Re: [Mailman-Users] authors

2010-12-09 Thread Mark Sapiro
Glenn Sieb wrote: > >On 12/9/10 5:36 PM, Mark Sapiro wrote: >> grep "From " $var_prefix/archives/private/$list.mbox/$list.mbox | \ >> awk '{print $2}' | sort -u > >Wouldn't > >grep "^From "... > >(or egrep, depending on your toolkit) > >work better? Actually, "^From " is what I intended, but I

Re: [Mailman-Users] authors

2010-12-09 Thread Glenn Sieb
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On 12/9/10 5:36 PM, Mark Sapiro wrote: > grep "From " $var_prefix/archives/private/$list.mbox/$list.mbox | \ > awk '{print $2}' | sort -u Actually.. a google netted me this... awk 'empty{if (/^From /) print $2; empty=0} /^$/{empty=1}' $list.mbox |

Re: [Mailman-Users] authors

2010-12-09 Thread Glenn Sieb
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On 12/9/10 5:36 PM, Mark Sapiro wrote: > grep "From " $var_prefix/archives/private/$list.mbox/$list.mbox | \ > awk '{print $2}' | sort -u Wouldn't grep "^From "... (or egrep, depending on your toolkit) work better? Otherwise you grab any sentence

Re: [Mailman-Users] authors

2010-12-09 Thread Mark Sapiro
Troy Knabe wrote: >I have an administrator requesting a list of everyone who has sent an >email to their list. Is there any way to quickly obtain that information? >This is a public list that lots of non-members send to. grep "From " $var_prefix/archives/private/$list.mbox/$list.mbox | \ awk

[Mailman-Users] authors

2010-12-09 Thread Troy Knabe
I have an administrator requesting a list of everyone who has sent an email to their list. Is there any way to quickly obtain that information? This is a public list that lots of non-members send to. Thanks -Troy -- Mailman-Users mailing