ##*. part removes the longest match of anything that ends with . from each array element, leaving just the extension
If you're hell-bent on a one-liner, just join the two commands with &&.
passing ls output to external programs is not recommended, following bash solution may help you here.
for file in prefix*; do echo ${file##*.}; done
Adding a non-one liner form of solution too now.
for file in prefix*
do
echo ${file##*.}
done
Here is a very simple Awk one-liner to achieve this :
awk -F. '{$0=FILENAME; printf $NF" "; nextfile}' /some/dir/prefix*
It essentially does the following :
-F.: Set the field separator FS to a .. This way $NF represents the extension.
$0=FILENAME: Ignore the current record and set it to FILENAME, reparse everything this way.
print $NF; nextfile : print the extension and go to the next file.
The problem with this is that the file still reads a record of the current file. If that file is empty this will fail.
To make this work with empty files, you could use the gawk extension BEGINFILE
awk -F. 'BEGINFILE{$0=FILENAME; printf $NF" "; nextfile}' /some/dir/prefix*
Or you can loop over all the arguments :
awk -F. 'BEGIN{for(i in ARGV){$0=ARGV[i]; printf $NF" "};exit}' /some/dir/prefix*
One approach with awk:
ls /some/dir/prefix* | awk -F"." '{printf "%s ", $2} END {print ""}'
It might qualify as being "nicer" because there's only one command the output is piped through?!
file.csv:
XA90;"standard"
XA100;"this is
the multi-line"
XA110;"other standard"
I want to grep the "XA100" entry like this:
grep XA100 file.csv
to obtain this result:
XA100;"this is
the multi-line"
but grep return only one line:
XA100;"this is
source.csv contains 3 entries.
The "XA100" entry contain a multi-line field.
And grep doesn't seem to be the right tool to "grep" CSV file including multilines fields.
Do you know the way to make the job ?
Edit: the real world file contains many columns. The researched term can be in any column (not at begin of line, nor at the begin of field). All fields are encapsulated by ". Any field can contain a multi-line, from 1 line to any, and this cannot be predicted.
Give this line a try:
awk '/^XA100;/{p=1}p;p&&/"$/{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'm on a Mac, and I want to find a field in a CSV file adjacent to a search string
This is going to be a single file with a hard path; here's a sample of it:
84:a5:7e:6c:a6:b0, AP-ATC-151g84
84:a5:7e:6c:a6:b1, AP-A88-131g84
84:a5:7e:73:10:32, AP-AG7-133g56
84:a5:7e:73:10:30, AP-ADC-152g81
84:a5:7e:73:10:31, AP-D78-152e80
so if my search string is "84:a5:7e:73:10:32"
I want to get returned "AP-AG7-133g56"
I had been working within an Applescript, but maybe a shell script will do.
I just need the proper syntax for opening the file and having awk search it. Again, I'm weak conceptually on how shell commands run, how they must be executed, etc
This errors, gives me ("command not found"):
set the_file to "/Users/Paw/Desktop/AP-Decoder 3.app/Contents/Resources/BSSIDtable.csv"
set the_val to "70:56:81:cb:a2:dc"
do shell script "'awk $1 ~ the_val {print $2} the_file'"
Thank you for coddling me...
This is a relatively simple:
awk '$1 == "70:56:81:cb:a2:dc," {print "The answer is "$2}' 'BSSIDtable.csv'
(the "The answer is " text can be omitted if you only wish to see only the data, but this shows you how to get more user-friendly output if desired).
The comma is included since awk uses white space for separators so the comma becomes part of column 1.
If the thing you're looking for is in a shell variable, you can use -v to provide that to awk as an awk variable:
lookfor="70:56:81:cb:a2:dc,"
awk -v mac=$lookfor '$1 == mac {print "The answer is "$2}' 'BSSIDtable.csv'
As an aside, your AppleScript solution is probably not working because the $1/$2 are being interpreted as shell variable rather than awk variables. If you insist on using AppleScript, you will have to figure out how to construct a shell command that quotes the awk commands correctly.
My advice is to just use the shell directly, the number of people proficient in that almost certainly far outnumber those proficient in AppleScript :-)
if sed is available (normaly on mac, event if not tagged in OP)
simple but read all the file
sed -n 's/84:a5:7e:73:10:32,[[:blank:]]*//p' YourFile
quit after first occurence (so average of 50% faster on huge file)
sed -n -e '/84:a5:7e:73:10:32,[[:blank:]]*/!b' -e 's///p;q' YourFile
awk
awk '/^84:a5:7e:73:10:32/ {print $2}'
# OR using a variable for batch interaction
awk -v Src='84:a5:7e:73:10:32' '$1 == Src {print $2}'
# OR assuming that case is unknow
awk -v Src='84:a5:7e:73:10:32' 'BEGIN{IGNORECASE=1} $1 == Src {print $2}'
by default it take $0 as compare test if a regex is present, just add the ^ to take first field content
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 would like to write a loop creating various output files with the first column of each input file, respectively.
So I wrote
for i in $(\ls -d /home/*paired.isoforms.results)
do
awk -F"\t" {print $1}' $i > $i.transcript_ids.txt
done
As an example if there were 5 files in the home directory named
A_paired.isoforms.results
B_paired.isoforms.results
C_paired.isoforms.results
D_paired.isoforms.results
E_paired.isoforms.results
I would like to print the first column of each of these files into a seperate output file, i.e. I would like to have 5 output files called
A.transcript_ids.txt
B.transcript_ids.txt
C.transcript_ids.txt
D.transcript_ids.txt
E.transcript_ids.txt
or any other name as long as it is 5 different names and I can still link them back to the original files.
I understand, that there is a problem with the double usage of $ in both the awk and the loop command, but I don't know how to change that.
Is it possible to write a command like this in a loop?
This should do the job:
for file in /home/*paired.isoforms.results
do
base=${file##*/}
base=${base%%_*}
awk -F"\t" '{print $1}' $file > $base.transcript_ids.txt
done
I assume that there can be spaces in the first field since you set the delimiter explicitly to tab. This runs awk once per file. There are ways to do it running awk once for all files, but I'm not convinced the benefit is significant. You could consider using cut instead of awk '{print $1}', too. Note that using ls as you did is less satisfactory than using globbing directly; it runs foul of file names with oddball characters (spaces, tabs, etc) in the name.
You can do that entirely in awk:
awk -F"\t" '{split(FILENAME,a,"_"); out=a[1]".transcript_ids.txt"; print $1 > out}' *_paired.isoforms.results
If your input files don't have names as indicated in the question, you'd have to split on something else ( as well as use a different pattern match for the input files ).
My original answer is actually doing extra name resolution every time something is printed. Here's a version that only updates the output filename when FILENAME changes:
awk -F"\t" 'FILENAME!=lf{split(FILENAME,a,"_"); out=a[1]".transcript_ids.txt"; lf=FILENAME} {print $1 > out}' *_paired.isoforms.results