In zabbix-agent.conf I have lines:
# Example: Server=127.0.0.1,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=127.0.0.1
I want to replace line
Server=127.0.0.1
with my
Server=zabbix.mydomain.com
But if I do
sed -i -e 's/Server=127.0.0.1/Server=zabbix.mydomain.com/g' /etc/zabbix/zabbix_agentd.conf
it found also line with commented example and replace string in it. I get:
#<----->Example: Server=zabbix.mydomain.com,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=zabbix.mydomain.com
How to replace only one line?
You need to
Match the text at the start of string
Escape the dots
Remove g flag since the match will only be found at the string start.
Also, you do not need the -e option, you can use
sed -i 's/^Server=127\.0\.0\.1/Server=zabbix.mydomain.com/' /etc/zabbix/zabbix_agentd.conf
See the online demo:
#!/bin/bash
s='# Example: Server=127.0.0.1,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=127.0.0.1'
sed 's/^Server=127\.0\.0\.1/Server=zabbix.mydomain.com/g' <<< "$s"
Output:
Example: Server=127.0.0.1,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=zabbix.mydomain.com
This might work for you (GNU sed):
sed -i '/^Server=127\.0\.0\.1/cServer=zabbix.mydomain.com' file
Change line beginning Server=127.0.0.1 to Server=zabbix.mydomain.com.
The other answers are almost fine but they would also replace a line like:
Server=127.0.0.10
A complete solution with any sed could be:
sed -i 's/^Server=127\.0\.0\.1$/Server=zabbix.mydomain.com/' /etc/zabbix/zabbix_agentd.conf
^ and $ anchor the string to the beginning and end of line, respectively. Dots need backslash escape, else they stand for any character.
I have a file, which has multiple lines.
For example:
a
ab#
ad.
a12fs
b
c
...
I want to use sed or awk delete the line, if the line include symbols or numbers. (For example, I want to delete: ab#, ad., a12fs.... lines)
or in another words, I just want to keep the line which include [a-z][A-Z] .
I know how to delete number line,
sed '/[0-9]/d' file.txt
but I do not know how to delete symbols lines.
Or there has any easy way to do that?
To keep blank lines:
grep '^[[:alpha:]]*$' file
sed '/[^[:alpha:]]/d' file
awk '/^[[:alpha:]]*$/' file
To remove blank lines:
grep '^[[:alpha:]]+$' file
sed -E -n '/^[[:alpha:]]+$/p' file
awk '/^[[:alpha:]]+$/' file
grep works well too and is even simpler: just do the reverse: keep the lines that interest you, which are way easier to define
grep -i '^[a-z]*$' file.txt
(match lines containing only letters and empty lines, and -i option makes grep case-insensitive)
to remove empty lines as well:
grep -i '^[a-z]+$' file.txt
caution when using Windows text files, as there's a carriage return at the end of the line, so nothing would match depending on grep versions (tested on windows here and it works)
but just in case:
grep -iP '^[a-z]*\r?$'
(note the P option to enable perl expressions or \r is not recognized)
You can use this sed:
sed '/^[A-Za-z0-9]\+$/!d' file
(OR)
sed '/[^A-Za-z0-9]/d' file
$ awk '!/[^[:alpha:]]/' file.txt
a
b
c
"The Unix Programming Environment" states that '$' used in regular expression in sed means end-of-the line which is fine for me, because
cat file | sed 's/$/\n/'
is interpretted as "add newline at the end of each line".
The question arises, when I try to use the command:
cat file | sed '$d'
Shouldn't this line remove each line instead of the last one? In this context, dollar sign means end of the LAST line. What am I getting wrong?
$ is treated as regex anchor when used in pattern in s command e.g.
s/$/\n
However in $d, $ is not a regex anchor, it is address notation that means the last line of the input, which is deleted using the d command.
Also note that cat is unnecessary in your last command. It can be used as:
sed '$d' file
In the second usage, there is no regular expression. The $ there is an address, meaning the last line.
Note that regex in sed must be inside the delimiters(;,:, ~, etc) other than quotes.
/regex/
ex:
sed '/foo/s/bar/bux/g' file
or
~regex~
ex:
sed 's~dd~s~' file
but not 'regex'. So $ in '$d' won't be considered as regex by sed. '$d' acts like an address which points out the last line.
I want to replace VERSION variable with $version_1 and PROG_VERSION with $version_2 within a same file. I used sed for that
sed s/VERSION/$version_1/g | sed s/PROG_VERSION/$version_2/g
The problem is sed also replaces a portion of PROG_VERSION with PROG_(value of $version_1).
How should I prevent sed from making unwanted changes.
The problem is sed also replaces a portion of PROG_VERSION with
PROG_(value of $version_1). How should I prevent sed from making
unwanted changes.
Use word boundaries in sed to prevent unwanted replacements:
sed "s/\bVERSION\b/$version_1/g"
OR:
sed "s/\<VERSION\>/$version_1/g"
For Mac OSX: For some strange reason none of above syntax works and you need to use this weird syntax:
sed "s/[[:<:]]VERSION[[:>:]]/$version_1/g"
I think the easiest would be to switch the order of the commands.
sed "s/PROG_VERSION/$version_2/g; s/VERSION/$version_1/g" file
How to insert a newline before a pattern within a line?
For example, this will insert a newline behind the regex pattern.
sed 's/regex/&\n/g'
How can I do the same but in front of the pattern?
Given this sample input file, the pattern to match on is the phone number.
some text (012)345-6789
Should become
some text
(012)345-6789
This works in bash and zsh, tested on Linux and OS X:
sed 's/regexp/\'$'\n/g'
In general, for $ followed by a string literal in single quotes bash performs C-style backslash substitution, e.g. $'\t' is translated to a literal tab. Plus, sed wants your newline literal to be escaped with a backslash, hence the \ before $. And finally, the dollar sign itself shouldn't be quoted so that it's interpreted by the shell, therefore we close the quote before the $ and then open it again.
Edit: As suggested in the comments by #mklement0, this works as well:
sed $'s/regexp/\\\n/g'
What happens here is: the entire sed command is now a C-style string, which means the backslash that sed requires to be placed before the new line literal should now be escaped with another backslash. Though more readable, in this case you won't be able to do shell string substitutions (without making it ugly again.)
Some of the other answers didn't work for my version of sed.
Switching the position of & and \n did work.
sed 's/regexp/\n&/g'
Edit: This doesn't seem to work on OS X, unless you install gnu-sed.
In sed, you can't add newlines in the output stream easily. You need to use a continuation line, which is awkward, but it works:
$ sed 's/regexp/\
&/'
Example:
$ echo foo | sed 's/.*/\
&/'
foo
See here for details. If you want something slightly less awkward you could try using perl -pe with match groups instead of sed:
$ echo foo | perl -pe 's/(.*)/\n$1/'
foo
$1 refers to the first matched group in the regular expression, where groups are in parentheses.
On my mac, the following inserts a single 'n' instead of newline:
sed 's/regexp/\n&/g'
This replaces with newline:
sed "s/regexp/\\`echo -e '\n\r'`/g"
echo one,two,three | sed 's/,/\
/g'
You can use perl one-liners much like you do with sed, with the advantage of full perl regular expression support (which is much more powerful than what you get with sed). There is also very little variation across *nix platforms - perl is generally perl. So you can stop worrying about how to make your particular system's version of sed do what you want.
In this case, you can do
perl -pe 's/(regex)/\n$1/'
-pe puts perl into a "execute and print" loop, much like sed's normal mode of operation.
' quotes everything else so the shell won't interfere
() surrounding the regex is a grouping operator. $1 on the right side of the substitution prints out whatever was matched inside these parens.
Finally, \n is a newline.
Regardless of whether you are using parentheses as a grouping operator, you have to escape any parentheses you are trying to match. So a regex to match the pattern you list above would be something like
\(\d\d\d\)\d\d\d-\d\d\d\d
\( or \) matches a literal paren, and \d matches a digit.
Better:
\(\d{3}\)\d{3}-\d{4}
I imagine you can figure out what the numbers in braces are doing.
Additionally, you can use delimiters other than / for your regex. So if you need to match / you won't need to escape it. Either of the below is equivalent to the regex at the beginning of my answer. In theory you can substitute any character for the standard /'s.
perl -pe 's#(regex)#\n$1#'
perl -pe 's{(regex)}{\n$1}'
A couple final thoughts.
using -ne instead of -pe acts similarly, but doesn't automatically print at the end. It can be handy if you want to print on your own. E.g., here's a grep-alike (m/foobar/ is a regex match):
perl -ne 'if (m/foobar/) {print}'
If you are finding dealing with newlines troublesome, and you want it to be magically handled for you, add -l. Not useful for the OP, who was working with newlines, though.
Bonus tip - if you have the pcre package installed, it comes with pcregrep, which uses full perl-compatible regexes.
In this case, I do not use sed. I use tr.
cat Somefile |tr ',' '\012'
This takes the comma and replaces it with the carriage return.
To insert a newline to output stream on Linux, I used:
sed -i "s/def/abc\\\ndef/" file1
Where file1 was:
def
Before the sed in-place replacement, and:
abc
def
After the sed in-place replacement. Please note the use of \\\n. If the patterns have a " inside it, escape using \".
Hmm, just escaped newlines seem to work in more recent versions of sed (I have GNU sed 4.2.1),
dev:~/pg/services/places> echo 'foobar' | sed -r 's/(bar)/\n\1/;'
foo
bar
echo pattern | sed -E -e $'s/^(pattern)/\\\n\\1/'
worked fine on El Captitan with () support
In my case the below method works.
sed -i 's/playstation/PS4/' input.txt
Can be written as:
sed -i 's/playstation/PS4\nplaystation/' input.txt
PS4
playstation
Consider using \\n while using it in a string literal.
sed : is stream editor
-i : Allows to edit the source file
+: Is delimiter.
I hope the above information works for you 😃.
in sed you can reference groups in your pattern with "\1", "\2", ....
so if the pattern you're looking for is "PATTERN", and you want to insert "BEFORE" in front of it, you can use, sans escaping
sed 's/(PATTERN)/BEFORE\1/g'
i.e.
sed 's/\(PATTERN\)/BEFORE\1/g'
You can also do this with awk, using -v to provide the pattern:
awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
This checks if a line contains a given pattern. If so, it appends a new line to the beginning of it.
See a basic example:
$ cat file
hello
this is some pattern and we are going ahead
bye!
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
hello
this is some
pattern and we are going ahead
bye!
Note it will affect to all patterns in a line:
$ cat file
this pattern is some pattern and we are going ahead
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' d
this
pattern is some
pattern and we are going ahead
sed -e 's/regexp/\0\n/g'
\0 is the null, so your expression is replaced with null (nothing) and then...
\n is the new line
On some flavors of Unix doesn't work, but I think it's the solution to your problem.
echo "Hello" | sed -e 's/Hello/\0\ntmow/g'
Hello
tmow
This works in MAC for me
sed -i.bak -e 's/regex/xregex/g' input.txt sed -i.bak -e 's/qregex/\'$'\nregex/g' input.txt
Dono whether its perfect one...
After reading all the answers to this question, it still took me many attempts to get the correct syntax to the following example script:
#!/bin/bash
# script: add_domain
# using fixed values instead of command line parameters $1, $2
# to show typical variable values in this example
ipaddr="127.0.0.1"
domain="example.com"
# no need to escape $ipaddr and $domain values if we use separate quotes.
sudo sed -i '$a \\n'"$ipaddr www.$domain $domain" /etc/hosts
The script appends a newline \n followed by another line of text to the end of a file using a single sed command.
In vi on Red Hat, I was able to insert carriage returns using just the \r character. I believe this internally executes 'ex' instead of 'sed', but it's similar, and vi can be another way to do bulk edits such as code patches. For example. I am surrounding a search term with an if statement that insists on carriage returns after the braces:
:.,$s/\(my_function(.*)\)/if(!skip_option){\r\t\1\r\t}/
Note that I also had it insert some tabs to make things align better.
Just to add to the list of many ways to do this, here is a simple python alternative. You could of course use re.sub() if a regex were needed.
python -c 'print(open("./myfile.txt", "r").read().replace("String to match", "String to match\n"))' > myfile_lines.txt
sed 's/regexp/\'$'\n/g'
works as justified and detailed by mojuba in his answer .
However, this did not work:
sed 's/regexp/\\\n/g'
It added a new line, but at the end of the original line, a \n was added.