sed replace range of lines with a single word

Question

I tried my best to get this and had little luck, I have following file and I want to replace first “<Connector port="8080"” xml block with ‘xxxx’

Before :

<!-- A "Connector" represents an endpoint by which requests are received
     and responses are returned. Documentation at :
     Java HTTP Connector: /docs/config/http.html (blocking & non-blocking)
     Java AJP  Connector: /docs/config/ajp.html
     APR (HTTP/AJP) Connector: /docs/apr.html
     Define a non-SSL HTTP/1.1 Connector on port 8080
-->
<Connector port="8080" protocol="HTTP/1.1" 
           connectionTimeout="20000" 
           redirectPort="8443" />
<!-- A "Connector" using the shared thread pool-->
<!--
<Connector executor="tomcatThreadPool"
           port="8080" protocol="HTTP/1.1" 
           connectionTimeout="20000" 
           redirectPort="8443" />
-->

After :

<!-- A "Connector" represents an endpoint by which requests are received
     and responses are returned. Documentation at :
     Java HTTP Connector: /docs/config/http.html (blocking & non-blocking)
     Java AJP  Connector: /docs/config/ajp.html
     APR (HTTP/AJP) Connector: /docs/apr.html
     Define a non-SSL HTTP/1.1 Connector on port 8080
-->
xxxx
<!-- A "Connector" using the shared thread pool-->
<!--
<Connector executor="tomcatThreadPool"
           port="8080" protocol="HTTP/1.1" 
           connectionTimeout="20000" 
           redirectPort="8443" />
-->

I got the matching string to print using following sed command :

sed -n '/<Connector port="8080"/,/>/p' filename

However I’m unable to develop this to get above output.

Be aware that replacing an element in XML with regular expressions (i.e. `sed`) is not possible for all general XML documents. With `sed`, you can only solve this problem in a subset of XML documents. For general XML processing, use a proper XML processing technology such as XSLT. — ndim, Nov 14 '13 at 18:30
@ndim: True. However, in this case it is not general xml, it is a very specific case, which does in deed work. — carlpett, Nov 14 '13 at 19:43

pobrelkey · Accepted Answer · 2013-11-14T19:58:57.053

2

This sed should do the trick on your sample input (but gets overzealous if you have more than one <Connector port="8080" section):

sed '/<Connector port="8080"/,/>/{ s/<Connector.*/xxxx/; t; d  }'

But handling this robustly calls for an XML parser. Example:

#!/usr/bin/env ruby
require 'rexml/document'
include REXML
d = Document.new(File.read(ARGV[0]))
e = d.get_elements('//Connector[@port="8080"]').first
if !e.nil?
    e.parent.insert_after(e, Text.new('xxxx'))
    e.parent.delete(e)
end
File.open(ARGV[0],'w'){|f| f.print(d) }

edited Nov 14 '13 at 19:58

answered Nov 14 '13 at 19:43

pobrelkey

5,853
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Be aware the difference between _parsing_ and _searching_ xml. I'm all for not trying to parse xml using regex, but in this case it is not actually parsing that is asked for. – carlpett Nov 14 '13 at 19:44

score 0 · Answer 2 · answered Nov 14 '13 at 21:44

0

sed  '/<Connector port="8080"/,/>/ {
   N
   />/ s/.*/xxxx/
  }' YourFile

load the section of connector before changing it to xxxx, this allow you to eventually test something on it (but not better than pobrelkey if not)

answered Nov 14 '13 at 21:44

NeronLeVelu

9,908
1
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43

score 0 · Answer 3 · answered Nov 15 '13 at 14:12

Thanks all for the answers , they worked well .Appriciate it: Also i got following from one of my firend which also a answer to this using awk

awk '/<Connector port/ {
print "abc"
getline
do {
 getline
} while (index($0, ">") == 0)
getline
} 
// { print $0}' < 1.txt

sed replace range of lines with a single word

3 Answers3