H5P66AA #ABA: adding another monitor to my HP Clubhouse model 500-023W
I have a monitor already connected, but I would add another and making them work together and be able to
to print from a monitor. Not sure if this can be done with my HP. Thank you for your help. In addition, what type of cables
will it take to connect?
Hello
It supports 2 monitors at the same time using DVI:
* Built-in video is not available if you have installed a graphics card or an Intel processor ending with 'P'.
Integrated to enable graphics to a processor that supports graphics.
Integrated graphics: Intel Graphics Media Accelerator HD (DX10.1)
Integrated graphics card or PCI Express x 16 slot are usable at once; they can not be used simultaneously.
Using two DVI ports at the same time is supported.
Tags: HP Desktops
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http://i49.Tinypic.com/2a98fht.jpg
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http://i45.Tinypic.com/2v3sodk.jpg
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'JACK' middle_name, 'JR' SUFFIX FROM DUAL UNION ALL */ SELECT 4498 WE_PID, 77127945 We_addr_id, 'PIGGEE' first_name, 'AARON' last_name, NULL middle_name, NULL SUFFIX FROM DUAL UNION ALL SELECT 4499 WE_PID, 77127945 We_addr_id, 'PIGGEE' first_name, 'AARON' last_name, NULL middle_name, 'JR' SUFFIX FROM DUAL UNION ALL SELECT 4450 WE_PID, 77127945 We_addr_id, 'PIGGEE' first_name, 'AARON' last_name, NULL middle_name, 'JR' SUFFIX FROM DUAL UNION ALL SELECT 4451 WE_PID, 77127945 We_addr_id, 'PIGGEE' first_name, 'AARON' last_name, NULL middle_name, 'JR' SUFFIX FROM DUAL UNION ALL SELECT 250 WE_PID, 77127950 We_addr_id, 'RODNEY' first_name, 'MONTGOMERY' last_name, NULL middle_name, NULL SUFFIX FROM DUAL UNION ALL SELECT 251 WE_PID, 77127950 We_addr_id, 'RODNEY' first_name, 'MONTGOMERY' last_name, NULL middle_name, NULL SUFFIX FROM DUAL UNION ALL SELECT 252 WE_PID, 77127950 We_addr_id, 'RODNEY' first_name, 'MONTGOMERY' last_name, NULL middle_name, 'SR' SUFFIX FROM DUAL UNION ALL SELECT 253 WE_PID, 77127950 We_addr_id, 'RODNEY' first_name, 'MONTGOMERY' last_name, NULL middle_name, 'III' SUFFIX FROM DUAL), got_min_we_pid AS (SELECT b.we_pid, b.we_addr_id, b.last_name, b.first_name, b.middle_name, b.suffix, MIN(b.we_pid) OVER ( PARTITION BY b.first_name, b.last_name, b.middle_name, b.suffix ) AS min_we_pid FROM merge_names b), possible_matches AS ( SELECT CONNECT_BY_ROOT min_we_pid AS min_we_pid, min_we_pid AS leaf_we_pid FROM got_min_we_pid WHERE CONNECT_BY_ISLEAF = 1 START WITH we_pid = min_we_pid CONNECT BY first_name = PRIOR first_name AND we_addr_id = PRIOR we_addr_id AND last_name = PRIOR last_name AND SUBSTR ('x' || middle_name, 1, LENGTH ('x' || PRIOR middle_name)) = 'x' || PRIOR middle_name AND 'x' || middle_name > 'x' || PRIOR middle_name AND 'x' || suffix = 'x' || PRIOR suffix), got_match_cnt AS (SELECT DISTINCT min_we_pid, leaf_we_pid, COUNT (DISTINCT leaf_we_pid) OVER (PARTITION BY min_we_pid) AS match_cnt FROM possible_matches) SELECT DISTINCT we_pid, we_addr_id, first_name, last_name, middle_name, suffix -- ***** CHANGED ***** , MIN (we_pid) OVER (PARTITION BY NVL (p.leaf_we_pid, c.min_we_pid)) AS merge_wepid -- ***** CHANGED ***** FROM GOT_MIN_we_pid c LEFT OUTER JOIN got_match_cnt p ON p.min_we_pid = c.min_we_pid AND p.match_cnt = 1 ORDER BY c.we_pid, c.last_name, c.first_name, c.middle_name;
Published by: new learning on July 25, 2012 05:50o/p required for the person PIGGEE AARON JR and RODNEY MONTGOMERY ALl other Person Mergings Are working Great. WE_PID WE_ADDR_ID FIRST_NAME LAST_NAME MIDDLE_NAME SUFFIX MERGE_WEPID 4450 77127945 PIGGEE AARON JR 4498 4451 77127945 PIGGEE AARON JR 4498 4498 77127945 PIGGEE AARON 4498 4499 77127945 PIGGEE AARON JR 4498 250 77127950 RODNEY MONTGOMERY 250 251 77127950 RODNEY MONTGOMERY 250 252 77127950 RODNEY MONTGOMERY SR 252 253 77127950 RODNEY MONTGOMERY III 253
Published by: new learning on July 25, 2012 06:49
Published by: new learning on July 25, 2012 06:51Hello
This example gets the results you requested on the sample data you posted:
WITH got_min_we_pid AS ( SELECT merge_names.* , MIN (we_pid) OVER ( PARTITION BY first_name , last_name , middle_name , suffix ) AS min_we_pid FROM merge_names ) , possible_matches AS ( SELECT CONNECT_BY_ROOT min_we_pid AS min_we_pid , min_we_pid AS leaf_we_pid FROM got_min_we_pid WHERE CONNECT_BY_ISLEAF = 1 START WITH we_pid = min_we_pid CONNECT BY first_name = PRIOR first_name AND last_name = PRIOR last_name AND SUBSTR ( 'x' || middle_name , 1 , LENGTH ('x' || PRIOR middle_name) ) = 'x' || PRIOR middle_name AND 'x' || middle_name >= 'x' || PRIOR middle_name -- ***** CHANGED ***** AND ( 'x' || suffix = 'x' || PRIOR suffix -- ***** CHANGED ***** OR PRIOR suffix IS NULL -- ***** NEW ***** ) -- ***** NEW ***** AND ( 'x' || middle_name != 'x' || PRIOR middle_name -- ***** NEW ***** OR 'x' || suffix != 'x' || PRIOR suffix -- ***** NEW ***** ) -- ***** NEW ***** ) , got_match_cnt AS ( SELECT DISTINCT min_we_pid , leaf_we_pid , COUNT (DISTINCT leaf_we_pid) OVER (PARTITION BY min_we_pid) AS match_cnt FROM possible_matches ) SELECT we_pid, we_addr_id, first_name, last_name, middle_name, suffix , MIN (we_pid) OVER (PARTITION BY NVL (p.leaf_we_pid, c.min_we_pid)) AS merge_wepid FROM GOT_MIN_we_pid c LEFT OUTER JOIN got_match_cnt p ON p.min_we_pid = c.min_we_pid AND p.match_cnt = 1 ORDER BY c.last_name , c.first_name , c.middle_name ;It will give the results you want in view of all the data of your table, or data that can be added to your table in the future? It depends on what you want. It is impossible to everything that publish you current data, and it is impossible to display all of the data that you can obtain in the future. The best is you specify exactly what you want. For example:
"I want to group the lines and assign the we_pid lowest in the group like the merge_we_pid for all members of the group. Every row belongs to a group exaclty. The following rules define what a group is:
(a) last_name, first_name, middle_name and suffix identical lines are always in the same group.
(b) the lines with different name or last_name are never in the same group.
(c) two lines with different middle_names and/or suffixes can be in the same group, subject to the following conditions
(c1) One of the lines must be 'more complete' than the other
(c2) If row is "less complete" that line B, then A must not be "less complete" any line C less than B or C is "less complete" than the other, whereby all 3 ranks fall in the same group.
Line A is considered to be "less full" line B if and only if
(1) A middle_name matches the beginning of B middle_name (first name of B can be longer), and
(2) if A suffix is not NULL, then B has the same suffix. »I think that the example above describes what you want now. I guess that a suffix simply cannot be less than a complete suffix; that would deal with "II" as a less complete form of 'III'
What you asked for in your previous thread differs only in the role of suffix. The query I posted yesterday treated suffix such as first and last name: it had to be exactly the same for all rows in the group, and suffix played no role in the condition (c) or the definition of the "complete".
In the above query, if CONNECT BY condition tests a child line is more comprehensive than its parent. In the version I posted today, I changed the CONNECT BY clause to match the new definition of the "complete". Obviously, there are requirement suffix had to change. Yesterday, the B line could be more complete that the line has only if its middle_name took longer; now, they can be the same. But we cannot tolerate a situation where a line is it's own parent: to avoid to connect BY loops, add a condition that the middle_name or the suffix must be different.Published by: Frank Kulash, July 25, 2012 10:07
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