file xz.txt
123
456
789
I want to merge
sed -i '1d' xz.txt
sed -i '1a abc' xz.txt
I tried
sed -i -e '1d' -e '1a abc' xz.txt
expect to get
456
abc
789
but I got
456
789
sed (GNU sed) 4.7
but it doesn't work, any help?
Sed goes line by line, first command 1d - deleted 1st line, 1st line is gone, there is no more 1st line, that is why second command 1a abc didn't work. Here is how it should be:
$ sed '1d; 2a abc' f
456
abc
789
What is going on is that the delete statement automatically ends the processing sequence:
[1addr]a\
text Write text to standard output as described previously (yes there is a new-line here)
[2addr]d: Delete the pattern space and start the next cycle
Source: Posix)
As the a command does not modify the pattern space but just writes to stdout, you can simply do
[POSIX]$ sed -e '1a\
abc' -e '1d'
[GNU]$ sed -e '1a abc' -e '1d'
However, the easiest is just to use the replace command c:
[POSIX]$ sed -e '1c\
abc'
[GNU]$ sed -e '1c abc`
Note: The reason the commands a and c write directly to the output and not to the pattern space is most likely that it would mess up the address ranging using line-numbers.
Related
I have a file with the following content.
test1
test2
test3
test4
test5
If I want to concatenate all lines into one line separated by commas, I can use vi and run the following command:
:%s/\n/,/g
I then get this, which is what I want
test1,test2,test3,test4,test5,
I'm trying to use sed to do the same thing but I'm missing some unknown command/option to make it work. When I look at the file in vi and search for "\n" or "$", it finds the newline or end of line. However, when I tell sed to look for a newline, it pretends it didn't find one.
$ cat test | sed --expression='s/\n/,/g'
test1
test2
test3
test4
test5
$
If I tell sed to look for end of line, it finds it and inserts the comma but it doesn't concatenate everything into one line.
$ cat test | sed --expression='s/$/,/g'
test1,
test2,
test3,
test4,
test5,
$
What command/option do I use with sed to make it concatenate everything into one line and replace the end of line/newline with a comma?
sed reads one line at a time, so, unless you're doing tricky things, there's never a newline to replace.
Here's the trickiness:
$ sed -n '1{h; n}; H; ${g; s/\n/,/gp}' test.file
test1,test2,test3,test4,test5
h, H, g documented at https://www.gnu.org/software/sed/manual/html_node/Other-Commands.html
When using a non-GNU sed, as found on MacOS, semi-colons before the closing braces are needed.
However, paste is really the tool for this job
$ paste -s -d, test.file
test1,test2,test3,test4,test5
If you really want the trailing comma:
printf '%s,\n' "$(paste -sd, file)"
tr instead of sed for this one:
$ tr '\n' ',' < input.txt
test1,test2,test3,test4,test5,
Just straight up translate newlines to commas.
Based on how can i replace each newline n with a space using sed:
sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/,/g' <file>
testing:
$ cat file.txt
test1
test2
test3
test4
test5
$ sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/,/g' file.txt
test1,test2,test3,test4,test5
Of course, if the question would have been more generic: How do I replace \n with any character using sed then one should only replace the , with ones desired char:
export CHAR_TO_REPLACE=','
export FILE_TO_PROCESS=<filename>
sed -e ':a' -e 'N' -e '$!ba' -e "s/\n/${CHAR_TO_REPLACE}/g" $FILE_TO_PROCESS
This answer is to satisfy the requirement of using sed. Otherwise, you can use alternatives like tr, awk etc.
This might work for you (GNU sed):
sed 'H;1h;$!d;x;y/\n/,/' file
Append all lines but the first to the hold space (the first replaces the hold space).
If it is not the last line of the file, delete it.
Otherwise, swap to the hold space and translate all newlines to commas.
My sed command does not work as expected.
sed -E ':a;N;$!ba;s/[^\.](\[[0-9]]) \n\n/\1 /g'
I want this :
blabla[3]
foofoo
barbar.[4]
blabla
To become :
blabla[3] foofoo
barbar.[4]
blabla
That is, I just want the new lines to be deleted when there is no dot before "[".
But my sed command does not take into account my [^\.].
Without [^\.], I get :
blabla[3] foofoo
barbar.[4] blabla
You may use this sed (should work with non-gnu sed also):
sed -E -e ':a' -e 'N;$!ba' -e 's/((^|[^.])\[[0-9]+]) *\n\n/\1 /g' file
blabla[3] foofoo
barbar.[4]
blabla
Code Demo
Note that we are matching (^|[^.]) to allow [<digits>] to appear at the line start as well. It is also important to keep this part in capture group so that we don't miss out on this char in replacement.
is it possible to use the pipe to redirect the output of the previous command, to sed, and let sed use this as input(pattern or string) to access a file?
I know if you only use sed, you can use something like
sed -i '1 i\anything' file
But can I do something like
head -1 file1 | sed -i '1 i\OutputFromPreviousCmd' file2
This way, I don't need to manually copy the output and change the sed command everytime
Update:
Added the files I meant
head -3 file1.txt
Side A,Age(us),mm:ss.ms_us_ns_ps
84 Vendor Specific, 0000000009096, 0349588242
84 Vendor Specific, 0000000011691, 0349591828
head -3 file2.txt
84 Vendor Specific, 0000000000418, 0349575322
83 Vendor Specific, 0000000002099, 0349575343
83 Vendor Specific, 0000000001628, 0349576662
I'd like to grab the first line of file1 and insert it to file2, so the result should be :
head -3 file2.txt
Side A,Age(us),mm:ss.ms_us_ns_ps
84 Vendor Specific, 0000000000418, 0349575322
83 Vendor Specific, 0000000002099, 0349575343
83 Vendor Specific, 0000000001628, 0349576662
head -1 file1 | sed '1s/^/1i /' | sed -i -f- file2
This takes your one line of output, prepends the sed 1i command, the pipes that sed command stream to sed using -f- to take sed commands from stdin.
For example:
$ echo bob > bob.txt
$ echo alice | sed '1s/^/1i /' | sed -i -f- bob.txt
$ more bob.txt
alice
bob
This looks like pipes and not commands ending in > temp ; mv temp file2, but sed is doing that nonetheless when -i is used.
This might work for you (GNU sed):
head -1 file1 | sed -i '1e cat /dev/stdin' file2
Insert the first line of file1 into the start of file2.
But why not use cat?:
cat <(head -1 file1) file2
I use below to replace text on line numbers 29 to 32:
sed -i '29,32 s/^ *#//' file
How can I further add line numbers i.e. line 35 to 38, line 43 & line 45 in above command?
With GNU sed. m is here a label.
sed -i '29,32bm;35,38bm;43bm;45bm;b;:m;s/^ *#//' file
From man sed:
b label: Branch to label; if label is omitted, branch to end of script.
: label: Label for b and t commands.
This might work for you (GNU sed):
cat <<\! | sed 's:$:s/^ *#//:' | sed -f - -i file
29,32
35,38
43,45
!
Create a here-document with the line ranges you desire and pass these to a sed script that creates another sed script which is run against the desired file.
The here-document could be replaced by a file:
sed 's:$:s/^ *#//:' lineRangeFile | sed -f - -i file
I need to replace the pattern ### with the current line number.
I managed to Print in the next line with both AWK and SED.
sed -n "/###/{p;=;}" file prints to the next line, without the p;, it replaces the whole line.
sed -e "s/###/{=;}/g" file used to make sense in my head, since the =; returns the line number of the matched pattern, but it will return me the the text {=;}
What am i Missing? I know this is a silly question. I couldn't find the answer to this question in the sed manual, it's not quite clear.
If possible, point me what was i missing, and what to make it work. Thank you
Simple awk oneliner:
awk '{gsub("###",NR,$0);print}'
Given the limitations of the = command, I think it's easier to divide the job in two (actually, three) parts. With GNU sed you can do:
$ sed -n '/###/=' test > lineno
and then something like
$ sed -e '/###/R lineno' test | sed '/###/{:r;N;s/###\([^\n]*\n\)\([^\n]*\)/\2\1/;tr;:c;s/\n\n/\n/;tc}'
I'm afraid there's no simple way with sed because, as well as the = command, the r and GNU extension R commands don't read files into the pattern space, but rather directly append the lines to the output, so the contents of the file cannot be modified in any way. Hence piping to another sed command.
If the contents of test are
fooo
bar ### aa
test
zz ### bar
the above will produce
fooo
bar 2 aa
test
zz 4 bar
This might work for you (GNU sed):
sed = file | sed 'N;:a;s/\(\(.*\)\n.*\)###/\1\2/;ta;s/.*\n//'
An alternative using cat:
cat -n file | sed -E ':a;s/^(\s*(\S*)\t.*)###/\1\2/;ta;s/.*\t//'
As noted by Lev Levitsky this isn't possible with one invocation of sed, because the line number is sent directly to standard out.
You could have sed write a sed-script for you, and do the replacement in two passes:
infile
a
b
c
d
e
###
###
###
a
b
###
c
d
e
###
Find the lines that contain the pattern:
sed -n '/###/=' infile
Output:
6
7
8
11
15
Pipe that into a sed-script writing a new sed-script:
sed 's:.*:&s/###/&/:'
Output:
6s/###/6/
7s/###/7/
8s/###/8/
11s/###/11/
15s/###/15/
Execute:
sed -n '/###/=' infile | sed 's:.*:&s/^/& \&/:' | sed -f - infile
Output:
a
b
c
d
e
6
7
8
a
b
11
c
d
e
15
is this ok ?
kent$ echo "a
b
c
d
e"|awk '/d/{$0=$0" "NR}1'
a
b
c
d 4
e
if match pattern "d", append line number at the end of the line.
edit
oh, you want to replace the pattern not append the line number... take a look the new cmd:
kent$ echo "a
b
c
d
e"|awk '/d/{gsub(/d/,NR)}1'
a
b
c
4
e
and the line could be written like this as well: awk '1+gsub(/d/,NR)' file
one-liner to modify the FILE in place, replacing LINE with the corresponding line number:
seq 1 `wc -l FILE | awk '{print $1}'` | xargs -IX sed -i 'X s/LINE/X/' FILE
Following on from https://stackoverflow.com/a/53519367/29924
If you try this on osx the version of sed is different and you need to do:
seq 1 `wc -l FILE | awk '{print $1}'` | xargs --verbose -IX sed -i bak "X s/__line__/X/" FILE
see https://markhneedham.com/blog/2011/01/14/sed-sed-1-invalid-command-code-r-on-mac-os-x/