How to replace a whole line (between 2 words) using sed? - shell

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.

Related

How to get paragraphs of text by index number

I am wondering if there is a way to get paragraphs of text (source file would be a pyx file) by number as sed does with lines
sed -n ${i}p
At this moment I'd be interested to use awk with:
awk '/custom-pyx-tag\(/,/\)custom-pyx-tag/'
but I can't find documentation or examples about that.
I'm also trying to trim "\r\n" with gsub(/\r\n/,"; ") int the same awk command but it doesn't work, and I can't really figure out why.
Any hint would be very appreciated, thanks
EDIT:
This is just one example and not my exact need but I would need to know how to do it for a multipurpose project
Let's take the case that I have exported the ID3Tags of a huge collection of audio files and these have been stored in a pyx-like format, so in the end I will have a nice big file with this pattern repeating for each file in the collection:
audio-genre(
blablabla
)audio-genre
audio-artist(
bla.blabla
)audio-artist
audio album(
bla-bla-bla
)audio-album
audio-track-num(
0x
)audio-track-num
audio-track-title(
bla.bla-bla
)audio-track-title
audio-lyrics(
blablablablabla
bla.bla.bla.bla
blah-blah-blah
blabla-blabla
)audio-lyrics
...
Now if I want to extract the artist of the 1234th audio file I can use:
awk '/audio-artist\(/, /)audio-artist/' | sed '/audio-artist/d' | sed -n 1234p
so being one line it can be obtained with sed, but I don't know how to get an entire paragraph given its index, for example if I want to get the lyrics of the 6543th file how could I do it?
In the end it is just a question of whether there is a command equivalent to
sed -n $ {num} p
but to be used for paragraphs
awk -v indx=1024
'BEGIN {
RS=""
}
{ split($0,arr,"audio-artist");
for (i=2;i<=length(arr);i=i+2)
{ gsub("[()]","",arr[i]);
arts[cnt+=1]=arr[i]
}
}
END {
print arts[indx]
}' audioartist
One liner:
awk -v indx=1234 'BEGIN {RS=""} NR==1 { split($0,arr,"audio-artist");for (i=2;i<=length(arr);i=i+2) { gsub("[()]","",arr[i]);arts[cnt+=1]=arr[i] } } END { print arts[indx] }' audioartist
Using awk, and the file called audioartist, we consume the file as one line by setting the records separator (RS) to "". We then split the whole file into an array arr, based on the separator audio-artist. We look through the array arr starting from 2 in steps of 2 till the end of the array and strip out the opening and closing brackets, creating another array called arts with an incrementing count as the index and the stripped artist as the value. At the end we print the arts index specified by the passed indx variable (in this case 1234).

How can I retrieve the matching records from mentioned file format in bash

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

Removing lines between tags in a text file

I have many text files containing annotations. The original text is marked with lines containing the words:
START OF TEXT OF PASSAGE 1
END OF TEXT OF PASSAGE 1
Obviously I can search each document for the phrase START OF TEXT and delete everything up to it. Then search for END OF TEXT and start selecting text for deletion until I get to the next START OF TEXT.
I have come up with this design so far:
#!/bin/bash
a="START OF PROJECT"
b="END OF PROJECT"
while read line; do
if line contains a; do
while read line; do
'if line does not contain b'
'append the line to output.txt'; fi
done
done
fi
done
Perhaps there is an easier way using sed, awk, grep and pipes?
'for every document' 'loop through it doing this' ('find the original text between START and END' | >> output.txt)
Unfortunately I am poor at bash and ignorant of sed/awk.
The reason for this is that I am assembling a huge text document that is a concatenation of thousands of marked up documents – each of which contains some annotated passages.
In Python:
import re
with open('in.txt') as f, open('out.txt', 'w') as output:
output.write('\n'.join(re.findall(r'START OF TEXT(.*?)END OF TEXT', f.read())))
This reads the input, searches for all matches that begin and end with the necessary markers, captures the text of interest in a group, joins all those groups on a linefeed, and writes that to the result file.
Pretty easy to do with awk. You would create a script (I'll call it yank.awk) containing this:
#!/usr/bin/awk
/START OF PROJECT/ { capture = 1; next }
/END OF PROJECT/ { capture = 0 }
capture == 1 { print }
and then run it like so:
yank.awk in.txt > output.txt
Could also do with sed and grep:
sed -ne '/START OF PROJECT/,/END OF PROJECT/p' in.txt | grep -vE '(START|END) OF PROJECT' > output.txt
(Another Python solution)
You can have itertools.groupby group lines together based on a boolean value - just use a global flag to keep track of whether you are in a block or not, and then use groupby to group the lines that are in or out of blocks. Then just discard the ones that are not blocks:
sample_lines = """
lskdjflsdkjf
sldkjfsdlkjf
START OF TEXT
Asdlkfjlsdkfj
Bsldkjf
Clsdkjf
END OF TEXT
sldkfjlsdkjf
sdlkjfdklsjf
sdlkfjdlskjf
START OF TEXT
Dsdlkfjlsdkfj
Esldkjf
Flsdkjf
END OF TEXT
sldkfjlsdkjf
sdlkjfdklsjf
sdlkfjdlskjf
""".splitlines()
from itertools import groupby
in_block = False
def is_in_block(line):
global in_block
if line.startswith("END OF TEXT"):
in_block = False
ret = in_block
if line.startswith("START OF TEXT"):
in_block = True
return ret
for lines_are_text,lines in groupby(sample_lines, key=is_in_block):
if lines_are_text:
print(list(lines))
gives:
['Asdlkfjlsdkfj', 'Bsldkjf', 'Clsdkjf']
['Dsdlkfjlsdkfj', 'Esldkjf', 'Flsdkjf']
See that first group has the lines that start with A, B, and C, and the second group is made up of those lines starting with D, E, and F.
It sounds like the specific solution you need is:
awk '/END OF TEXT OF PASSAGE/{f=0} f; /START OF TEXT OF PASSAGE/{f=1}' file
See https://stackoverflow.com/a/18409469/1745001 for other ways to select text from files.
Use Perl's Flip-Flop Operator to Print Text Between Markers
Given a corpus like:
START OF TEXT OF PASSAGE 1
foo
END OF TEXT OF PASSAGE 1
START OF TEXT OF PASSAGE 2
bar
END OF TEXT OF PASSAGE 2
you can use the Perl flip-flop operator to process within a range of lines. For example, from the shell prompt:
$ perl -ne 'if (/^START OF TEXT/ ... /^END OF TEXT/) {
next if /^(?:START|END)/;
print;
}' /tmp/corpus
foo
bar
Basically, this short Perl script loops through your input. When it finds your start and end tags, it throws away the tags themselves and prints everything else in between.
Usage Notes
The line breaks between passages in the corpus are for readability. It doesn't matter if your real corpus has no line breaks between passages, so long as the text markers always start at the beginning of the line as shown in your original post. If that assumption doesn't hold true, then you will need to adjust the regular expressions used to identify the start and end of your passages.
You can pass multiple files to the Perl script. Again, it makes no practical difference as long as you don't exceed the length limit of your shell.
If you want the final output to go to somewhere other than standard output, just use shell redirection. For example:
perl -ne 'if (/^START OF TEXT/ ... /^END OF TEXT/) {
next if /^(?:START|END)/;
print;
}' /tmp/file1 /tmp/file2 /tmp/file3 > /tmp/output
You can use sed as follows:
sed -n '/^START OF TEXT/,/^END OF TEXT/{/^\(START\|END\) OF TEXT/!p}' infile
or, with extended regular expressions (-r):
sed -rn '/^START OF TEXT/,/^END OF TEXT/{/^(START|END) OF TEXT/!p}' infile
-n prevents sed from printing as a default. The rest works as follows:
/^START OF TEXT/,/^END OF TEXT/ { # For lines between these two matches
/^\(START\|END\) OF TEXT/!p # If the line does NOT match, print it
}
This works with GNU sed and might require some tweaking to run with other seds.

Delete lines before and after a match in bash (with sed or awk)?

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

Replace text with sed

A program creates HTML files from a database. There are headings and stuff in between the headings.
There are not a set amount of headings.
After each heading the program places the text:
$WHITE*("5")$
$WHITE*("20")$
$HRULE$
I need every occurrence of these 4 lines to be replaced with:
$WHITE*("20")$
$HRULE$
$WHITE*("10")$
I am not fussed what program is used :)
I have tried:
sed 's:\$WHITE\*(\"5\")\$\n\n\$WHITE\*(\"20\")\$\n\$HRULE\$:\$WHITE\*(\"20\")\$\
\$HRULE$\
\$WHITE*("10")$:g'
and various other permutations
If that'S your input file, and this is the spec, you can do:
sed -n '3,$p;$a$WHITE*("10")$' INPUTFILE
But I assume that's not the case, so you might want to rephrase your question and/or giving some more detailes.
More specific solution with sed:
sed '/^\$WHITE\*("5")\$$/,/^$/d;/\$HRULE\$/ a$WHITE*("10")$' INPUTFILE
(Searches for the $WHITE*("5")$ line and deletes it till (including!) the next empty line. Then searches for the next $HRULE$ line and appends an $WHITE*("10")$ line.
awk solution:
awk '/\$WHITE\*\("5"\)\$/ { getline ; next }
/\$WHITE\*\("20"\)\$/ { print ;
getline ;
if ($0 ~ /\$HRULE\$/) { print ;
print "$WHITE*(\"10\")$" ;
}
else { print }
}
1 ' INPUTFILE
This reads the file and prints every line - that's why the 1 is there, except if it finds the $WHITE*("5") pattern it drops it, reads the next line and drops that too. if it finds the $WHITE*("20") prints it. Reads the next line and if its $HRULE$ then prints that and the appended $WHITE*("10") line. Else just prints the line.
HTH
UPDATE #2
From the sed faq, section 4.23.3
If you need to match a static block of text (which may occur any number of times throughout a file), where the contents of the block are known in advance, then this script is easy to use
UPDATE #1
Python?
$ cat input
first line
second line
3rd line
$WHITE*("5")$
$WHITE*("20")$
$HRULE$
some more lines
yet another
$WHITE*("5")$
$WHITE*("20")$
$HRULE$
THE END
the script:
#!/usr/bin/env python
## Use these 3 lines for python version < 2.5
#fd=open('input')
#text=fd.read()
#fd.close()
## Use these 2 lines for python version >= 2.5
with open('input') as fd:
text=fd.read()
old="""$WHITE*("5")$
$WHITE*("20")$
$HRULE$
"""
new="""$WHITE*("20")$
$HRULE$
$WHITE*("10")$
"""
print text.replace(old,new)
output:
first line
second line
3rd line
$WHITE*("20")$
$HRULE$
$WHITE*("10")$
some more lines
yet another
$WHITE*("20")$
$HRULE$
$WHITE*("10")$
THE END
Try something like
sed -e '${p;};/$WHITE\*("5")\$/,/$HRULE\$/{H;/$HRULE\$/{g;s/$HRULE\$//;s/20/10/;s/5/20/;s/\n/&$HRULE$/2p;s/.*//p;x;d;};d;};' white.txt
Crude, but it should work.
This might work for you:
sed '/^\$WHITE\*(\"5\")\$/{N;N;N;s/.*\n\n\(\(\$WHITE\*(\"\)20\(\")\$\s*\)\n\$HRULE\$\s*$\)/\1\n\210\3/}' file
Explanation:
Match on first string $WHITE*("5")$, read the next 3 lines and match on remainder. Use grouping and back references to formulate output lines.

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