I'm trying to insert this text...
"error.emailNotActivated":"This email address has not been activated yet."
... at line number 5 using sed.
Here is my command so far
translated="This email address has not been activated yet.";
sed -i '' -e '5i\'$'\n''"error.emailNotActivated":'"\"$translated\"" local.strings;
I unfortunately keep getting the error message "invalid command code T".
It seems that sed is interpreting the colon as part of a command.
Any suggestions how i can avoid this?
EDIT:
Seems like an update error (working with old file d'oh...)
the above expression works fine as do the other suggestions.
Why are you fighting with sed for this? It's trivial in awk:
awk -v line='"error.emailNotActivated":"'"$translated"'"' '
NR==5{print line} {print}
' file
or:
awk -v line="\"error.emailNotActivated\":\"${translated}\"" '
NR==5{print line} {print}
' file
Are you looking for something like this?
$ seq 1 5 > file
$ cat file
1
2
3
4
5
$ translated="\"error.emailNotActivated\":\"This email address has not been activated yet.\""
$ echo $translated
"error.emailNotActivated":"This email address has not been activated yet."
$ sed -i "5i $translated" file
$ cat file
1
2
3
4
"error.emailNotActivated":"This email address has not been activated yet."
5
Related
On a Unix system I am trying to add a new line in a file using sed or perl but it seems I am missing something.
Supposing my file has multiple lines of texts, always ending like this {TNG:}}${1:F01.
I am trying to find a to way to add a new line after the }$, in this way {1 should always start on a new line.
I tried it by escaping $ sign using this:
perl -e '$/ = "\${"; while (<>) { s/\$}\{$/}\n{/; print; }' but it does not work.
Any ideas will be appreciated.
give this a try:
sed 's/{TNG:}}\$/&\n/' file > newfile
The sed will by default use BRE, that is, the {}s are literal characters. But we must escape the $.
kent$ cat f
{TNG:}}${1:F01.
kent$ sed 's/{TNG:}}\$/&\n/' f
{TNG:}}$
{1:F01.
With perl:
$ cat input.txt
line 1 {TNG:}}${1:F01
line 2 {TNG:}}${1:F01
$ perl -pe 's/TNG:\}\}\$\K/\n/' input.txt
line 1 {TNG:}}$
{1:F01
line 2 {TNG:}}$
{1:F01
(Read up on the -p and -n options in perlrun and use them instead of trying to do what they do in a one-liner yourself)
I have a list of domain names in a file. I have to get 'A Record' for each domain name.
Input:
:~$ cat domainlist.txt
google.com
yahoo.com
facebook.com
Required Output:
google.com
216.58.220.46
yahoo.com
98.139.183.24
206.190.36.45
98.138.253.109
facebook.com
66.220.146.36
So, I have this sed command which helps me to get A Records.
sed 's/^/dig +short A/e' domainlist.txt
But it just outputs the IP. The domain names are not printed with it.
Output:
:~$ sed 's/^/dig +short a/e' domainlist.txt
98.124.199.55
188.125.73.108
74.6.50.150
77.238.184.150
98.137.236.150
199.59.243.120
I could get the required output using bash. But I want to know is there any way to get this ouput using sed as it minimizes the code lines.
Here is what the e flag to the s command does (emphasis mine):
This command allows one to pipe input from a shell command into pattern space. If a substitution was made, the command that is found in pattern space is executed and pattern space is replaced with its output. A trailing newline is suppressed; results are undefined if the command to be executed contains a nul character. This is a GNU sed extension.
So you remove what you'd like to print, but the fix is simple:
$ sed 'p;s/^/dig +short A/e' domainlist.txt
google.com
98.124.199.55
yahoo.com
74.6.50.150
98.137.236.150
77.238.184.150
188.125.73.108
facebook.com
199.59.243.120
Just print the pattern space before replacing it.
I don't have dig on my system but this is the job xargs was created to do:
xargs -n 1 -I % sh -c 'echo %; dig +short A %;' < domainlist.txt
with awk
awk '{cmd="dig "$0" +short";print $0; while((cmd|getline var) > 0){ print var};close(cmd) }' domainlist.txt
I am new to bash and having a tough time figuring this out.
Using sed, could anyone help me in finding only even numbers in a given file?
I figured out how to find all numbers starting from [0,2,4,6,8] using this:
sed -n 's/^[0-9]*[02468] /&/w even' <file
But this doesn't guarantee that the number is even for sure.
I am having trouble in finding if the matched number ends with either [0,2,4,6,8] for it to be even for sure.
So can any one help me out with this?
Your regex looks a bit weird and I am not sure what you want to do, but this should help:
sed -r -n 's/^[0-9]*?[02468] /even/g'
-r to enable extended regex, *? to make it non-greedy, and /g to perform replacement globally for all lines in file.
Your command should work fine assuming that there is a space after all even numbers and that they are all at the beginning of the lines:
$ echo 'foo
1231
2220
1254 ' | sed -n '/[0-9]*[02468] /p'
2220
1254
Also note that, as you don't actually do a substitution, you don't need the s command. Use an address (pattern) specifier and w command (like I did above with the p command).
To make sure that the even digit is the last, but is not necessarily followed by a space, you can do something like
$ echo 'foo
1231
2220
1254 ' | sed -n '/[0-9]*[02468]\($\|[^0-9]\)/p'
2220
1254
Actually, your case looks more like a use case for grep, not sed, because you do filtering rather than editing. Everything becomes easier with GNU grep, as you can do
$ echo 'foo
1231
2220
1254 ' | grep -P '\d*[02468](?!\d)'
2220
1254
Just append > even to the command to make it write to the file even.
$ cat file
1
2
3
498
57
12345678
$ awk '$0%2' file
1
3
57
$ awk '!($0%2)' file
2
498
12345678
Why don't you find the numbers ending with [02468] ?
I have the following sed command. I need to execute the below command in single line
cat File | sed -n '
/NetworkName/ {
N
/\n.*ims3/ p
}' | sed -n 1p | awk -F"=" '{print $2}'
I need to execute the above command in single line. can anyone please help.
Assume that the contents of the File is
System.DomainName=shayam
System.Addresses=Fr6
System.Trusted=Yes
System.Infrastructure=No
System.NetworkName=AS
System.DomainName=ims5.com
System.DomainName=Ram
System.Addresses=Fr9
System.Trusted=Yes
System.Infrastructure=No
System.NetworkName=Peer
System.DomainName=ims7.com
System.DomainName=mani
System.Addresses=Hello
System.Trusted=Yes
System.Infrastructure=No
System.NetworkName=Peer
System.DomainName=ims3.com
And after executing the command you will get only peer as the output. Can anyone please help me out?
You can use a single nawk command. And you can lost the useless cat
nawk -F"=" '/NetworkName/{n=$2;getline;if($2~/ims3/){print n} }' file
You can use sed as well as proposed by others, but i prefer less regex and less clutter.
The above save the value of the network name to "n". Then, get the next line and check the 2nd field against "ims3". If matched, then print the value of "n".
Put that code in a separate .sh file, and run it as your single-line command.
cat File | sed -n '/NetworkName/ { N; /\n.*ims3/ p }' | sed -n 1p | awk -F"=" '{print $2}'
Assuming that you want the network name for the domain ims3, this command line works without sed:
grep -B 1 ims3 File | head -n 1 | awk -F"=" '{print $2}'
So, you want the network name where the domain name on the following line includes 'ims3', and not the one where the following line includes 'ims7' (even though the network names in the example are the same).
sed -n '/NetworkName/{N;/ims3/{s/.*NetworkName=\(.*\)\n.*/\1/p;};}' File
This avoids abuse of felines, too (not to mention reducing the number of commands executed).
Tested on MacOS X 10.6.4, but there's no reason to think it won't work elsewhere too.
However, empirical evidence shows that Solaris sed is different from MacOS sed. It can all be done in one sed command, but it needs three lines:
sed -n '/NetworkName/{N
/ims3/{s/.*NetworkName=\(.*\)\n.*/\1/p;}
}' File
Tested on Solaris 10.
You just need to put -e pretty much everywhere you'd break the command at a newline or have a semicolon. You don't need the extra call to sed or awk or cat.
sed -n -e '/NetworkName/ {' -e 'N' -e '/\n.*ims3/ s/[^\n]*=\(.*\).*/\1/P' -e '}' File
There are often times I will grep -n whatever file to find what I am looking for. Say the output is:
1234: whatev 1
5555: whatev 2
6643: whatev 3
If I want to then just extract the lines between 1234 and 5555, is there a tool to do that? For static files I have a script that does wc -l of the file and then does the math to split it out with tail & head but that doesn't work out so well with log files that are constantly being written to.
Try using sed as mentioned on
http://linuxcommando.blogspot.com/2008/03/using-sed-to-extract-lines-in-text-file.html. For example use
sed '2,4!d' somefile.txt
to print from the second line to the fourth line of somefile.txt. (And don't forget to check http://www.grymoire.com/Unix/Sed.html, sed is a wonderful tool.)
The following command will do what you asked for "extract the lines between 1234 and 5555" in someFile.
sed -n '1234,5555p' someFile
If I understand correctly, you want to find a pattern between two line numbers. The awk one-liner could be
awk '/whatev/ && NR >= 1234 && NR <= 5555' file
You don't need to run grep followed by sed.
Perl one-liner:
perl -ne 'if (/whatev/ && $. >= 1234 && $. <= 5555) {print}' file
Line numbers are OK if you can guarantee the position of what you want. Over the years, my favorite flavor of this has been something like this:
sed "/First Line of Text/,/Last Line of Text/d" filename
which deletes all lines from the first matched line to the last match, including those lines.
Use sed -n with "p" instead of "d" to print those lines instead. Way more useful for me, as I usually don't know where those lines are.
Put this in a file and make it executable:
#!/usr/bin/env bash
start=`grep -n $1 < $3 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[0]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
echo "couldn't find start pattern!" 1>&2
exit 1
fi
stop=`tail -n +$start < $3 | grep -n $2 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[1]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
echo "couldn't find end pattern!" 1>&2
exit 1
fi
stop=$(( $stop + $start - 1))
sed "$start,$stop!d" < $3
Execute the file with arguments (NOTE that the script does not handle spaces in arguments!):
Starting grep pattern
Stopping grep pattern
File path
To use with your example, use arguments: 1234 5555 myfile.txt
Includes lines with starting and stopping pattern.
If I want to then just extract the lines between 1234 and 5555, is
there a tool to do that?
There is also ugrep, a GNU/BSD grep compatible tool but one that offers a -K option (or --range) with a range of line numbers to do just that:
ugrep -K1234,5555 -n '' somefile.log
You can use the usual GNU/BSD grep options and regex patterns (but it also offers a lot more such as -K.)
If you want lines instead of line ranges, you can do it with perl: eg. if you want to get line 1, 3 and 5 from a file, say /etc/passwd:
perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd