Let's say I have a file file.txt:
hello
world
hi
there
this
wow
I know sed can operate on one line like:
sed -n '2p' file.txt
or a range :
sed -n '2,4p' file.txt
but how to operate on one specific line plus a range ?
sed -n '1line and 4,5' file.txt or in words print the first line and the range from 4th to 5th line ??
Could you please try following. Written and tested with shown samples in GNU sed. This will print line which has string hello and will print lines from 2nd line to 4th line.
sed -n '/hello/p;2,4p' Input_file
OR to give single line number along with a lines range try following:
sed -n '1p;2,4p' Input_file
Explanation: Using -n option to stop printing for all lines first. Then checking condition if line contains /hello then printing that line by p option. After that giving 2nd statement separated by ; to print a range from 2nd line to 4th line number.
Related
I have a markdown file that looks something like this:
markdown.md
# Title1
line 1
line 2
line 3
# Title2
line 1
line 2
line 3
I'd like to be able to delete one of the paragraphs by searching for the title. I would need to delete the title, the following line, and then every subsequent line that is not blank.
The desired output would be:
# Title2
line 1
line 2
line 3
I was doing some reading about using {} to group multiple commands together but I can't seem to quite get the syntax right.
cat markdown.md | sed '/^# Title1.*/,+1d {/^\s*$/d}'
My thinking was this would delete the line beginning with '# Title1', then the following line with ,+1d, then subsequent lines until a blank line, but i see the following error:
sed: 1: "/^# Title1.*/,+1d { ...": extra characters at the end of d command
I've tried a few variations but no luck. Any help would be appreciated!
This is the kind of sed puzzle that makes me wish for a slightly different tool.
sed -n -e '/Title1/!{p;d;};n;' -e ':a' -e 'n;/./ba'
Loosely translated: "Don't print anything. If it doesn't contain 'Title1', then all right, print it, then start over with the next line. But if it does contain 'Title1', then grab the next line (which will be blank), enter a loop, and keep grabbing new lines until you come to the next empty line."
Using GNU sed
$ sed -z 's/# Title1[^#]*//' input_file
# Title2
line 1
line 2
line 3
This might work for you (GNU sed):
sed '/^# /h;G;/\n# Title1/!P;d' file
If a line begins # , make a copy.
Append the copy to each line and if that line does not contain \n# Title1, print it.
Delete all lines.
Alternative:
sed '/^# Title1/{:a;N;/\n#/!s/\n//;ta;D}' file
I need to find the line with first occurrence of a pattern, then I need to replace the whole line with a completely new one.
I found this command that replaces the first occurrence of a pattern, but not the whole line:
sed -e "0,/something/ s//other-thing/" <in.txt >out.txt
If in.txt is
one two three
four something
five six
something seven
As a result I get in out.txt:
one two three
four other-thing
five six
something seven
However, when I try to modify this code to replace the whole line, as follows:
sed -e "0,/something/ c\COMPLETE NEW LINE" <in.txt >out.txt
This is what I get in out.txt:
COMPLETE NEW LINE
five six
something seven
Do you have any idea why the first line is lost?
The c\ command deletes all lines between and inclusive the first matching address through the second matching address, when used with 2 addresses, and prints out the text specified following the c\ upon matching the second address. If there is no line matching the second address in the input, it just deletes all lines (inclusively) between the first matching address through the last line. Since you want to replace one line only, you shouldn't use the c\ command on an address range. The c\ is immediately followed by a new-line character in normal usage.
The 0,/regexp/ address range is a GNU sed extension, which will try to match regexp in the first input line too, which is different from 1,/regexp/ in that aspect. So, the correct command in GNU sed could be
sed '0,/something/{/something/c\
COMPLETE NEW LINE
}' < in.txt
or simplified as pointed out by Sundeep
sed '0,/something/{//c\
COMPLETE NEW LINE
}' < in.txt
or a one-liner,
sed -e '0,/something/{//cCOMPLETE NEW LINE' -e '}' < in.txt
if a literal new-line character is not desirable.
This one-liner also works as pointed out by potong:
sed '0,/something/!b;//cCOMPLETE NEW LINE' in.txt
This might work for you (GNU sed):
sed '1!b;:a;/something/!{n;ba};cCOMPLETE NEW LINE' file
Set up a loop that will only operate from the first line.
Within in the loop, if the key word is not found in the current line, print the current line, fetch the next and repeat until the end of the file or a match is found.
When a match is found, change the contents of the current line to the required result.
N.B. The c command terminates any further processing of sed commands in the same way the d command does.
If there are lines in the input following the key word match, the negation of address at the start of the sed cycle will capture these lines and result in their printing and no further processing.
An alternative:
sed 'x;/./{x;b};x;/something/h;//cCOMPLETE NEW LINE' file
Or (specific to GNU and bash):
sed $'0,/something/{//cCOMPLETE NEW LINE\n}' file
Just use awk:
$ awk '!done && sub(/something/,"other-thing"){done=1} {print}' file
one two three
four other-thing
five six
something seven
$ awk '!done && sub(/.*something.*/,"other-thing"){done=1} {print}' file
one two three
other-thing
five six
something seven
$ awk '!done && /something/{$0="other-thing"; done=1} {print}' file
one two three
other-thing
five six
something seven
and look what you can trivially do if you want to replace the Nth occurrence of something:
$ awk -v n=1 '/something/ && (++cnt == n){$0="other-thing"} {print}' file
one two three
other-thing
five six
something seven
$ awk -v n=2 '/something/ && (++cnt == n){$0="other-thing"} {print}' file
one two three
four something
five six
other-thing
I am struggling to use sed to work through 'testfile.txt' and every time it encounters a line that starts delete_me: abc it will then:
leave the line delete_me: abc intact
but delete all the lines that follow until the next blank line is reached in the file.
eg. I want this input:
delete_me: abc
sSAsaAaSA
AsaSAsaSAsa
asASAsS
^--- <blank line>
...to be changed to just this one line:
delete_me: abc
I have tried:
sed '/delete_me/ {n;d}' jil_testfile.txt
# deletes only the first line after 'delete_me'
sed '/delete_me/,/^$/d' jil_testfile.txt
# nearly works but deletes the 'delete_me' line too which I want to stay preserved.
Any suggestions please?
This might work for you (GNU sed):
sed -n ':a;/delete_me/{p;:b;n;//ba;bb};p' file
Print lines as normal until the first occurrence of delete_me. Print this line and do not print any further lines unless that line contains delete_me.
As the spec has changed since I wrote the first solution, here is new one:
sed -n '/delete_me/{p;:a;n;/^$/b;ba};p' file
I would like to delete the first 100 lines of a text file using sed. I know how delete to the first line by using:
sed '1d' filename
or the 100th line by typing
sed '100d' filename
How do I specify a range? I thought something like this would work:
sed '1:100d' filename
However, this obviously didn't work. Can someone show me how to specify a range? Thanks in advance for your help.
This should work in gnu sed
sed '1,100d' file
awk can also be used to print data based on conditions related to rows.
Like: Following will print the lines (Records in terms of awk) whose number is greater than 100.
awk 'NR>100' inputfile
One can also use other conditions like:
awk 'NR==100' inpuftile #this will print the 100th line
awk 'NR<100' inputfile #this will print 1-99th line
awk 'NR>100' inputfile #this will print from 101st line onwards
awk 'NR>=100' inputfile #this will print from 100th onwards
try: following too:
sed -n '1,100p' Input_file
I need to replace a pattern in a file, only if it is followed by an empty line. Suppose I have following file:
test
test
test
...
the following command would replace all occurrences of test with xxx
cat file | sed 's/test/xxx/g'
but I need to only replace test if next line is empty. I have tried matching a hex code, but that doesn ot work:
cat file | sed 's/test\x0a/xxx/g'
The desired output should look like this:
test
xxx
xxx
...
Suggested solutions for sed, perl and awk:
sed
sed -rn '1h;1!H;${g;s/test([^\n]*\n\n)/xxx\1/g;p;}' file
I got the idea from sed multiline search and replace. Basically slurp the entire file into sed's hold space and do global replacement on the whole chunk at once.
perl
$ perl -00 -pe 's/test(?=[^\n]*\n\n)$/xxx/m' file
-00 triggers paragraph mode which makes perl read chunks separated by one or several empty lines (just what OP is looking for). Positive look ahead (?=) to anchor substitution to the last line of the chunk.
Caveat: -00 will squash multiple empty lines into single empty lines.
awk
$ awk 'NR==1 {l=$0; next}
/^$/ {gsub(/test/,"xxx", l)}
{print l; l=$0}
END {print l}' file
Basically store previous line in l, substitute pattern in l if current line is empty. Print l. Finally print the very last line.
Output in all three cases
test
xxx
xxx
...
This might work for you (GNU sed):
sed -r '$!N;s/test(\n\s*)$/xxx\1/;P;D' file
Keep a window of 2 lines throughout the length of the file and if the second line is empty and the first line contains the pattern then make a substitution.
Using sed
sed -r ':a;$!{N;ba};s/test([^\n]*\n(\n|$))/xxx\1/g'
explanation
:a # set label a
$ !{ # if not end of file
N # Add a newline to the pattern space, then append the next line of input to the pattern space
b a # Unconditionally branch to label. The label may be omitted, in which case the next cycle is started.
}
# simply, above command :a;$!{N;ba} is used to read the whole file into pattern.
s/test([^\n]*\n(\n|$))/xxx\1/g # replace the key word if next line is empty (\n\n) or end of line ($)