How can I do so? I have multiple e.g. foobar patterns in my file, how can I add after e.g. the 4th one some_text?
Does the following work for you?
sed ':a;$!{N;ba};s/\(foobar\)/\1\nsome_text/4' inputfile
For the input:
$ cat inputfile
line
foobar
foobar
foobar
line
foobar
line
foobar
line
foobar
line
This would generate
$ sed ':a;$!{N;ba};s/\(foobar\)/\1\nsome_text/4' inputfile
line
foobar
foobar
foobar
line
foobar
some_text
line
foobar
line
foobar
line
Use -e on FreeBSD:
sed -e :a -e '$!N' -e '$!ba' -e 's/\(foobar\)/\1\nsome_text/4' inputfile
an awk version:
awk '/foo/{x++} x==4{sub(/foo/,"&some_text")}1' file
example:
kent$ cat file
foo
foo
trash
foo
trash
foo
foo
foo
kent$ awk '/foo/{x++} x==4{sub(/foo/,"&some_text")}1' file
foo
foo
trash
foo
trash
foosome_text
foo
foo
Related
I have a file with a certain line, let's say...
AAA BBB CCC
I need to replace that entire line, after finding it, so I did:
q1=`grep -Hnm 1 "AAA" FILE | cut -d : -f 2`
That outputs me the line number of the first occurrence (in q1), because it has more than one occurrence, now, here comes my problem... In a previous step I was using this sed to replace a certain line in the file:
sed -e '3s/.*/WHATEVER/' FILE
To replace (in the example, line 3) the full line with WHATEVER, but now if I try to use $q1 instead of the "3" indicating the line number it doesn't work:
sed -e '$q1s/.*/WHATEVER/' FILE
It's probably a stupid syntax mistake, any help is welcome; thanks in advance
Try:
sed -e "${q1}s/.*/WHATEVER/" FILE
I'd use awk for this:
awk '/AAA/ && !r {print "WHATEVER"; r=1; next} {print}' <<END
a
b
AAA BBB CCC
d
e
AAA foo bar
f
END
a
b
WHATEVER
d
e
AAA foo bar
f
If you want to replace the first occurrence of a string in a file, you could use this awk script:
awk '/occurrence/ && !f++ {print "replacement"; next}1' file
The replacement will only be printed the first time, as !f++ will only evaluate to true once (on subsequent evaluations, f will be greater than zero so !f will be false. The 1 at the end is always true, so for each line other than the matched one, awk does the default action and prints the line.
Testing it out:
$ cat file
blah
blah
occurrence 1 and some other stuff
blah
blah
some more stuff and occurrence 2
blah
$ awk '/occurrence/ && !f++ {print "replacement"; next}1' file
blah
blah
replacement
blah
blah
some more stuff and occurrence 2
blah
The "replacement" string could easily be set to the value of a shell variable in the following way:
awk -v rep="$r" '/occurrence/ && !f++ {print rep; next}1' file
where $r is a shell variable.
Using the same file as above and the example variable in your comment:
$ q2="x=\"Second\""
$ awk -v rep="$q2" '/occurrence/ && !f++ {print rep; next}1' file
blah
blah
x="Second"
stuff
blah
blah
some more stuff and occurrence 2
blah
sed "${q1} c\
WHATEVER" YourFile
but you can directly use
sed '/YourPatternToFound/ {s/.*/WHATEVER/
:a
N;$!ba
}' YourFile
I am trying to use sed script to remove the content of an array in a file. I have tried to delete the content to only leave the brackets (). However I can't get the sed script to work over multiple lines.
I am trying to change the current state of the file:
LIST = ( "content"
"content1"
"content3")
to this:
LIST = ()
However the sed script I am using only changes the file to this:
LIST = ()
"content"
"content1"
"content2"
sed -e 's/LIST=\([^)]*\)/LIST=() /g' filename
I should also mention there are other sets of brackets in the file which I don't want affected.
e.g
LISTNUMBER2("CONTENT")
should not be emptied.
this sed one-liner works for your example:
sed -n '1!H;1h;${x;s/(.*)/()/;p}'
test:
kent$ echo 'LIST = ( "content"
"content1"
"content3")'|sed -n '1!H;1h;${x;s/(.*)/()/;p}'
LIST = ()
if you could use awk, this one-liner works for your example too:
awk -v RS="" '{sub(/\(.*\)/,"()")}1'
test:
kent$ echo 'LIST = ( "content"
"content1"
"content3")'|awk -v RS="" '{sub(/\(.*\)/,"()")}1'
LIST = ()
EDIT for OP's comment
multi brackets situation:
awk
awk -v RS="\0" -v ORS="" '{gsub(/LIST\s*=\s*\([^)]*\)/,"LIST = ()")}1' file
test:
kent$ cat file
LISTKEEP2("CONTENT")
LIST = ( "content"
"content1"
"content3")
LISTNUMBER2("CONTENT")
kent$ awk -v RS="\0" -v ORS="" '{gsub(/LIST\s*=\s*\([^)]*\)/,"LIST = ()")}1' file
LISTKEEP2("CONTENT")
LIST = ()
LISTNUMBER2("CONTENT")
sed:
sed -nr '1!H;1h;${x;s/(LIST\s*=\s*\()[^)]*\)/\1)/;p}' file
kent$ sed -nr '1!H;1h;${x;s/(LIST\s*=\s*\()[^)]*\)/\1)/;p}' file
LISTKEEP2("CONTENT")
LIST = ()
LISTNUMBER2("CONTENT")
Another sed solution:
sed '/LIST = (/{:next;/)/{s/(.*)/()/;b;};N;b next;}'
Here's a version that would not change any block containing a certain string ("keepme" in this example, but could be anything):
sed '/LIST = (/{:next;/)/{/keepme/b;s/(.*)/()/;b;};N;b next;}'
Since this does the keepme test after it finds the closing parenthesis that tag can be anywhere in the block.
Is it possible to encapsulate the following pseudocode using sed?
for line in lines:
if line == "foo":
print "FOO"
else:
print "- " + line
Here's the first thing I tried:
> echo 'foo
> bar
> baz' | sed -e 's/^foo$/FOO/' -e 's/^/- /'
- FOO
- bar
- baz
This is incorrect since both substitutions are applied to the first line.
Is it possible to tell sed to perform a maximum of one substitution per line?
You can limit what lines a substitution affects, by prefixing it with a pattern:
sed -e '/^foo$/! s/^/- /' -e '/^foo$/ s//FOO/' infile
A better alternative is to use the t branch command which will go to the next line if the previous substitution succeeded:
sed 's/^foo$/FOO/; t; s/^/- /' infile
Or the more portable:
sed -e 's/^foo$/FOO/' -e t -e 's/^/- /' infile
Output in both cases:
FOO
- bar
- baz
My text looks like this:
cat
catch
cat_mouse
catty
I want to replace "cat" with "dog".
When I do
sed "s/cat/dog/"
my result is:
dog
catch
cat_mouse
catty
How do I replace with sed if only part of the word matches?
There's a mistake :
You lack the g modifier
sed 's/cat/dog/g'
g
Apply the replacement to all matches to the regexp, not just the first.
See
http://www.gnu.org/software/sed/manual/html_node/The-_0022s_0022-Command.html
http://sed.sourceforge.net/sedfaq3.html#s3.1.3
If you want to replace only cat by dog only if part of the word matches :
$ perl -pe 's/cat(?=.)/dog/' file.txt
cat
dogch
dog_mouse
dogty
I use Positive Look Around, see http://www.perlmonks.org/?node_id=518444
If you really want sed :
sed '/^cat$/!s/cat/dog/' file.txt
bash-3.00$ cat t
cat
catch
cat_mouse
catty
To replace cat only if it is part of a string
bash-3.00$ sed 's/cat\([^$]\)/dog\1/' t
cat
dogch
dog_mouse
dogty
To replace all occurrences of cat:
bash-3.00$ sed 's/cat/dog/' t
dog
dogch
dog_mouse
dogty
awk solution for this
awk '{gsub("cat","dog",$0); print}' temp.txt
it's possible to change N (for example second occurrence) in file using one-line sed/awk except such method?:
line_num=`awk '/WHAT_TO_CHANGE/ {c++; if (c>=2) {c=NR;exit}}END {print c}' INPUT_FILE` && sed "$line_num,$ s/WHAT_TO_CHANGE/REPLACE_TO/g" INPUT_FILE > OUTPUT_FILE
Thanks
To change the Nth occurence in a line you can use this:
$ echo foo bar foo bar foo bar foo bar | sed 's/foo/FOO/2'
foo bar FOO bar foo bar foo bar
So all you have to do is to create a "one-liner" of your text, e.g. using tr
tr '\n' ';'
do your replacement and then convert it back to a multiline again using
tr ';' '\n'
This awk solution assumes that WHAT_TO_CHANGE occurs only once per line. The following replaces the second 'one' with 'TWO':
awk -v n=2 '/one/ { if (++count == n) sub(/one/, "TWO"); } 1' file.txt