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Shell Programming and Scripting

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# 1  
Old 11-02-2011
Filtering a list
Hi all
When I run the system command "
Code:
rpm -qa |grep xmpp

" it lists many files like
Code:
      xmpp-3.2.10.20111024-3
      xmpp-3.2.10.201110_asd
      xmpp-3.2.10.201

and I want to uninstall all the rpms by using rpm -e. How can I do this???
NOTE:
The number of rpms will vary dynamically.



Thanks in advance

Ananth

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# 2  
Old 11-02-2011
Since you have rpm I assume you're using Linux, which means it's highly likely that you also have xargs, which you could use like this:
Code:
rpm -qa | grep xmpp | xargs rpm -e

But this could introduce a problem if the database is still locked for reading, making the removal fail. A failsafe way would be writing the contents to a temporary file & using that later, like this:
Code:
rpm -qa | grep xmpp > to_remove
rpm -e $( cat to_remove )

# 3  
Old 11-08-2011
Hi for that
code
rpm -e $(cat t0_remove) will not work since we cant uninstall all in a single command.
How can I uninstall all in the to_remove list???


Thanks in advance
Ananth
# 4  
Old 11-08-2011
If one experiments a bit with the commands I've mentioned, you'd get this:
Code:
cat to_remove | xargs rpm -e

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