[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

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

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 sentences

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 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 mistyped it.