I'm new to Unix in all its forms, so please go easy on me!
I have a bash script that will pipe an ls command with arbitrary filenames into sed, which will use an arbitrary replacement pattern on the files, and then this will be piped into awk for some processing. The catch is, awk needs to know both the original file name and the new one.
I've managed everything except getting the original file names into awk. For instance, let's say my files are test.* and my replacement pattern is 's:es:ar;', which would change every occurrence of "test" to "tart". For testing purposes I'm just using awk to print what it's receiving:
ls "$#" | sed "$pattern" | awk '{printf "0: %s\n1: %s\n2: %s\n", $0,$1,$2}'
where test.* is in $# and the pattern is stored in $pattern.
Clearly, this doesn't get me to where I want to be. The output is obviously
0: tart.c
1: tart.c
2:
If I could get sed to output "test.c tart.c", then I'd have two parameters for awk. I've played around with the pattern to no avail, even hardcoding "test.c" into the replacement. But of course that just gave me amateur results like "ttest.c art.c". Is it possible for sed to remember the input, then work it into the beginning of the output? Do I even have the right ideas? Thanks in advance!
Two ways to change the first t in a b in the duplicated field.
Duplicate (& replays the matched part), change first word and swap (remember 2 strings with a space in between):
echo test.c | sed -r 's/.*/& &/;s/t/b/;s/([^ ]*) (.*)/\2 \1/'
or with more magic (copy original value to buffer, make the change, insert value from buffer as the first line and replace eond of line with a space)
echo test.c | sed 'h;s/t/b/;x;G;s/\n/ /'
Use Perl instead of sed:
echo test.c | perl -lne 'print "$_ ", s/es/ar/r'
-l removes the newline from input and adds it after each print. The /r modifier to the substitution returns the modified string instead of changing the variable (Perl 5.14+ needed).
Old answer, not working for s/t/b/2 or s/.*/replaced/2:
You can duplicate the contents of the line with s/.*/& &/, then just tell sed that it should only apply the second substitution (this works at least in GNU sed):
echo test.c | sed 's/.*/& &/; s/es/ar/2'
$ echo 'foo' | awk '{old=$0; gsub(/o/,"e"); print old, $0}'
foo fee
Related
if I have the following:
>AB ABABABA
>AC ACACACA
how do I shift everything onto a newline after the space i.e.
>AB
ABABABABA
>AC
ACACACACA
I have tried:
cat file | sed 's/ /\n/g'
cat file | tr ' ' '\n'
however I get the exact same output.
** UPDATE **
Upon inspecting the file using less and nano, the output was different to using cat. The file contains some terminal escape characters that aren't displayed in cat, but are in less. (how does this even happen?)
This was a terrible bug to spot and everyone has actually posted corrected answers based on the output of cat. So thank you for your help. Could the mods close this one?
It seems that you need to replace (any kind of) space with a newline
perl -pe 's/\s+/\n/' data.txt
This produces the required output in my tests. The -p sets up the loop over input (opening files or using STDIN) and sets $_ to the current line. It also prints $_ each time after processing.
If there are multiple spaces, each to be replaced by \n, add /g modifier.
If there may be more to do you can also capture patterns and replace them
perl -pe 's/\s+(.*)/\n$1/' data.txt
Following the observation in the answer by glenn jackman and looking "more closely" it appears that the first word on the line need be copied to the next line. Then the above is modified to
perl -pe 's/^>(\S+)\K\s+/\n$1';
The \K is a particular form of the positive lookbehind, which asserts that the pattern preceeds the current match position but it discards all previous matches (so you don't have to capture and copy them). You can find it in perlre. Without it the >(\S+) would be consumed so it would have to be copied back in the replacement part, as />$1\n$1/.
Are you trying to move the content from before the space onto the next line as well?
As in >A BC becomes:
>A
ABC
Then one can use sed like this:
$ sed 'h;s/^>\([^ ]*\) /\1/;x;s/ .*/ /;G' file
>AB
ABABABABA
>AC
ACACACACA
Breakdown:
h; # Copy pattern space to hold buffer
s/^>\([^ ]*\) /\1/; # Convert >A BC to ABC
x; # eXchange hold buffer and pattern space
s/ .*/ /; # Remove everything after, but including the
# first space: >A BC -> >A
G # Append hold buffer to pattern space
Looking more closely it looks like you want to repeat the first word on the next line: to transform this
>foo bar
>baz qux
into this
>foo
foobar
>baz
bazqux
If that's true, you can do
sed -r 's/^>([^ ]+) />\1\n\1/' file # or
perl -pe 's/^>(\S+) />$1\n$1/' file
sed is for simple substitutions on individual lines, that is all. For anything else you should be using awk, e.g.:
$ awk '{print $1 ORS substr($1,2) $2}' file
>AB
ABABABABA
>AC
ACACACACA
I am in need to extract all characters after a pattern match.
For example ,
NAME=John
Age=16
I need to extract all characters after "=". Output should be like
John
16
I cant go with perl or Jython for this purpose because of some restrictions.
I tried with grep , but to my knowledge I came as shown below only
echo "NAME=John" |grep -o -P '=.{0,}'
You were pretty close:
grep -oP '(?<=\w=)\w+' file
makes it.
Explanation
it looks for any word after word= and prints it.
-o stands for "Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line".
-P stands for "Interpret PATTERN as a Perl regular expression".
(?<=\w=)\w+ means: match only \w+ following word=. More info in [Regex tutorial - Lookahead][1] and in [this nice explanation by sudo_O][2].
Test
$ cat file
NAME=John
Age=16
$ grep -oP '(?<=\w=)\w+' file
John
16
One sed solution
sed -ne 's/.*=//gp' <filename>
another awk solution
awk -F= '$0=$2' <filename>
Explanation:
in sed we remove anything from the beginning of a line till a = and print the rest.
in awk we break the string in 2 parts, separated by =, now after that $0=$2 is making replacing the whole string with the second portion
I would like ignore all lines which occur before a match in bash (also ignoring the matched line. Example of input could be
R1-01.sql
R1-02.sql
R1-03.sql
R1-04.sql
R2-01.sql
R2-02.sql
R2-03.sql
and if I match R2-01.sql in this already sorted input I would like to get
R2-02.sql
R2-03.sql
Many ways possible. For example: assuming that your input is in list.txt
PATTERN="R2-01.sql"
sed "0,/$PATTERN/d" <list.txt
because, the 0,/pattern/ works only on GNU sed, (e.g. doesn't works on OS X), here is an tampered solution. ;)
PATTERN="R2-01.sql"
(echo "dummy-line-to-the-start" ; cat - ) < list.txt | sed "1,/$PATTERN/d"
This will add one dummy line to the start, so the real pattern must be on line the 1 or higher, so the 1,/pattern/ will works - deleting everything from the line 1 (dummy one) up to the pattern.
Or you can print lines after the pattern and delete the 1st, like:
sed -n '/pattern/,$p' < list.txt | sed '1d'
with awk, e.g.:
awk '/pattern/,0{if (!/pattern/)print}' < list.txt
or, my favorite use the next perl command:
perl -ne 'print unless 1../pattern/' < list.txt
deletes the 1.st line when the pattern is on 1st line...
another solution is reverse-delete-reverse
tail -r < list.txt | sed '/pattern/,$d' | tail -r
if you have the tac command use it instead of tail -r The interesant thing is than the /pattern/,$d' works on the last line but the1,/pattern/d` doesn't on the first.
How to ignore all lines before a match occurs in bash?
The question headline and your example don't quite match up.
Print all lines from "R2-01.sql" in sed:
sed -n '/R2-01.sql/,$p' input_file.txt
Where:
-n suppresses printing the pattern space to stdout
/ starts and ends the pattern to match (regular expression)
, separates the start of the range from the end
$ addresses the last line in the input
p echoes the pattern space in that range to stdout
input_file.txt is the input file
Print all lines after "R2-01.sql" in sed:
sed '1,/R2-01.sql/d' input_file.txt
1 addresses the first line of the input
, separates the start of the range from the end
/ starts and ends the pattern to match (regular expression)
$ addresses the last line in the input
d deletes the pattern space in that range
input_file.txt is the input file
Everything not deleted is echoed to stdout.
This is a little hacky, but it's easy to remember for quickly getting the output you need:
$ grep -A99999 $match $file
Obviously you need to pick a value for -A that's large enough to match all contents; if you use a too-small value the output will be silently truncated.
To ensure you get all output you can do:
$ grep -A$(wc -l $file) $match $file
Of course at that point you might be better off with the sed solutions, since they don't require two reads of the file.
And if you don't want the matching line itself, you can simply pipe this command into tail -n+1 to skip the first line of output.
awk -v pattern=R2-01.sql '
print_it {print}
$0 ~ pattern {print_it = 1}
'
you can do with this,but i think jomo666's answer was better.
sed -nr '/R2-01.sql/,${/R2-01/d;p}' <<END
R1-01.sql
R1-02.sql
R1-03.sql
R1-04.sql
R2-01.sql
R2-02.sql
R2-03.sql
END
Perl is another option:
perl -ne 'if ($f){print} elsif (/R2-01\.sql/){$f++}' sql
To pass in the regex as an argument, use -s to enable a simple argument parser
perl -sne 'if ($f){print} elsif (/$r/){$f++}' -- -r=R2-01\\.sql file
This can be accomplished with grep, by printing a large enough context following the $match. This example will output the first matching line followed by 999,999 lines of "context".
grep -A999999 $match $file
For added safety (in case the $match begins with a hyphen, say) you should use -e to force $match to be used as an expression.
grep -A999999 -e '$match' $file
I have a properties file, which, when unmodified has the following line:
worker.list=
I would like to use sed to append to that line a value so that after sed has run, the line in the file reads:
worker.list=test
But, when I run the script a second time, I want sed to pick up that a value has already been added, and thus adds a separator:
worker.list=test,test
That's the bit that stumps me (frankly sed scares me with its power, but that's my problem!)
Rich
Thats easy! If you're running GNU sed, you can write it rather short
sed -e '/worker.list=/{s/$/,myValue/;s/=,/=/}'
That'll add ',myValue' to the line, and then remove the comma (if any) after the equal sign.
If you're stuck on some other platform you need to break it apart like so
sed -e '/worker.list=/{' -e 's/$/,myValue/' -e 's/=,/=/' -e '}'
It's a pretty stupid script in that it doesn't know about existance of values etc (I suppose you CAN do a more elaborate parsing, but why should you?), but I guess that's the beauty of it. Oh and it'll destroy a line like this
worker.list=,myval
which will turn into
worker.list=myval,test
If that's a problem let me know, and I'll fix that for you.
HTH.
you can also use awk. Set field delimiter to "=". then what you want to append is always field number 2. example
$ more file
worker.list=
$ awk -F"=" '/worker\.list/{$2=($2=="")? $2="test" : $2",test"}1' OFS="=" file
worker.list=test
$ awk -F"=" '/worker\.list/{$2=($2=="")? $2="test" : $2",test"}1' OFS="=" file >temp
$ mv temp file
$ awk -F"=" '/worker\.list/{$2=($2=="")? $2="test1" : $2",test1"}1' OFS="=" file
worker.list=test,test1
or the equivalent of the sed answer
$ awk -F"=" '/worker\.list/{$2=",test1";sub("=,","=")}1' OFS="=" file
[Editorial insertion: Possible duplicate of the same poster's earlier question?]
Hi, I need to extract from the file:
first
second
third
using the grep command, the following line:
second
third
How should the grep command look like?
Instead of grep, you can use pcregrep which supports multiline patterns
pcregrep -M 'second\nthird' file
-M allows the pattern to match more than one line.
Your question abstract "bash grep newline", implies that you would want to match on the second\nthird sequence of characters - i.e. something containing newline within it.
Since the grep works on "lines" and these two are different lines, you would not be able to match it this way.
So, I'd split it into several tasks:
you match the line that contains "second" and output the line that has matched and the subsequent line:
grep -A 1 "second" testfile
you translate every other newline into the sequence that is guaranteed not to occur in the input. I think the simplest way to do that would be using perl:
perl -npe '$x=1-$x; s/\n/##UnUsedSequence##/ if $x;'
you do a grep on these lines, this time searching for string ##UnUsedSequence##third:
grep "##UnUsedSequence##third"
you unwrap the unused sequences back into the newlines, sed might be the simplest:
sed -e 's/##UnUsedSequence##/\n'
So the resulting pipe command to do what you want would look like:
grep -A 1 "second" testfile | perl -npe '$x=1-$x; s/\n/##UnUsedSequence##/ if $x;' | grep "##UnUsedSequence##third" | sed -e 's/##UnUsedSequence##/\n/'
Not the most elegant by far, but should work. I'm curious to know of better approaches, though - there should be some.
I don't think grep is the way to go on this.
If you just want to strip the first line from any file (to generalize your question), I would use sed instead.
sed '1d' INPUT_FILE_NAME
This will send the contents of the file to standard output with the first line deleted.
Then you can redirect the standard output to another file to capture the results.
sed '1d' INPUT_FILE_NAME > OUTPUT_FILE_NAME
That should do it.
If you have to use grep and just don't want to display the line with first on it, then try this:
grep -v first INPUT_FILE_NAME
By passing the -v switch, you are telling grep to show you everything but the expression that you are passing. In effect show me everything but the line(s) with first in them.
However, the downside is that a file with multiple first's in it will not show those other lines either and may not be the behavior that you are expecting.
To shunt the results into a new file, try this:
grep -v first INPUT_FILE_NAME > OUTPUT_FILE_NAME
Hope this helps.
I don't really understand what do you want to match. I would not use grep, but one of the following:
tail -2 file # to get last two lines
head -n +2 file # to get all but first line
sed -e '2,3p;d' file # to get lines from second to third
(not sure how standard it is, it works in GNU tools for sure)
So you just don't want the line containing "first"? -v inverts the grep results.
$ echo -e "first\nsecond\nthird\n" | grep -v first
second
third
Line? Or lines?
Try
grep -E -e '(second|third)' filename
Edit: grep is line oriented. you're going to have to use either Perl, sed or awk to perform the pattern match across lines.
BTW -E tell grep that the regexp is extended RE.
grep -A1 "second" | grep -B1 "third" works nicely, and if you have multiple matches it will even get rid of the original -- match delimiter
grep -E '(second|third)' /path/to/file
egrep -w 'second|third' /path/to/file
you could use
$ grep -1 third filename
this will print a string with match and one string before and after. Since "third" is in the last string you get last two strings.
I like notnoop's answer, but building on AndrewY's answer (which is better for those without pcregrep, but way too complicated), you can just do:
RESULT=`grep -A1 -s -m1 '^\s*second\s*$' file | grep -s -B1 -m1 '^\s*third\s*$'`
grep -v '^first' filename
Where the -v flag inverts the match.