I have a text file containing this :-
# Comment
# Comment
# Comment
property1
# Comment
# Comment
property2
I wanted to use unix command (awk/sed etc.) to search for a pattern with property2 and then delete all the comments before it. Hence, after operation output should be :-
# Comment
# Comment
# Comment
property1
This is what I tried (using awk command) :-
awk -v pat='^property2' -v comment='^#' '$1~pat{p=NR} p && NR>=p-3{del=($1~comment)} del{next} 1' test.txt
Basically, the logic I tried to use was :-
Search for property2
and then loop over previous 3 lines
Search if it is a comment (starts with #)
Delete those lines (including the searched pattern and the comments above).
Can someone help me achieve this? Thanks.
This, using any awk, might be what you're trying to do but it's not clear from your question:
$ awk -v RS= -v ORS='\n\n' -F'\n' '$NF != "property2"' file
# Comment
# Comment
# Comment
property1
You could use a scriptable editor such as ed to:
Search for the first match of property2 (anchored to the beginning of the line)
Search backwards from there for a line that does not start with #
From the line after this one until one that starts with property2, delete those lines
write the file out to disk
quit the editor
One way to write that would be:
#!/bin/sh
printf '%s\n' \
'/^property2' \
'?^[^#]' \
'+1,/^property2/d' \
'w' \
'q' \
| ed input > /dev/null
I've dropped the stdout of ed to /dev/null because it will report the lines that it matches along the way, which we're not interested in. ed will make the changes to the file "in-place". This ed-script will fail if there is not a non-empty, non-commented line before property2 (the backwards search will fail).
In your sample input, this will delete the blank line between the stanzas as well, which seems to match your desired output.
It is not clear what you are trying to do; maybe this is it:
Mac_3.2.57$cat test.txt
# Comment1
# Comment2
# Comment3
property1
# Comment4
# Comment5
property2
Mac_3.2.57$awk '{if(NR==FNR){{if($0!~/^#/&&startFound==1){startFound=0;end=NR};if($0~/^#/&&startFound==0){startFound=1;start=NR}}}else {if(FNR<start||FNR>=end){print}}}' test.txt test.txt
# Comment1
# Comment2
# Comment3
property1
property2
Mac_3.2.57$
This might work for you (GNU sed):
sed -E '/^#/{:a;N;/^property[^2]/Mb;/^property2/M!ba
:b;/^#|^property2/!P;s/[^\n]*\n//;tb;d}' file
If a line is not a comment, let it be.
Otherwise, accumulate the lines in the pattern space.
If a subsequent line begins with a property that is not 2, print the accumulate lines and repeat.
If a subsequent line does not begin with property2, continue accumulating lines.
Otherwise, remove comments and print any lines other than the last which is deleted.
Using gnu-sed with the -z commandline option to use NUL delimited records reading the whole input, and replace the match with an empty string:
sed -zE 's/(^|\n)#[^\n]*(\n#[^\n]*)*\nproperty2//g' test.txt
The pattern matches:
(^|\n)# Either match a newline or assert the start of the string
[^\n]* Match optional characters other than a newline
(\n#[^\n]*)* Optionally repeat matching a newline # and optional chars other than a newline
\nproperty2 Match a newline and property2
Output
# Comment
# Comment
# Comment
property1
Related
I have a file 0.txt containing the following value fields contents in parentheses:
(bread,milk,),
(rice,brand B,),
(pan,eggs,Brandc,),
I'm looking in OS and elsewhere for how to prepend the letter x to the beginning of each value between commas so that my output file becomes (using bash unix):
(xbread,xmilk,),
(xrice,xbrand B,),
(xpan,xeggs,xBrand C,),
the only thing I've really tried but not enough is:
awk '{gsub(/,/,",x");print}' 0.txt
for all purposes the prefix should not be applied to the last commas at the end of each line.
With awk
awk 'BEGIN{FS=OFS=","}{$1="(x"substr($1,2);for(i=2;i<=NF-2;i++){$i="x"$i}}1'
Explanation:
# Before you start, set the input and output delimiter
BEGIN{
FS=OFS=","
}
# The first field is special, the x has to be inserted
# after the opening (
$1="(x"substr($1,2)
# Prepend 'x' from field 2 until the previous to last field
for(i=2;i<=NF-2;i++){
$i="x"$i
}
# 1 is always true. awk will print in that case
1
The trick is to anchor the regexp so that it matches the whole comma-terminated substring you want to work with, not just the comma (and avoids other “special” characters in the syntax).
awk '{ gsub(/[^,()]+,/, "x&") } 1' 0.txt
sed -r 's/([^,()]+,)/x\1/g' 0.txt
Suppose I have text as:
This is a sample text.
I have 2 sentences.
text is present there.
I need to replace whole text between two 'text' words. The required solution should be
This is a sample text.
I have new sentences.
text is present there.
I tried using the below command but its not working:
sed -i 's/text.*?text/text\
\nI have new sentence/g' file.txt
With your shown samples please try following. sed doesn't support lazy matching in regex. With awk's RS you could do the substitution with your shown samples only. You need to create variable val which has new value in it. Then in awk performing simple substitution operation will so the rest to get your expected output.
awk -v val="your_new_line_Value" -v RS="" '
{
sub(/text\.\n*[^\n]*\n*text/,"text.\n"val"\ntext")
}
1
' Input_file
Above code will print output on terminal, once you are Happy with results of above and want to save output into Input_file itself then try following code.
awk -v val="your_new_line_Value" -v RS="" '
{
sub(/text\.\n*[^\n]*\n*text/,"text.\n"val"\ntext")
}
1
' Input_file > temp && mv temp Input_file
You have already solved your problem using awk, but in case anyone else will be looking for a sed solution in the future, here's a sed script that does what you needed. Granted, the script is using some advanced sed features, but that's the fun part of it :)
replace.sed
#!/usr/bin/env sed -nEf
# This pattern determines the start marker for the range of lines where we
# want to perform the substitution. In our case the pattern is any line that
# ends with "text." — the `$` symbol meaning end-of-line.
/text\.$/ {
# [p]rint the start-marker line.
p
# Next, we'll read lines (using `n`) in a loop, so mark this point in
# the script as the beginning of the loop using a label called `loop`.
:loop
# Read the next line.
n
# If the last read line doesn't match the pattern for the end marker,
# just continue looping by [b]ranching to the `:loop` label.
/^text/! {
b loop
}
# If the last read line matches the end marker pattern, then just insert
# the text we want and print the last read line. The net effect is that
# all the previous read lines will be replaced by the inserted text.
/^text/ {
# Insert the replacement text
i\
I have a new sentence.
# [print] the end-marker line
p
}
# Exit the script, so that we don't hit the [p]rint command below.
b
}
# Print all other lines.
p
Usage
$ cat lines.txt
foo
This is a sample text.
I have many sentences.
I have many sentences.
I have many sentences.
I have many sentences.
text is present there.
bar
$
$ ./replace.sed lines.txt
foo
This is a sample text.
I have a new sentence.
text is present there.
bar
Substitue
sed -i 's/I have 2 sentences./I have new sentences./g'
sed -i 's/[A-Z]\s[a-z].*/I have new sentences./g'
Insert
sed -i -e '2iI have new sentences.' -e '2d'
I need to replace whole text between two 'text' words.
If I understand, first text. (with a dot) is at the end of first line and second text at the beginning of third line. With awk you can get the required solution adding values to var s:
awk -v s='\nI have new sentences.\n' '/text.?$/ {s=$0 s;next} /^text/ {s=s $0;print s;s=""}' file
This is a sample text.
I have new sentences.
text is present there.
I have files that need to be removed from comments and white space until keyword . Line number varies . Is it possible to limit multiple continued sed substitutions based on Keyword ?
This removes all comments and white spaces from file :
sed -i -e 's/#.*$//' -e 's/;.*$//' -e '/^$/d' file
For example something like this :
# string1
# string2
some string
; string3
; string4
####
<Keyword_Keep_this_line_and_comments_white_space_after_this>
# More comments that need to be here
; etc.
sed -i '1,/keyword/{/^[#;]/d;/^$/d;}' file
I would suggest using awk and setting a flag when you reach your keyword:
awk '/Keyword/ { stop = 1 } stop || !/^[[:blank:]]*([;#]|$)/' file
Set stop to true when the line contains Keyword. Do the default action (print the line) when stop is true or when the line doesn't match the regex. The regex matches lines whose first non-blank character is a semicolon or hash, or blank lines. It's slightly different to your condition but I think it does what you want.
The command prints to standard output so you should redirect to a new file and then overwrite the original to achieve an "in-place edit":
awk '...' input > tmp && mv tmp input
Use grep -n keyword to get the line number that contains the keyword.
Use sed -i -e '1,N s/#..., when N is the line number that contains the keyword, to only remove comments on the lines 1 to N.
XYZNA0000778800Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
XYZNA0000778900Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
I have above file format from which I want to find a matching record. For example, match a number(7789) on line starting with XYZ and once matched look for a matching number (7345) in lines below starting with 1 until it reaches to line starting with 9. retrieve the entire line record. How can I accomplish this using shell script, awk, sed or any combination.
Expected Output:
XYZNA0000778900Z
17345000012300324000000004000000000000000
With sed one can do:
$ sed -n '/^XYZ.*7789/,/^9$/{/^1.*7345/p}' file
17345000012300324000000004000000000000000
Breakdown:
sed -n ' ' # -n disabled automatic printing
/^XYZ.*7789/, # Match line starting with XYZ, and
# containing 7789
/^1.*7345/p # Print line starting with 1 and
# containing 7345, which is coming
# after the previous match
/^9$/ { } # Match line that is 9
range { stuff } will execute stuff when it's inside range, in this case the range is starting at /^XYZ.*7789/ and ending with /^9$/.
.* will match anything but newlines zero or more times.
If you want to print the whole block matching the conditions, one can use:
$ sed -n '/^XYZ.*7789/{:s;N;/\n9$/!bs;/\n1.*7345/p}' file
XYZNA0000778900Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
This works by reading lines between ^XYZ.*7779 and ^9$ into the pattern
space. And then printing the whole thing if ^1.*7345 can be matches:
sed -n ' ' # -n disables printing
/^XYZ.*7789/{ } # Match line starting
# with XYZ that also contains 7789
:s; # Define label s
N; # Append next line to pattern space
/\n9$/!bs; # Goto s unless \n9$ matches
/\n1.*7345/p # Print whole pattern space
# if \n1.*7345 matches
I'd use awk:
awk -v rid=7789 -v fid=7345 -v RS='\n9\n' -F '\n' 'index($1, rid) { for(i = 2; i < $NF; ++i) { if(index($i, fid)) { print $i; next } } }' filename
This works as follows:
-v RS='\n9\n' is the meat of the whole thing. Awk separates its input into records (by default lines). This sets the record separator to \n9\n, which means that records are separated by lines with a single 9 on them. These records are further separated into fields, and
-F '\n' tells awk that fields in a record are separated by newlines, so that each line in a record becomes a field.
-v rid=7789 -v fid=7345 sets two awk variables rid and fid (meant by me as record identifier and field identifier, respectively. The names are arbitrary.) to your search strings. You could encode these in the awk script directly, but this way makes it easier and safer to replace the values with those of a shell variables (which I expect you'll want to do).
Then the code:
index($1, rid) { # In records whose first field contains rid
for(i = 2; i < $NF; ++i) { # Walk through the fields from the second
if(index($i, fid)) { # When you find one that contains fid
print $i # Print it,
next # and continue with the next record.
} # Remove the "next" line if you want all matching
} # fields.
}
Note that multi-character record separators are not strictly required by POSIX awk, and I'm not certain if BSD awk accepts it. Both GNU awk and mawk do, though.
EDIT: Misread question the first time around.
an extendable awk script can be
$ awk '/^9$/{s=0} s&&/7345/; /^XYZ/&&/7789/{s=1} ' file
set flag s when line starts with XYZ and contains 7789; reset when line is just 9, and print when flag is set and contains pattern 7345.
This might work for you (GNU sed):
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^XYZ[^\n]*7789/!b;/7345/p' file
Use the option -n for the grep-like nature of sed. Gather up records beginning with XYZ and ending in 9. Reject any records which do not have 7789 in the header. Print any remaining records that contain 7345.
If the 7345 will always follow the header,this could be shortened to:
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^XYZ[^\n]*7789.*7345/p' file
If all records are well-formed (begin XYZ and end in 9) then use:
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^[^\n]*7789.*7345/p' file
I'm trying to delete two lines either side of a pattern match from a file full of transactions. Ie. find the match then delete two lines before it, then delete two lines after it and then delete the match. The write this back to the original file.
So the input data is
D28/10/2011
T-3.48
PINITIAL BALANCE
M
^
and my pattern is
sed -i '/PINITIAL BALANCE/,+2d' test.txt
However this is only deleting two lines after the pattern match and then deleting the pattern match. I can't work out any logical way to delete all 5 lines of data from the original file using sed.
an awk one-liner may do the job:
awk '/PINITIAL BALANCE/{for(x=NR-2;x<=NR+2;x++)d[x];}{a[NR]=$0}END{for(i=1;i<=NR;i++)if(!(i in d))print a[i]}' file
test:
kent$ cat file
######
foo
D28/10/2011
T-3.48
PINITIAL BALANCE
M
x
bar
######
this line will be kept
here
comes
PINITIAL BALANCE
again
blah
this line will be kept too
########
kent$ awk '/PINITIAL BALANCE/{for(x=NR-2;x<=NR+2;x++)d[x];}{a[NR]=$0}END{for(i=1;i<=NR;i++)if(!(i in d))print a[i]}' file
######
foo
bar
######
this line will be kept
this line will be kept too
########
add some explanation
awk '/PINITIAL BALANCE/{for(x=NR-2;x<=NR+2;x++)d[x];} #if match found, add the line and +- 2 lines' line number in an array "d"
{a[NR]=$0} # save all lines in an array with line number as index
END{for(i=1;i<=NR;i++)if(!(i in d))print a[i]}' #finally print only those index not in array "d"
file # your input file
sed will do it:
sed '/\n/!N;/\n.*\n/!N;/\n.*\n.*PINITIAL BALANCE/{$d;N;N;d};P;D'
It works this way:
if sed has only one string in pattern space it joins another one
if there are only two it joins the third one
if it does natch to pattern LINE + LINE + LINE with BALANCE it joins two following strings, deletes them and goes at the beginning
if not, it prints the first string from pattern and deletes it and goes at the beginning without swiping the pattern space
To prevent the appearance of pattern on the first string you should modify the script:
sed '1{/PINITIAL BALANCE/{N;N;d}};/\n/!N;/\n.*\n/!N;/\n.*\n.*PINITIAL BALANCE/{$d;N;N;d};P;D'
However, it fails in case you have another PINITIAL BALANCE in string which are going to be deleted. However, other solutions fails too =)
For such a task, I would probably reach for a more advanced tool like Perl:
perl -ne 'push #x, $_;
if (#x > 4) {
if ($x[2] =~ /PINITIAL BALANCE/) { undef #x }
else { print shift #x }
}
END { print #x }' input-file > output-file
This will remove 5 lines from the input file. These lines will be the 2 lines before the match, the matched line, and the two lines afterwards. You can change the total number of lines being removed modifying #x > 4 (this removes 5 lines) and the line being matched modifying $x[2] (this makes the match on the third line to be removed and so removes the two lines before the match).
A more simple and easy to understand solution might be:
awk '/PINITIAL BALANCE/ {print NR-2 "," NR+2 "d"}' input_filename \
| sed -f - input_filename > output_filename
awk is used to make a sed-script that deletes the lines in question and the result is written on the output_filename.
This uses two processes which might be less efficient than the other answers.
This might work for you (GNU sed):
sed ':a;$q;N;s/\n/&/2;Ta;/\nPINITIAL BALANCE$/!{P;D};$q;N;$q;N;d' file
save this code into a file grep.sed
H
s:.*::
x
s:^\n::
:r
/PINITIAL BALANCE/ {
N
N
d
}
/.*\n.*\n/ {
P
D
}
x
d
and run a command like this:
`sed -i -f grep.sed FILE`
You can use it so either:
sed -i 'H;s:.*::;x;s:^\n::;:r;/PINITIAL BALANCE/{N;N;d;};/.*\n.*\n/{P;D;};x;d' FILE