johnirl has asked for the wisdom of the Perl Monks concerning the following question:
please notice that the last value in each hash element is not included in the resultinput: %hash = ( "view0" => "buffersize, size, value", "view1" => "buffersize, size, value"); output: view0 full outer join view1 on view0.size = view1.size and view0.buffe +rsize = view1.buffersize
again the last element value in each hash element is omittedinput: %hash = ( "view0" => "buffersize, size, value", "view1" => "buffersize, size, value", "view2" => "buffersize, size, value"); output: view0 full outer join view1 on view0.size = view1.size and view0.buffe +rsize = view1.buffersize full outer join view2 on view1.size = view2. +size and view1.buffersize = view2.buffersize
Thanks. And if anything is unclear please ask. It does take a while to wrap your head around what is needed.input: %hash = ( "view0" => "buffersize, size, value", "view1" => "buffersize, size, value", "view2" => "buffersize, size, value" "view3" => "buffersize, size, value"); output: view0 full outer join view1 on view0.size = view1.size and view0.buffe +rsize = view1.buffersize full outer join view2 on view1.size = view2. +size and view1.buffersize = view2.buffersize full outer join view3 on + view2.size = view3.size and view2.buffersize = view3.buffersize
j o h n i r l .
Sum day soon I'Il lern how 2 spelI (nad tYpe)
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How to build up complex SQL queries
by blssu (Pilgrim) on Sep 10, 2002 at 13:57 UTC | |
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Re: Using hash elements to create a string
by RollyGuy (Chaplain) on Sep 10, 2002 at 12:54 UTC | |
by johnirl (Monk) on Sep 10, 2002 at 13:17 UTC | |
by RollyGuy (Chaplain) on Sep 10, 2002 at 14:53 UTC | |
by johnirl (Monk) on Sep 11, 2002 at 08:30 UTC | |
by RollyGuy (Chaplain) on Sep 11, 2002 at 13:40 UTC |