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Print lines between a regExp & a blank line

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# 1  
Old 11-30-2011
Print lines between a regExp & a blank line


I have a file, say files_list, as below (o/p of ls -R cmd)

$ cat files_list
<blank line 1>
<blank line 2>
<blank line 3>

I want to read this file in a loop and produce an o/p like
Output file

I was able to extract all sub-dir in a file, say sub_dir_file
$ cat sub_dir_file

I thought of looping through the list of sub-dir and using below sed between
for line in cat sub_dir_file
sed -n '/remote/dir/path/$line, <blank line of this block i.e. blank line1>/p' files_list
.........more formatting..........

Am I on the right path or can you please correct the sed /from/,/to pattern/p command?

# 2  
Old 11-30-2011

please give this a try

ls -R | gawk ' /:$/ { dir=gensub(/^(.*):$/, "\\1/", "g"); next; } { print dir $NF }'

# 3  
Old 11-30-2011
You are on the right path, however, note that a slash is a metacharacter for sed. You have to escape your path slashes with backslash. Like:
sed -n '/\path\/to\/somewhere/,/^$/p' files_list

Yeah, and the /^$/ is an empty line. A pattern /^[ \t]*$/ would be even better, if you suspect there may be some whitespace chars hidden.

However, isn't this too much work? Have you looked at find(1) yet?

find /remote/dir/path/to/file -type f

# 4  
Old 11-30-2011
Similar awk solution..
ls -R | awk -F'[ :]' '$0~/^[.]*\/.*/ {r=$1;next} r!="" && $0!="" {print r "/" $0}'

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