Replace a line starting with '-' hyphen with a character repeated n times - linux

I have a text file that has some lines like this (hyphen repeated)
-------------------------------------------------------
I need to replace these lines with character 'B' repeated 1500 times. For example, like
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
Any suggestions using 'sed' or 'awk' command?

With awk:
$ awk '/^-+$/ {s = sprintf("% 1500s", ""); gsub(/ /,"B",s); print s; next} 1' file
Or, maybe a bit more efficient if you have many such lines:
$ awk 'BEGIN {s = sprintf("% 1500s", ""); gsub(/ /,"B",s)} \
/^-+$/ {print s; next} 1' file

I think
perl -pe 'my $bb = "B"x1500; s/^-+$/$bb/g'
should do it.

printf+sed variant:
$ cat file
1111
----
2222
$ sed -r 's/^-+$/'"$(printf -- "%.1s" B{1..150})/" file
1111
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
2222
Here sed is used only for replacement.
printf is used for generating 1500 times B. (The above scriptlet has 150 instead of 1500, because it required too much scrolling.)

Related

how to modify a text file that every line has same number of columns?

I've got a text file which includes several lines. Every line has words which are separated with a comma. The number of words in lines are not the same. I would like with the help of the awk command to make every line have same number of column. For example, if the text file is as follows:
word1, text, help, test
number, begin
last, line, line
I would like the output be as the following which every line has same size in column with an extra null word:
word1, text, help, test
number, begin, null, null
last, line, line, null
I tried the following code:
awk '{print $0,Null}' file.txt
$ awk 'BEGIN {OFS=FS=", "}
NR==FNR {max=max<NF?NF:max; next}
{for(i=NF+1;i<=max;i++) $i="null"}1' file{,}
first scan to find the max number of columns and fill the missing entries in the second round. If the first line contains all the columns (header perhaps), you can change to
$ awk 'BEGIN {OFS=FS=", "}
NR==1 {max=NF}
{for(i=NF+1;i<=max;i++) $i="null"}1' file
file{,} is expanded by bash to file file, a neat trick not to repeat the filename (and eliminates possible typos).
Passing twice through the input file, using getline on first pass:
awk '
BEGIN {
OFS=FS=", "
while(getline < ARGV[1]) {
if (NF > max) {max = NF}
}
close(ARGV[1])
}
{ for(i=NF+1; i<=max; i++) $i="null" } 1
' file.txt
Alternatively, keeping it simple by running awk twice...
#!/bin/bash
infile="file.txt"
maxfields=$(awk 'BEGIN {FS=", "} {if (NF > max) {max = NF}} END{print max}' "$infile" )
awk -v max="$maxfields" 'BEGIN {OFS=FS=", "} {for(i=NF+1;i<=max;i++) $i="null"} 1' "$infile"
Use these Perl one-liners. The first one goes through the file and finds the max number of fields to use. The second one goes through the file and prints the input fields, padded at the end by the null strings:
export num_fields=$( perl -F'/,\s+/' -lane 'print scalar #F;' in_file | sort -nr | head -n1 )
perl -F'/,\s+/' -lane 'print join ", ", map { defined $F[$_] ? $F[$_] : "null" } 0..( $ENV{num_fields} - 1 );' in_file > out_file
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-n : Loop over the input one line at a time, assigning it to $_ by default.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.
-a : Split $_ into array #F on whitespace or on the regex specified in -F option.
-F'/,\s+/' : Split into #F on comma with whitespace.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches

How to filter a the required content from a string in linux?

I had a string like:-
sometext sometext BASEDIR=/someword/someword/someword/1342.32 sometext sometext.
Could someone tell me, how to filter this number 1342.32, from the above string in linux??
$ echo "sometext BASEDIR=/someword/1342.32 sometext." |
sed "s/[^0-9.]//g"
> 1342.32.
The sed command searches for anything not in the set "0123456789" or ".", and replaces it with nothing (deletes it). It does this in global mode, so it doesn't stop on the first match.
This is enough if you're just trying to read it. If you're trying to feed the number into another command and need a real number, you will need to clean it up:
$ ... | cut -f 1-2 -d "."
> 1342.32
cut splits the input on the delemiter, then selects fields 1 and 2 (numbered from one). So "1.2.3.4" would return "1.2".
If sometext is always delimited from the surrounding fields by a white space, try this
cat log.txt | awk '{for (i=1;i<=NF;i++) {if ($i ~
/BASEDIR/) {print i,$i}}}' | awk -F/ '{for (i=1;i<=NF;i++) {if ($i ~
/^[0-9][0-9]*$/) {print $i}}}'
The code snippet above assumes that your data is contained in a file called log.txt and organised in records(read this awk-wise)
This works also if digits appear in sometext before BASEDIR as well as if the input has additional lines:
sed -n 's,.*BASEDIR=\(/\w*\)*/\([0-9.]*\).*,\2,p'
-n do not output lines without BASEDIR…
\(/\w*\)* group of / and someword, repeated
\([0-9.]*\) group of repeated digit or decimal point
\2 replacement of everything matched (the entire line) with the 2nd group
p print the result

How do I use cut to output initial line in different format?

I have a file like this:
A: 1.2.3.4 Z: xxx R: a Q W
A: 5.6.7.8 Z: yyy R: b X Y
A: 9.10.11.12 Z: zzz R: c L N X
The desired output after running command like cut or sed (something which exists in 99% of Linux environments):
1.2.3.4:xxx:a:Q-W
5.6.7.8:yyy:b:X-Y
9.10.11.12:zzz:c:L-N-X
One way with awk:
$ awk '{printf "%s:%s:%s:",$2,$4,$6;for(i=7;i<NF;i++)printf "%s-",$i;print $NF}' file
1.2.3.4:xxx:a:Q-W
5.6.7.8:yyy:b:X-Y
9.10.11.12:zzz:c:L-N-X
Explanation:
The script will run for every line in the file:
printf "%s:%s:%s:",$2,$4,$6; # print the 2nd, 4th, 6th field separated by a :
for(i=7;i<NF;i++) # from the 7th field to the penultimate field
printf "%s-",$i; # print the field value followed by a -
print $NF # print the last field (followed by a newline)
Through sed,
$ sed 's/^[^[:space:]]\+[[:space:]]\+\([^[:space:]]\+\)[[:space:]]\+[^:]*:[[:space:]]\+\([^[:space:]]\+\)[[:space:]]\+[^:]*:[[:space:]]\+\([^[:space:]]\+\)[[:space:]]\+/\1:\2:\3:/;s/[[:space:]]\+/-/g' file
1.2.3.4:xxx:a:Q-W
5.6.7.8:yyy:b:X-Y
9.10.11.12:zzz:c:L-N-X
A shorter version of awk would be like
$ awk -F[\ :]+ '{print $2":"$4":"$6":"$7"-"$8}' input
1.2.3.4:xxx:a:Q-W
5.6.7.8:yyy:b:X-Y
9.10.11.12:zzz:c:L-N
-F[\ :]+ sets the field seperator as space or :
sed 's/ *[A-Z]: */:/g;s/...//;s/\(:[a-z]\) */\1:/;s/ */-/g' YourFile
without more spec on xxx, yyy, ...

Exchange columns in bash

I have this file:
$ cat file
1515523 A45678BF141 A11269151
2234545 A45678BE145 A87979746
5432568 A45678B2123 A40629187
7234573 A45678B4154 A98879129
8889568 A45678B5123 A13409137
9234511 A45678B9176 A23589941
3904568 A45678B7123 A52329165
3234555 A45678B1169 A23589497
9643568 A45678B6123 A39969112
1234547 A45678B2132 A40579243
and this script:
cat file | awk '{FS = " "} {print $1" "$3" "$5}'| awk '{
n = split($3, a, "");
s = "";
for (i = 1; i <= n; i += 2) s = s a[i+1] a[i];
print $1, substr($2, length($2)-3, 4), s
}'| cut -d" " -f3,1 > output
And when I open the output with vi, I have:
1515523 F141 11621915^M
2234545 E145 78797964^M
5432568 2123 04261978^M
7234573 4154 89781992^M
8889568 5123 31041973^M
9234511 9176 32859914^M
3904568 7123 25231956^M
3234555 1169 32854979^M
9643568 6123 93691921^M
1234547 2132 04752934^M
I don't know why I obtain ^M, because when I intend to run the awk snippet:
cat imei | awk '{FS=" "} {print $2","$1}'
the output is mistaken, i.e., it does not exchange the columns, as it does not print the second column. Any ideas on what may be happening?
There are carriage returns (^M or Control-M) in the data file. It probably came from a Windows machine at some point.
When you print $2","$1 (which concatenates $2 with a string containing a comma and then $1 — it took me a couple of looks to see what it was really doing), the carriage return makes the second column overwrite the first.
Look at the data file with od -c or similar tools to see the carriage returns in it.
You can use dos2unix or tr or various other techniques to convert the file from DOS/Windows format to Unix format.
Also, given the data format shown, I'd expect not to use -F " " (or the FS = " ", which is equivalent), so that you have columns $1, $2, and $3, which is more obvious than working with columns 1, 3, 5 as shown. You could set OFS to double-blank if you wanted the output with two blanks between columns.
$ dos2unix file
$ awk '{split($3,a,""); print $1, substr($2,8), a[3]a[2]a[5]a[4]a[7]a[6]a[9]a[8]}' file
1515523 F141 11621915
2234545 E145 78797964
5432568 2123 04261978
7234573 4154 89781992
8889568 5123 31041973
9234511 9176 32859914
3904568 7123 25231956
3234555 1169 32854979
9643568 6123 93691921
1234547 2132 04752934
Since you are using awk you do not need a dos2unix.
simply insert
gsub(/\r/,"");
as a first statement in your awk Script
It cleans up each line read in. Subsequent matching or processing does not get any 'carriage return' characters.
How about a perl 'one liner' (with a continuation line)
$ dos2unix file
$ perl -lane \
'$xxxx = substr($F[1],-4);
#c = split(//,$F[2]);
print "$F[0] $xxxx $c[2]$c[1]$c[4]$c[3]$c[6]$c[5]$c[8]$c[7]"' file

Using awk to print all columns from the nth to the last

This line worked until I had whitespace in the second field.
svn status | grep '\!' | gawk '{print $2;}' > removedProjs
is there a way to have awk print everything in $2 or greater? ($3, $4.. until we don't have anymore columns?)
I suppose I should add that I'm doing this in a Windows environment with Cygwin.
Print all columns:
awk '{print $0}' somefile
Print all but the first column:
awk '{$1=""; print $0}' somefile
Print all but the first two columns:
awk '{$1=$2=""; print $0}' somefile
There's a duplicate question with a simpler answer using cut:
svn status | grep '\!' | cut -d\ -f2-
-d specifies the delimeter (space), -f specifies the list of columns (all starting with the 2nd)
You could use a for-loop to loop through printing fields $2 through $NF (built-in variable that represents the number of fields on the line).
Edit:
Since "print" appends a newline, you'll want to buffer the results:
awk '{out = ""; for (i = 2; i <= NF; i++) {out = out " " $i}; print out}'
Alternatively, use printf:
awk '{for (i = 2; i <= NF; i++) {printf "%s ", $i}; printf "\n"}'
awk '{out=$2; for(i=3;i<=NF;i++){out=out" "$i}; print out}'
My answer is based on the one of VeeArr, but I noticed it started with a white space before it would print the second column (and the rest). As I only have 1 reputation point, I can't comment on it, so here it goes as a new answer:
start with "out" as the second column and then add all the other columns (if they exist). This goes well as long as there is a second column.
Most solutions with awk leave an space. The options here avoid that problem.
Option 1
A simple cut solution (works only with single delimiters):
command | cut -d' ' -f3-
Option 2
Forcing an awk re-calc sometimes remove the added leading space (OFS) left by removing the first fields (works with some versions of awk):
command | awk '{ $1=$2="";$0=$0;} NF=NF'
Option 3
Printing each field formatted with printf will give more control:
$ in=' 1 2 3 4 5 6 7 8 '
$ echo "$in"|awk -v n=2 '{ for(i=n+1;i<=NF;i++) printf("%s%s",$i,i==NF?RS:OFS);}'
3 4 5 6 7 8
However, all previous answers change all repeated FS between fields to OFS. Let's build a couple of option that do not do that.
Option 4 (recommended)
A loop with sub to remove fields and delimiters at the front.
And using the value of FS instead of space (which could be changed).
Is more portable, and doesn't trigger a change of FS to OFS:
NOTE: The ^[FS]* is to accept an input with leading spaces.
$ in=' 1 2 3 4 5 6 7 8 '
$ echo "$in" | awk '{ n=2; a="^["FS"]*[^"FS"]+["FS"]+";
for(i=1;i<=n;i++) sub( a , "" , $0 ) } 1 '
3 4 5 6 7 8
Option 5
It is quite possible to build a solution that does not add extra (leading or trailing) whitespace, and preserve existing whitespace(s) using the function gensub from GNU awk, as this:
$ echo ' 1 2 3 4 5 6 7 8 ' |
awk -v n=2 'BEGIN{ a="^["FS"]*"; b="([^"FS"]+["FS"]+)"; c="{"n"}"; }
{ print(gensub(a""b""c,"",1)); }'
3 4 5 6 7 8
It also may be used to swap a group of fields given a count n:
$ echo ' 1 2 3 4 5 6 7 8 ' |
awk -v n=2 'BEGIN{ a="^["FS"]*"; b="([^"FS"]+["FS"]+)"; c="{"n"}"; }
{
d=gensub(a""b""c,"",1);
e=gensub("^(.*)"d,"\\1",1,$0);
print("|"d"|","!"e"!");
}'
|3 4 5 6 7 8 | ! 1 2 !
Of course, in such case, the OFS is used to separate both parts of the line, and the trailing white space of the fields is still printed.
NOTE: [FS]* is used to allow leading spaces in the input line.
I personally tried all the answers mentioned above, but most of them were a bit complex or just not right. The easiest way to do it from my point of view is:
awk -F" " '{ for (i=4; i<=NF; i++) print $i }'
Where -F" " defines the delimiter for awk to use. In my case is the whitespace, which is also the default delimiter for awk. This means that -F" " can be ignored.
Where NF defines the total number of fields/columns. Therefore the loop will begin from the 4th field up to the last field/column.
Where $N retrieves the value of the Nth field. Therefore print $i will print the current field/column based based on the loop count.
awk '{ for(i=3; i<=NF; ++i) printf $i""FS; print "" }'
lauhub proposed this correct, simple and fast solution here
This was irritating me so much, I sat down and wrote a cut-like field specification parser, tested with GNU Awk 3.1.7.
First, create a new Awk library script called pfcut, with e.g.
sudo nano /usr/share/awk/pfcut
Then, paste in the script below, and save. After that, this is how the usage looks like:
$ echo "t1 t2 t3 t4 t5 t6 t7" | awk -f pfcut --source '/^/ { pfcut("-4"); }'
t1 t2 t3 t4
$ echo "t1 t2 t3 t4 t5 t6 t7" | awk -f pfcut --source '/^/ { pfcut("2-"); }'
t2 t3 t4 t5 t6 t7
$ echo "t1 t2 t3 t4 t5 t6 t7" | awk -f pfcut --source '/^/ { pfcut("-2,4,6-"); }'
t1 t2 t4 t6 t7
To avoid typing all that, I guess the best one can do (see otherwise Automatically load a user function at startup with awk? - Unix & Linux Stack Exchange) is add an alias to ~/.bashrc; e.g. with:
$ echo "alias awk-pfcut='awk -f pfcut --source'" >> ~/.bashrc
$ source ~/.bashrc # refresh bash aliases
... then you can just call:
$ echo "t1 t2 t3 t4 t5 t6 t7" | awk-pfcut '/^/ { pfcut("-2,4,6-"); }'
t1 t2 t4 t6 t7
Here is the source of the pfcut script:
# pfcut - print fields like cut
#
# sdaau, GNU GPL
# Nov, 2013
function spfcut(formatstring)
{
# parse format string
numsplitscomma = split(formatstring, fsa, ",");
numspecparts = 0;
split("", parts); # clear/initialize array (for e.g. `tail` piping into `awk`)
for(i=1;i<=numsplitscomma;i++) {
commapart=fsa[i];
numsplitsminus = split(fsa[i], cpa, "-");
# assume here a range is always just two parts: "a-b"
# also assume user has already sorted the ranges
#print numsplitsminus, cpa[1], cpa[2]; # debug
if(numsplitsminus==2) {
if ((cpa[1]) == "") cpa[1] = 1;
if ((cpa[2]) == "") cpa[2] = NF;
for(j=cpa[1];j<=cpa[2];j++) {
parts[numspecparts++] = j;
}
} else parts[numspecparts++] = commapart;
}
n=asort(parts); outs="";
for(i=1;i<=n;i++) {
outs = outs sprintf("%s%s", $parts[i], (i==n)?"":OFS);
#print(i, parts[i]); # debug
}
return outs;
}
function pfcut(formatstring) {
print spfcut(formatstring);
}
Would this work?
awk '{print substr($0,length($1)+1);}' < file
It leaves some whitespace in front though.
Printing out columns starting from #2 (the output will have no trailing space in the beginning):
ls -l | awk '{sub(/[^ ]+ /, ""); print $0}'
echo "1 2 3 4 5 6" | awk '{ $NF = ""; print $0}'
this one uses awk to print all except the last field
This is what I preferred from all the recommendations:
Printing from the 6th to last column.
ls -lthr | awk '{out=$6; for(i=7;i<=NF;i++){out=out" "$i}; print out}'
or
ls -lthr | awk '{ORS=" "; for(i=6;i<=NF;i++) print $i;print "\n"}'
If you need specific columns printed with arbitrary delimeter:
awk '{print $3 " " $4}'
col#3 col#4
awk '{print $3 "anything" $4}'
col#3anythingcol#4
So if you have whitespace in a column it will be two columns, but you can connect it with any delimiter or without it.
Perl solution:
perl -lane 'splice #F,0,1; print join " ",#F' file
These command-line options are used:
-n loop around every line of the input file, do not automatically print every line
-l removes newlines before processing, and adds them back in afterwards
-a autosplit mode – split input lines into the #F array. Defaults to splitting on whitespace
-e execute the perl code
splice #F,0,1 cleanly removes column 0 from the #F array
join " ",#F joins the elements of the #F array, using a space in-between each element
Python solution:
python -c "import sys;[sys.stdout.write(' '.join(line.split()[1:]) + '\n') for line in sys.stdin]" < file
I want to extend the proposed answers to the situation where fields are delimited by possibly several whitespaces –the reason why the OP is not using cut I suppose.
I know the OP asked about awk, but a sed approach would work here (example with printing columns from the 5th to the last):
pure sed approach
sed -r 's/^\s*(\S+\s+){4}//' somefile
Explanation:
s/// is the standard command to perform substitution
^\s* matches any consecutive whitespace at the beginning of the line
\S+\s+ means a column of data (non-whitespace chars followed by whitespace chars)
(){4} means the pattern is repeated 4 times.
sed and cut
sed -r 's/^\s+//; s/\s+/\t/g' somefile | cut -f5-
by just replacing consecutive whitespaces by a single tab;
tr and cut:
tr can also be used to squeeze consecutive characters with the -s option.
tr -s [:blank:] <somefile | cut -d' ' -f5-
If you don't want to reformat the part of the line that you don't chop off, the best solution I can think of is written in my answer in:
How to print all the columns after a particular number using awk?
It chops what is before the given field number N, and prints all the rest of the line, including field number N and maintaining the original spacing (it does not reformat). It doesn't mater if the string of the field appears also somewhere else in the line.
Define a function:
fromField () {
awk -v m="\x01" -v N="$1" '{$N=m$N; print substr($0,index($0,m)+1)}'
}
And use it like this:
$ echo " bat bi iru lau bost " | fromField 3
iru lau bost
$ echo " bat bi iru lau bost " | fromField 2
bi iru lau bost
Output maintains everything, including trailing spaces
In you particular case:
svn status | grep '\!' | fromField 2 > removedProjs
If your file/stream does not contain new-line characters in the middle of the lines (you could be using a different Record Separator), you can use:
awk -v m="\x0a" -v N="3" '{$N=m$N ;print substr($0, index($0,m)+1)}'
The first case will fail only in files/streams that contain the rare hexadecimal char number 1
This awk function returns substring of $0 that includes fields from begin to end:
function fields(begin, end, b, e, p, i) {
b = 0; e = 0; p = 0;
for (i = 1; i <= NF; ++i) {
if (begin == i) { b = p; }
p += length($i);
e = p;
if (end == i) { break; }
p += length(FS);
}
return substr($0, b + 1, e - b);
}
To get everything starting from field 3:
tail = fields(3);
To get section of $0 that covers fields 3 to 5:
middle = fields(3, 5);
b, e, p, i nonsense in function parameter list is just an awk way of declaring local variables.
All of the other answers given here and in linked questions fail in various ways given various possible FS values. Some leave leading and/or trailing white space, some convert every FS to the OFS, some rely on semantics that only apply when FS is the default value, some rely on negating FS in a bracket expression which will fail given a multi-char FS, etc.
To do this robustly for any FS, use GNU awk for the 4th arg to split():
$ cat tst.awk
{
split($0,flds,FS,seps)
for ( i=n; i<=NF; i++ ) {
printf "%s%s", flds[i], seps[i]
}
print ""
}
$ printf 'a b c d\n' | awk -v n=3 -f tst.awk
c d
$ printf ' a b c d\n' | awk -v n=3 -f tst.awk
c d
$ printf ' a b c d\n' | awk -v n=3 -F'[ ]' -f tst.awk
b c d
$ printf ' a b c d\n' | awk -v n=3 -F'[ ]+' -f tst.awk
b c d
$ printf 'a###b###c###d\n' | awk -v n=3 -F'###' -f tst.awk
c###d
$ printf '###a###b###c###d\n' | awk -v n=3 -F'###' -f tst.awk
b###c###d
Note that I'm using split() above because it's 3rg arg is a field separator, not just a regexp like the 2nd arg to match(). The difference is that field separators have additional semantics to regexps such as skipping leading and/or trailing blanks when the separator is a single blank char - if you wanted to use a while(match()) loop or any form of *sub() to emulate the above then you'd need to write code to implement those semantics whereas split() already implements them for you.
Awk examples looks complex here, here is simple Bash shell syntax:
command | while read -a cols; do echo ${cols[#]:1}; done
Where 1 is your nth column counting from 0.
Example
Given this content of file (in.txt):
c1
c1 c2
c1 c2 c3
c1 c2 c3 c4
c1 c2 c3 c4 c5
here is the output:
$ while read -a cols; do echo ${cols[#]:1}; done < in.txt
c2
c2 c3
c2 c3 c4
c2 c3 c4 c5
This would work if you are using Bash and you could use as many 'x ' as elements you wish to discard and it ignores multiple spaces if they are not escaped.
while read x b; do echo "$b"; done < filename
Perl:
#m=`ls -ltr dir | grep ^d | awk '{print \$6,\$7,\$8,\$9}'`;
foreach $i (#m)
{
print "$i\n";
}
UPDATE :
if you wanna use no function calls at all while preserving the spaces and tabs in between the remaining fields, then do :
echo " 1 2 33 4444 555555 \t6666666 " |
{m,g}awk ++NF FS='^[ \t]*[^ \t]*[ \t]+|[ \t]+$' OFS=
=
2 33 4444 555555 6666666
===================
You can make it a lot more straight forward :
svn status | [m/g]awk '/!/*sub("^[^ \t]*[ \t]+",_)'
svn status | [n]awk '(/!/)*sub("^[^ \t]*[ \t]+",_)'
Automatically takes care of the grep earlier in the pipe, as well as trimming out extra FS after blanking out $1, with the added bonus of leaving rest of the original input untouched instead of having tabs overwritten with spaces (unless that's the desired effect)
If you're very certain $1 does not contain special characters that need regex escaping, then it's even easier :
mawk '/!/*sub($!_"[ \t]+",_)'
gawk -c/P/e '/!/*sub($!_"""[ \t]+",_)'
Or if you prefer customizing FS+OFS to handle it all :
mawk 'NF*=/!/' FS='^[^ \t]*[ \t]+' OFS='' # this version uses OFS
This should be a reasonably comprehensive awk-field-sub-string-extraction function that
returns substring of $0 based on input ranges, inclusive
clamp in out of range values,
handle variable length field SEPs
has speedup treatments for ::
completely no inputs, returning $0 directly
input values resulting in guaranteed empty string ("")
FROM-field == 1
FS = "" that has split $0 out by individual chars
(so the FROM <(_)> and TO <(__)> fields behave like cut -c rather than cut -f)
original $0 restored, w/o overwriting FS seps with OFS
|
{m,g}awk '{
2 print "\n|---BEFORE-------------------------\n"
3 ($0) "\n|----------------------------\n\n ["
4 fld2(2, 5) "]\n [" fld2(3) "]\n [" fld2(4, 2)
5 "]<----------------------------------------------should be
6 empty\n [" fld2(3, 11) "]<------------------------should be
7 capped by NF\n [" fld2() "]\n [" fld2((OFS=FS="")*($0=$0)+11,
8 23) "]<-------------------FS=\"\", split by chars
9 \n\n|---AFTER-------------------------\n" ($0)
10 "\n|----------------------------"
11 }
12 function fld2(_,__,___,____,_____)
13 {
if (+__==(_=-_<+_ ?+_:_<_) || (___=____="")==__ || !NF) {
return $_
16 } else if (NF<_ || (__=NF<+__?NF:+__)<(_=+_?_:!_)) {
return ___
18 } else if (___==FS || _==!___) {
19 return ___<FS \
? substr("",$!_=$!_ substr("",__=$!(NF=__)))__
20 : substr($(_<_),_,__)
21 }
22 _____=$+(____=___="\37\36\35\32\31\30\27\26\25"\
"\24\23\21\20\17\16\6\5\4\3\2\1")
23 NF=__
24 if ($(!_)~("["(___)"]")) {
25 gsub("..","\\&&",___) + gsub(".",___,____)
27 ___=____
28 }
29 __=(_) substr("",_+=_^=_<_)
30 while(___!="") {
31 if ($(!_)!~(____=substr(___,--_,++_))) {
32 ___=____
33 break }
35 ___=substr(___,_+_^(!_))
36 }
37 return \
substr("",($__=___ $__)==(__=substr($!_,
_+index($!_,___))),_*($!_=_____))(__)
}'
those <TAB> are actual \t \011 but relabeled for display clarity
|---BEFORE-------------------------
1 2 33 4444 555555 <TAB>6666666
|----------------------------
[2 33 4444 555555]
[33]
[]<---------------------------------------------- should be empty
[33 4444 555555 6666666]<------------------------ should be capped by NF
[ 1 2 33 4444 555555 <TAB>6666666 ]
[ 2 33 4444 555555 <TAB>66]<------------------- FS="", split by chars
|---AFTER-------------------------
1 2 33 4444 555555 <TAB>6666666
|----------------------------
I wasn't happy with any of the awk solutions presented here because I wanted to extract the first few columns and then print the rest, so I turned to perl instead. The following code extracts the first two columns, and displays the rest as is:
echo -e "a b c d\te\t\tf g" | \
perl -ne 'my #f = split /\s+/, $_, 3; printf "first: %s second: %s rest: %s", #f;'
The advantage compared to the perl solution from Chris Koknat is that really only the first n elements are split off from the input string; the rest of the string isn't split at all and therefor stays completely intact. My example demonstrates this with a mix of spaces and tabs.
To change the amount of columns that should be extracted, replace the 3 in the example with n+1.
ls -la | awk '{o=$1" "$3; for (i=5; i<=NF; i++) o=o" "$i; print o }'
from this answer is not bad but the natural spacing is gone.
Please then compare it to this one:
ls -la | cut -d\ -f4-
Then you'd see the difference.
Even ls -la | awk '{$1=$2=""; print}' which is based on the answer voted best thus far is not preserve the formatting.
Thus I would use the following, and it also allows explicit selective columns in the beginning:
ls -la | cut -d\ -f1,4-
Note that every space counts for columns too, so for instance in the below, columns 1 and 3 are empty, 2 is INFO and 4 is:
$ echo " INFO 2014-10-11 10:16:19 main " | cut -d\ -f1,3
$ echo " INFO 2014-10-11 10:16:19 main " | cut -d\ -f2,4
INFO 2014-10-11
$
If you want formatted text, chain your commands with echo and use $0 to print the last field.
Example:
for i in {8..11}; do
s1="$i"
s2="str$i"
s3="str with spaces $i"
echo -n "$s1 $s2" | awk '{printf "|%3d|%6s",$1,$2}'
echo -en "$s3" | awk '{printf "|%-19s|\n", $0}'
done
Prints:
| 8| str8|str with spaces 8 |
| 9| str9|str with spaces 9 |
| 10| str10|str with spaces 10 |
| 11| str11|str with spaces 11 |
The top-voted answer by zed_0xff did not work for me.
I have a log where after $5 with an IP address can be more text or no text. I need everything from the IP address to the end of the line should there be anything after $5. In my case, this is actually within an awk program, not an awk one-liner so awk must solve the problem. When I try to remove the first 4 fields using the solution proposed by zed_0xff:
echo " 7 27.10.16. Thu 11:57:18 37.244.182.218" | awk '{$1=$2=$3=$4=""; printf "[%s]\n", $0}'
it spits out wrong and useless response (I added [..] to demonstrate):
[ 37.244.182.218 one two three]
There are even some suggestions to combine substr with this wrong answer, but that only complicates things. It offers no improvement.
Instead, if columns are fixed width until the cut point and awk is needed, the correct answer is:
echo " 7 27.10.16. Thu 11:57:18 37.244.182.218" | awk '{printf "[%s]\n", substr($0,28)}'
which produces the desired output:
[37.244.182.218 one two three]

Resources