I need to copy line 3 and later from file a to file b (starting from line 3)
File a.py
import a
print('file a')
Original File b.py
import b
print('file b')
After copy
import b
print('file a')
I tried to use sed -n 3r<(sed -n '3,$p' a.py) b.py, but it inserts the lines instead of replacing them.
Using sed
$ sed "3c $(sed -n '3p' a.py)" b.py
import b
print('file a')
Use command substitution to allow another sed command to be run as an arguments and c to replace the contents of line 3.
This might work for you (GNU sed):
sed -n 3p filea | sed -i -e '3r /dev/stdin' -e '3d' fileb
Output line 3 of filea to /dev/stdout.
Pipe that output to a second sed invocation that reads that line in on line 3 and then deletes the original line.
N.B. The -i amends fileb in situ.
I would do this:
sed -i '' '3,$d' b.py ; sed '1,2d' a.py >> b.py
This translates as "amputate lines 3 and below from b.py, then echo lines 3 and below from a.py and append them to b.py".
Related
I am trying to insert the contents of a file1.txt into file2.txt using sed. The content of file1.txt is just a single line, which is a path.
I want it to be added as a prefix to each line in file2.txt as well as add another / character.
$ cat file1.txt
/psot/rot8888/orce/db/tier/data/tine
$ cat file2.txt
o1_mf_users_abchwfg_.dbf
o1_mf_toptbs2_abchrq0_.dbf
o1_mf_toptbs1_abchrl2_.dbf
o1_mf_toptbs1_abchtlf_.dbf
Desired output should be like:
/psot/rot8888/orce/db/tier/data/tine/o1_mf_users_abchwfg_.dbf
/psot/rot8888/orce/db/tier/data/tine/o1_mf_toptbs2_abchrq0_.dbf
/psot/rot8888/orce/db/tier/data/tine/o1_mf_toptbs1_abchrl2_.dbf
/psot/rot8888/orce/db/tier/data/tine/o1_mf_toptbs1_abchtlf_.dbf
Tried command:
$ sed '/o1/ r file1.txt' file2.txt >> test.txt
$ cat test.txt
o1_mf_users_abchwfg_.dbf
/psot/rot8888/orce/db/tier/data/tine
o1_mf_toptbs2_abchrq0_.dbf
/psot/rot8888/orce/db/tier/data/tine
o1_mf_toptbs1_abchrl2_.dbf
/psot/rot8888/orce/db/tier/data/tine
o1_mf_toptbs1_abchtlf_.dbf
/psot/rot8888/orce/db/tier/data/tine
You can use pr for this without having to worry about sed metacharacters, delimiters, etc.
$ cat ip.txt
abcd.xyz
123.txt
foo_baz.txt
$ cat f1
/a/b/c/d/
$ pr -mts"$(< f1)" /dev/null ip.txt
/a/b/c/d/abcd.xyz
/a/b/c/d/123.txt
/a/b/c/d/foo_baz.txt
Where -m allows pasting files parallely and -s is the separator between the files to be merged. Here, /dev/null is used as a dummy for one of the files as only the separator has to be prefixed.
If you need to add some more characters after the contents of file containing the prefix:
$ cat ip.txt
abcd.xyz
123.txt
foo_baz.txt
$ cat f1
/a/b/c/d
$ pr -mts"$(< f1)"'/' /dev/null ip.txt
/a/b/c/d/abcd.xyz
/a/b/c/d/123.txt
/a/b/c/d/foo_baz.txt
This will work using any awk in any shell on every UNIX box:
$ awk 'NR==FNR{p=$0; next} {print p "/" $0}' file1 file2
/psot/rot8888/orce/db/tier/data/tine/o1_mf_users_abchwfg_.dbf
/psot/rot8888/orce/db/tier/data/tine/o1_mf_toptbs2_abchrq0_.dbf
/psot/rot8888/orce/db/tier/data/tine/o1_mf_toptbs1_abchrl2_.dbf
/psot/rot8888/orce/db/tier/data/tine/o1_mf_toptbs1_abchtlf_.dbf
This might work for you (GNU sed):
sed '1h;1d;G;s/\(.*\)\n\(.*\)/\2\1/' file1 file2
Copy file1 into the hold space and append it to each line in file2. Using regexp and back references, manipulate the two lines into one in the correct order.
Alternative:
sed 'x;s/.*/cat file1/e;G;s/\n//' file2
Insert file1 into the hold space, append the current line of file2, and remove the newline connecting them.
A third way:
sed 'r file1' file2 | sed -E 'N;s/(.*)\n(.*)/\2\/\1/'
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 have a sed command which will append a string on the end of a line. When I re-run the same command again the same content is getting append at the end of the line again and again.
I am looking for a command which will check if the content is already there or not then proceed.
Here is my sed command:
shell: sed -i '/only_from/s/$/ xx.xx.xx.xx\/24/' file.txt
this line works for your needs:
sed -i '/only_from/{/ xx\.xx\.xx\.xx\/24$/!s#$# xx.xx.xx.xx/24#}' file
E.g:
kent$ cat f
only_from foo bar
kent$ sed -i '/only_from/{/xx\.xx\.xx\.xx\/24$/!s#$# xx.xx.xx.xx/24#}' f
kent$ cat f
only_from foo bar xx.xx.xx.xx/24
kent$ sed -i '/only_from/{/xx\.xx\.xx\.xx\/24$/!s#$# xx.xx.xx.xx/24#}' f
kent$ cat f
only_from foo bar xx.xx.xx.xx/24
You can try this sed:
sed '/only_from/{ / xx\.xx\.xx\.xx\/24/ !s/$/ xx\.xx\.xx\.xx\/24/}' file
This might be a bit naive, but why don't you write something as simple as
sed -i '/only_from$/s/$/ xx.xx.xx.xx\/24/' file.txt
I have 'file1' with (say) 100 lines. I want to use sed or awk to print lines 23, 71 and 84 (for example) to 'file2'. Those 3 line numbers are in a separate file, 'list', with each number on a separate line.
When I use either of these commands, only line 84 gets printed:
for i in $(cat list); do sed -n "${i}p" file1 > file2; done
for i in $(cat list); do awk 'NR==x {print}' x=$i file1 > file2; done
Can a for loop be used in this way to supply line addresses to sed or awk?
This might work for you (GNU sed):
sed 's/.*/&p/' list | sed -nf - file1 >file2
Use list to build a sed script.
You need to do > after the loop in order to capture everything. Since you are using it inside the loop, the file gets overwritten. Inside the loop you need to do >>.
Good practice is to or use > outside the loop so the file is not open for writing during every loop iteration.
However, you can do everything in awk without for loop.
awk 'NR==FNR{a[$1]++;next}FNR in a' list file1 > file2
You have to >>(append to the file) . But you are overwriting the file. That is why, You are always getting 84 line only in the file2.
Try use,
for i in $(cat list); do sed -n "${i}p" file1 >> file2; done
With sed:
sed -n $(sed -e 's/^/-e /' -e 's/$/p/' list) input
given the example input, the inner command create a string like this: `
-e 23p
-e 71p
-e 84p
so the outer sed then prints out given lines
You can avoid running sed/awk in a for/while loop altgether:
# store all lines numbers in a variable using pipe
lines=$(echo $(<list) | sed 's/ /|/g')
# print lines of specified line numbers and store output
awk -v lineS="^($lines)$" 'NR ~ lineS' file1 > out
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/