I have few thousands of text lines like this:
go to <CITY>rome</CITY> <COUNTRY>italy</COUNTRY>
My desired output is to replace everything from the first tagged word (rome) to the last one (italy) and put tag:
go to <ADDRESS>rome italy</ADDRESS>
I can match the portion of the text line which is tagged with:
<.*>
This will greedily select all text from first < to last >. I would like then the tags removed and put <ADDRESS> and </ADDRESS> around the matched portion.
The possible tags are: <STREETNUM>, <STREET>, <CITY>, <STATE>, <ZIP> and <COUNTRY>. Any subset of these tags can appear and in any order. The tags are never nested.
I have searched SO and googled to no avail. Perhaps I can use a named capturing group and then apply search/replace regex on it but I don't know how. Any help would appreciated.
This sed line will do it:
sed 's/<CITY>\(.*\)<\/CITY>.*<COUNTRY>\(.*\)<\/COUNTRY>/<ADDRESS>\1 \2<\/ADDRESS> /g'
For example:
sed 's/<CITY>\(.*\)<\/CITY>.*<COUNTRY>\(.*\)<\/COUNTRY>/<ADDRESS>\1 \2<\/ADDRESS> /g' <<< "go to <CITY>rome</CITY> <COUNTRY>italy</COUNTRY>"
It prints:
go to <ADDRESS>rome italy</ADDRESS>
It basically captures what is inside the CITY tag and inside the COUNTRY tag and then replace them with the captured groups values enclose the ADDRESS tag
If you're using Linux, you can avoid escaping ( using the -E flag:
sed -E 's/<CITY>(.*)<\/CITY>.*<COUNTRY>(.*)<\/COUNTRY>/<ADDRESS>\1 \2<\/ADDRESS> /g'
UPDATE:
To achieve the expected result you could use several commands in the following order of operation:
Remove the go to text: sed 's/go to //g'
Remove all the tag characters: tr -d '</>'
Once all tag chars are removed, you can safely delete the words STREETNUM, STREET, CITY, STATE, ZIP and COUNTRY from the input:
sed -E 's/CITY|COUNTRY|STATE|ZIP|STREETNUM|STREET//g'
Take the output generated from the previous commands concatenation and output it inside the <ADDRESS></ADDRESS> tags:
xargs -i echo "go to <ADDRESS>{}</ADDRESS>"
The final command is the following, here $LINE should contain the line to process:
sed 's/go to //g' <<< "$LINE" | tr -d '</>' | sed -E 's/CITY|COUNTRY|STATE|ZIP|STREETNUM|STREET//g' | xargs -i echo "go to <ADDRESS>{}</ADDRESS>"
An example:
Running:
sed 's/go to //g' <<< "go to <STATE>Bolivar</STATE> <COUNTRY>Venezuela</COUNTRY> <STREETNUM>5</STREETNUM> " | tr -d '</>' | sed -E 's/CITY|COUNTRY|STATE|ZIP|STREETNUM|STREET//g' | xargs -i echo "go to <ADDRESS>{}</ADDRESS>"
Will print:
go to <ADDRESS>Bolivar Venezuela 5 </ADDRESS>
Related
I have yet another list of subdomain. I want to remove any Wildcard subdomain which include these special characters:
()!&$#*+?
Mostly, the data are prefixly random. Also, could be middle. Here's some sample of output data
(www.imgur.com
***************diet.blogspot.com
*-1.gbc.criteo.com
------------------------------------------------------------i.imgur.com
This has been quite an inconvenience while scanning through the list. As always, I'm trying sed to fix it:
sed -i "/[!()#$&?+]/d" foo.txt ###Didn't work
sed -i "/[\!\(\)\#\$\&\?\+]/d" ###Escaping char didn't work
Performing commands above still result in an unchanged list and the file still on original state. I'm thinking that; to fix this is to pipe series of sed command in order to remove it one by one:
cat foo.txt | sed -e "/!/d" -e "/#/d" -e "/\*/d" -e "/\$/d" -e "/(/d" -e "/)/d" -e "/+/d" -e "/\'/d" -e "/&/d" >> foo2.txt
cat foo.txt | sed -e "/\!/d" | sed -e "/\#/d" | sed -e "/\*/d" | sed -e "/\$/d" | sed -e "/\+/d" | sed -e "/\'/d" | sed -e "/\&/d" >> foo2.txt
If escaping all special char doesn't work, it must've been my false logic. Also tried with /g still doesn't increase my luck.
As a side note: I don't want - to be deleted as some valid subdomain can have - character:
line-apps.com
line-apps-beta.com
line-apps-rc.com
line-apps-dev.com
Any help would be cherished.
Using sed
$ sed '/[[:punct:]]/d' input_file
This should delete all lines with special characters, however, it would help if you provided sample data.
To do what you're trying to do in your answer (which adds [ and ] and more to the set of characters in your question) would be:
sed '/[][!?+,#$&*() ]/d'
or just:
grep -v '[][!?+,#$&*() ]'
Per POSIX to include ] in a bracket expression it must be the first character otherwise it indicates the end of the bracket expression.
Consider printing lines you want instead of deleting lines you do not want, though, e.g.:
grep '^[[:alnum:]_.-]$' file
to print lines that only contain letters, numbers, underscores, dashes, and/or periods.
this is my original string:
NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
I want to only add back slash to all the spaces before ':'
so, this is what I finally want:
NetworkManager/system\ connections/Wired\ 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
I need to do this in bash, so, sed, awk, grep are all ok for me.
I have tried following sed, but none of them work
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/ .*\(:.*$\)/\\ .*\1/g'
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/\( \).*\(:.*$\)/\\ \1.*\2/g'
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/ .*\(:.*$\)/\\ \1/g'
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/\( \).*\(:.*$\)/\\ \1\2/g'
thanks for answering my question.
I am still quite newbie to stackoverflow, I don't know how to control the format in comment.
so, I just edit my original question
my real story is:
when I do grep or use cscope to search keyword, for example "address1" under /etc folder.
the result would be like:
./NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
if I use vim to open file under cursor, suppose my vim cursor is now at word "NetworkManager",
then vim will understand it as
"./NetworkManager/system"
that's why I want to add "\" before space, so the search result would be more vim friendly:)
I did try to change cscope's source code, but very difficult to fully achieve this. so have to do a post replacement:(
If you only want to do the replacements if there is a : present in the string, you can check if there are at least 2 columns, setting the (output)field separator to a colon.
Data:
cat file michaelvandam#Michaels-MacBook-Pro
NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
NetworkManager/system connections/Wired 1.nmconnection 14 address1=10.1.10.71/24,10.1.10.1%
Example in awk:
awk 'BEGIN {FS=OFS=":"}{if(NF>1)gsub(" ","\\ ",$1)}1' file
Output
NetworkManager/system\ connections/Wired\ 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
NetworkManager/system connections/Wired 1.nmconnection 14 address1=10.1.10.71/24,10.1.10.1
This could be simply done in awk program, with your shown samples, please try following.
awk 'BEGIN{FS=OFS=":"} {gsub(/ /,"\\\\&",$1)} 1' Input_file
Explanation: Simple explanation would be, setting field separator and output field separator as : for this program. Then in main program using gsub(Global substitution) function of awk. Where substituting space with \ in 1st field only(as per OP's remarks it should be done before :) and printing line then.
An idea for a perl one liner in bash to use \G and \K (similar #CarySwoveland's comment).
perl -pe 's/\G[^ :]*\K /\\ /g' myfile
See this demo at tio.run or a pattern demo at regex101.
This might work for you (GNU sed):
sed -E ':a;s/^([^: ]*) /\1\n/;ta;s/\n/\\ /g' file
Replace spaces before : by newlines then replace newlines by \ 's.
Alternative using the hold space:
sed -E 's/:/\n:/;h;s/ /\\ /g;G;s/\n.*\n//' file
Split the line on the first :.
Amend the front section, remove the middle and append the unadulterated back section.
My answer is ugly and I think RavinderSingh13's answer is THE ONE, but I already took the time to write mine and it works (It's written step by step, but it's a one line command):
I got inspired by HatLess answer:
first get the text before the : with cut (I put the string in a file to make it easy to read, but this works on echo):
cut -d':' -f1 infile
Then replace spaces using sed:
cut -d':' -f1 infile | sed 's/\([a-z]\) /\1\\ /g'
Then echo the output with no new line:
echo -n "$(cut -d':' -f1 infile | sed -e 's/\([a-z]\) /\1\\ /g')"
Add the missing : and what comes after it:
echo -n "$(cut -d':' -f1 infile | sed -e 's/\([a-z]\) /\1\\ /g')" | cat - <(echo -n :) | cat - <(cut -d':' -f2 infile)
I have text like:
TEXT="I need to replace the hostname [[google.com]] with it's ip in side the text"
Is there a way to use something like below, but working?
sed -Ee "s/\[\[(.*)\]\]/`host -t A \1 | rev | cut -d " " -f1 | rev`/g" <<< $TEXT
looks like the value of \1 is not being passed to the shell command used inside sed.
Thanks
Backquote interpolation is performed by the shell, not by sed. This means that your backquotes will either be replaced by the output of a command before the sed command is run, or (if you correctly quote them) they will not be replaced at all, and sed will see the backquotes.
You appear to be trying to have sed perform a replacement, then have the shell perform backquote interpolation.
You can get the backquotes past the shell by quoting them properly:
$ echo "" | sed -e 's/^/`hostname`/'
`hostname`
However, in that case you will have to use the resulting string in a shell command line to cause backquote interpolation again.
Depending on how you feel about awk, perl, or python, I'd suggest you use one of them to do this job in a single pass. Alternatively, you could make a first pass extracting the hostnames into a command without backquotes, then execute the commands to get the IP addresses you want, then replace them in another pass.
It's got to be a two part command, one to get a variable that bash can use, the other to do a straight-up /s/ replacement with sed.
TEXT="I need to replace the hostname [[google.com]] with it's ip in side the text"
DOMAIN=$(echo $TEXT | sed -e 's/^.*\[\[//' -e 's/\]\].*$//')
echo $TEXT | sed -e 's/\[\[.*\]\]/'$(host -tA $DOMAIN | rev | cut -d " " -f1 | rev)'/'
But, more cleanly using how to split a string in shell and get the last field
TEXT="I need to replace the hostname [[google.com]] with it's ip in side the text"
DOMAIN=$(echo $TEXT | sed -e 's/^.*\[\[//' -e 's/\]\].*$//')
HOSTLOOKUP=$(host -tA $DOMAIN)
echo $TEXT | sed -e 's/\[\[.*\]\]/'${HOSTLOOKUP##* }/
The short version is that you can't mix sed and bash the way you're expecting to.
This works:
#!/bin/bash
txt="I need to replace the hostname [[google.com]] with it's ip in side the text"
host_name=$(sed -E 's/^[^[]*\[\[//; s/^(.*)\]\].*$/\1/' <<<"$txt")
ip_addr=$(host -tA "$host_name" | sed -E 's/.* ([0-9.]*)$/\1/')
echo "$txt" | sed -E 's/\[\[.*\]\]/'"$ip_addr/"
# I need to replace the hostname 172.217.4.174 with it's ip in side the text
Thank you all,
I made the below solution:
function host_to_ip () {
echo $(host -t A $1 | head -n 1 | rev | cut -d" " -f1 | rev)
}
function resolve_hosts () {
local host_placeholders=$(grep -o -e "##.*##" $1)
for HOST in ${host_placeholders[#]}
do
sed -i -e "s/$HOST/$(host_to_ip $(sed -Ee 's/##(.*)##/\1/g' <<< $HOST))/g" $1
done
}
Where resolve_hosts gets a text file as an argument
I just started using sed from doing regex. I wanted to extract XXXXXX from *****/XXXXXX> so I was following
sed -n "/^/*/(\S*\).>$/p"
If I do so I get following error
sed: 1: "/^//(\S).>$/p": invalid command code *
I am not sure what am I missing here.
Try:
$ echo '*****/XXXXXX>' | sed 's|.*/||; s|>.*||'
XXXXXX
The substitute command s|.*/|| removes everything up to the last / in the string. The substitute command s|>.*|| removes everything from the first > in the string that remains to the end of the line.
Or:
$ echo '*****/XXXXXX>' | sed -E 's|.*/(.*)>|\1|'
XXXXXX
The substitute command s|.*/(.*)>|\1| captures whatever is between the last / and the last > and saves it in group 1. That is then replaced with group 1, \1.
In my opinion awk performs better this task. Using -F you can use multiple delimiters such as "/" and ">":
echo "*****/XXXXXX>" | awk -F'/|>' '{print $1}'
Of course you could use sed, but it's more complicated to understand. First I'm removing the first part (delimited by "/") and after the second one (delimited by ">"):
echo "*****/XXXXXX>" | sed -e s/.*[/]// -e s/\>//
Both will bring the expected result: XXXXXX.
with grep if you have pcre option
$ echo '*****/XXXXXX>' | grep -oP '/\K[^>]+'
XXXXXX
/\K positive lookbehind / - not part of output
[^>]+ characters other than >
echo '*****/XXXXXX>' |sed 's/^.*\/\|>$//g'
XXXXXX
Start from start of the line, then proceed till lask / ALSO find > followed by EOL , if any of these found then replace it with blank.
I got my research result after using sed :
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | cut -f 1 - | grep "pattern"
But it only shows the part that I cut. How can I print all lines after a match ?
I'm using zcat so I cannot use awk.
Thanks.
Edited :
This is my log file :
[01/09/2015 00:00:47] INFO=54646486432154646 from=steve idfrom=55516654455457 to=jone idto=5552045646464 guid=100021623456461451463 n
um=6 text=hi my number is 0 811 22 1/12 status=new survstatus=new
My aim is to find all users that spam my site with their telephone numbers (using grep "pattern") then print all the lines to get all the information about each spam. The problem is there may be matches in INFO or id, so I use sed to get the text first.
Printing all lines after a match in sed:
$ sed -ne '/pattern/,$ p'
# alternatively, if you don't want to print the match:
$ sed -e '1,/pattern/ d'
Filtering lines when pattern matches between "text=" and "status=" can be done with a simple grep, no need for sed and cut:
$ grep 'text=.*pattern.* status='
You can use awk
awk '/pattern/,EOF'
n.b. don't be fooled: EOF is just an uninitialized variable, and by default 0 (false). So that condition cannot be satisfied until the end of file.
Perhaps this could be combined with all the previous answers using awk as well.
Maybe this is what you actually want? Find lines matching "pattern" and extract the field after text= up through just before status=?
zcat file* | sed -e '/pattern/s/.*text=\(.*\)status=[^/]*/\1/'
You are not revealing what pattern actually is -- if it's a variable, you cannot use single quotes around it.
Notice that \(.*\)status=[^/]* would match up through survstatus=new in your example. That is probably not what you want? There doesn't seem to be a status= followed by a slash anywhere -- you really should explain in more detail what you are actually trying to accomplish.
Your question title says "all line after a match" so perhaps you want everything after text=? Then that's simply
sed 's/.*text=//'
i.e. replace up through text= with nothing, and keep the rest. (I trust you can figure out how to change the surrounding script into zcat file* | sed '/pattern/s/.*text=//' ... oops, maybe my trust failed.)
The seldom used branch command will do this for you. Until you match, use n for next then branch to beginning. After match, use n to skip the matching line, then a loop copying the remaining lines.
cat file | sed -n -e ':start; /pattern/b match;n; b start; :match n; :copy; p; n ; b copy'
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | ***cut -f 1 - | grep "pattern"***
instead change the last 2 segments of your pipeline so that:
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | **awk '$1 ~ "pattern" {print $0}'**