Need help getting multiple substrings from one line using find | grep

pnkflydgr · 06-19-2012, 12:38 PM

My find | grep is this:

Code:

find ../xml | xargs grep -s '<parameter name="ldapPassword">'

which returns a string like this:
../xml/file.xml:<parameter name="ldapPassword">Password</parameter>

The result I need is this:
file = file.xml
password = Password

This is what I was using to get the password:

Code:

find ../xml | xargs grep -s '<parameter name="ldapPassword">' | sed -e 's/.*ldapPassword">//' -e 's/<\/parameter.*//'

But I'm not sure how to get multiple strings. Well I have an idea of how to do it, but it does not seem like the most efficient. Should I be using Perl instead?

pan64 · 06-19-2012, 01:13 PM

assuming you have only one password entry in a file you can have a grep -r <key> ../xml | sed 'commands' to collect the information. and command will look like:
s/:.*ldapPassword">/\npassword = /; s!</ldapPass.*$!\n!; s/^/file = /;
first will drop everything between : and ldapPassword"> and will insert password = , second will drop the end of the line and third will insert filename =

But I have not tested, just wrote it, I have no source files to check

grail · 06-20-2012, 04:18 AM

Assuming this is in a bash script and being assigned to variable(s), the first thing is obvious, you would need to assign to an array if only wanting to do a single pass.

The following would assume knowledge of where the items are:

Code:

find ../xml | awk -F"[/:<>]+" '/<parameter name="ldapPassword">/{print $3,$5}'