I am learning how to use sed, and saw that I can use the = to insert line numbers. However, this includes a newline:
file:
alpha
beta
gamma
running sed -n '=;p' file:
1
alpha
2
beta
3
gamma
Is it possible, in a single call to sed, to insert these line numbers on the same line? So:
1 alpha
2 beta
3 gamma
I know that it is possible to do this with other tools, but I am wondering about the specific functionality of sed. Is there a way to append perhaps another regular-expression substitution after the = to remove newlines?
This would be work for you.
sed '=' file | sed 'N; s/\n/ /'
Both orders:
echo -e "Jan J.\nKarel K.\nPetr P." |sed -rn "=;p;"| sed -r "N;s/\n/ /;"
1 Jan J.
2 Karel K.
3 Petr P.
echo -e "Jan J.\nKarel K.\nPetr P." |sed -rn "p;=;"| sed -r "N;s/\n/ /;"
Jan J. 1
Karel K. 2
Petr P. 3
echo -e "Jan J.\nKarel K.\nPetr P." |sed -rn "=;p;" | sed -r "N;s/\n/ /;" | sed -r 's/^([0-9]*) (.*) (.*)$/\2 \1 \3/'
Jan 1 J.
Karel 2 K.
Petr 3 P.
Related
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.
can please somebody help me with this?
I have this line
test.txt
siemplog1.nw.lan / 172.31.180.22
I tried this command sed -Ei "s/^[a-z A-Z].*([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}).*/\1/" test.txt
result should be 172.31.180.22 but I got this 2.31.180.22
thank you
The .* matches as many chars as it can (it is "greedy") and since [0-9]{1,3} can match just 1 digit, the 17 is matched by the .* and 2 is matched by [0-9]{1,3}.
You may stop the .* before any non-digit:
sed -Ei 's~.*[^0-9]([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}).*~\1~' test.txt
Or, before /:
sed -Ei 's~.*/ *([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}).*~\1~' test.txt
See online sed demo:
s='siemplog1.nw.lan / 172.31.180.22'
sed -E 's~.*/ *([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}).*~\1~' <<< "$s"
# => 172.31.180.22
If you string is always in this format, you might simplify the sed command to
sed -E 's~.*/ *([0-9.]+)~\1~p'
sed -E 's~.*/ *([0-9.]+).*~\1~p'
If you have space before ip
$echo siemplog1.nw.lan / 172.31.180.22 | sed -E "s/.* ([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}).*/\1/"
172.31.180.22
$
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/
I have a document containing many percent, plus, and pipe signs. I want to replace them with a code, for use in TeX.
% becomes \textpercent.
+ becomes \textplus.
| becomes \textbar.
This is the code I am using, but it does not work:
sed -i "s/\%/\\\textpercent /g" ./file.txt
sed -i "s/|/\\\textbar /g" ./file.txt
sed -i "s/\+/\\\textplus /g" ./file.txt
How can I replace these symbols with this code?
Test script:
#!/bin/bash
cat << 'EOF' > testfile.txt
1+2+3=6
12 is 50% of 24
The pipe character '|' looks like a vertical line.
EOF
sed -i -r 's/%/\\textpercent /g;s/[+]/\\textplus /g;s/[|]/\\textbar /g' testfile.txt
cat testfile.txt
Output:
1\textplus 2\textplus 3=6
12 is 50\textpercent of 24
The pipe character '\textbar ' looks like a vertical line.
This was already suggested in a similar way by #tripleee, and I see no reason why it should not work. As you can see, my platform uses the very same version of GNU sed as yours. The only difference to #tripleee's version is that I use the extended regex mode, so I have to either escape the pipe and the plus or put it into a character class with [].
nawk '{sub(/%/,"\\textpercent");sub(/\+/,"\\textplus");sub(/\|/,"\\textpipe"); print}' file
Tested below:
> echo "% + |" | nawk '{sub(/%/,"\\textpercent");sub(/\+/,"\\textplus");sub(/\|/,"\\textpipe"); print}'
\textpercent \textplus \textpipe
Use single quotes:
$ cat in.txt
foo % bar
foo + bar
foo | bar
$ sed -e 's/%/\\textpercent /g' -e 's/\+/\\textplus /g' -e 's/|/\\textbar /g' < in.txt
foo \textpercent bar
foo \textplus bar
foo \textbar bar
I can chain multiple sed substitutions and a awk operation to achieve this, but is there a single sed substitution that can do it?
Also is there any other tool that is more suitable for this parsing task?
You could try:
sed -r 's!rec:id=(.*?)&name=(.*?)&age=(.*?)!\1 \2 \3!' input_file
If you don't know the rec:id etc in advance but you know there's three, you could try:
sed -r 's![^=]+=(.*?)&[^=]+=(.*?)&[^=]+=(.*?)!\1 \2 \3!' input_file
If you don't know how many &name=value pairs you're after in advance but want to output all the values, you could try something like:
grep -P -o '(?<==)([^&]*)(?=&|$)' | xargs
where the -P means 'perl regex', the regex says "find the string followed by an & (or end of string) and preceded by and equals sign", the -o means to print just the matches (ie the 1, zz, and 21) each on their own line, and the | xargs moves these from their own line to one line and space separated (ie 1\nzz\n21 to 1 zz 21).
This might work for you:
echo "rec:id=1&name=zz&age=21" | sed 's/[^=]*=\([^&]*\)/\1 /g'
1 zz 21
However this leaves an extra space at the end, to solve this use:
echo "rec:id=1&name=zz&age=21"|sed 's/[^=]*=\([^&]*\)/\1 /g:;s/ $//'
1 zz 21
How about parsing the values directly into variables?
inbound="rec:id=1&name=zz&age=21"
eval $(echo $inbound | cut -c5- | tr \& "\n")
echo "Name:$name, ID:$id, Age:$age"
Or even better, though slightly more arcane:
inbound="rec:id=1&name=zz&age=21"
IFS=\& eval $(cut -c5- <<< $inbound)
echo "Name:$name, ID:$id, Age:$age"