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Extract URL from RSS Feed in AWK

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# 1  
Extract URL from RSS Feed in AWK

I have following data file;
<outline title="Matt Cutts" type="rss" version="RSS" xmlUrl="" htmlUrl=""/>
<outline title="Stone" text="Stone" type="rss" version="RSS" xmlUrl="" htmlUrl=""/>
<outline title="Stone" text="Stone" type="rss" version="RSS" ymlUrl="" htmlUrl=""/>
<outline title="Adam Leventhal's Weblog" text="Adam Leventhal's Weblog" type="rss" version="RSS" xmlUrl="" htmlUrl=""/>

I want to just extract the url in xmlUrl attribute and save it another file. I want to do it in awk.

Thanks for your time.

# 2  
exec 6<"file"
while read -r LINE<&6
  case "$LINE" in
      echo ${LINE%%\" *};;
exec 6<&-

# 3  
awk 'BEGIN{RS=FS}/^xmlUrl/{print $2}' FS='"' infile


This User Gave Thanks to Scrutinizer For This Post:
# 4  
Hi Scrutinizer,
Thanks a lot. It works. Please do a favor and explain your code in words please.

# 5  
Hi fahdmirza, this awk script changes the record separator to the value of the field separator so that every record becomes one field. Then it splits the new records in new fields separated by double quotes. The required values are then in the second new field of the new records that start with xmlUrl.

Last edited by Scrutinizer; 08-28-2010 at 06:33 AM..
This User Gave Thanks to Scrutinizer For This Post:
# 6  
RS=FS, good idea

# 7  
Hi Scrutinizer, thanks for the reply. Pardon my ignorance, but I have little confusion.

For example take the following line from the data:

<outline title="Matt Cutts" type="rss" version="RSS" xmlUrl="" htmlUrl=""/>

Now first your code makes the above full line (or record) as one field by doing RS=FS.

Then it matches the start of xmlUrl in above line, and now the field separater is ".

My question is how $2 contains the required url. Please explain.


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