The best solution to a problem ugly pt 2

I posted a very similar question earlier this week and got a GREAT solution of Frank Kulash [post earlier | https://forums.oracle.com/forums/message.jspa?messageID=10823001#10823001], but now I have a different condition, after trying again and again, I turn to the pros:

I'm looking for the best way to set up a query on a table that has a part of their construction as follows:

Loss of keys
--- ---
1 arm
step 2
3 eye
4 leg/eye
5 leg/eye/arm/foot
6 foot/eye
7 hands/feet

(I know, I know. Don't shoot the Messenger! "This is the chart that I have to deal with.)
I have to accept the value of an Apex shuttle object that returns the values selected in colon-delimited string. Two examples of a query string potential:
leg: eye
Eye

I need develop a query that returns records that respond to the < = l the minimum criteria. For example, if the application is:
eye: foot
My query must return only the records * 3 * and 6.
Crazy, right?
As you can see, the order doesn't matter, but {here} the records returned by the query cannot containing all of the values that were not specifically requested. They peuvent contain less, as long as they also have been expressly requested.

The database is 10.2
Thanks in advance!

Hello

Here's one way:

WITH      targets     AS
(
     SELECT     '%/' || REGEXP_SUBSTR ( :str
                                , '[^:]+'
                                , 1
                                , LEVEL
                                )
               || '/%'          AS target
     ,     target_cnt
     FROM     (
              SELECT  LENGTH (         :str       || 'ab')
                    - LENGTH (REPLACE (:str, ':') || 'a')     AS target_cnt
              FROM  dual
          )
     CONNECT BY     LEVEL     <= target_cnt
)
SELECT       l.key
FROM       limbs        l
JOIN       targets  t  ON   '/' || l.loss
                              || '/'     LIKE t.target
GROUP BY  l.key, l.loss
HAVING       COUNT (*)     = 1 + LENGTH (         l.loss      )
                     - LENGTH (REPLACE (l.loss, '/'))
;

You will notice that it is identical to the solution to your previous question, except that the HAVING clause has changed, and because the fact HAVING clause now refers to l.loss, I added l.loss to the GROUP BY clause.

This requires that the elements of: str and l.loss are unique. In other words, if the user enters "eyes: the eyes", it will match l.keys 4 and 6, because these keys are 2 times, and there are 2 items in l.loss. If this is a problem, the above query will have changed a little, but only a little.

The HAVING clause says now that the number of items that match to the must be the number of items in l.loss. How is calculated the number of items in l.loss is similar to the way in which the number of elements in the parameter: str is calculated. However, the earlier motion went to a little trouble in the case: str has been: NULL. I don't know what is necessary, and I guess that's not necessary for l.loss, either.

Tags: Database

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    Welcome to the community - just treat the VM as you would a physical machine - connect via RDP, SSH etc - which you could do with a Windows machine or lelievre

    See VMware Virtualize your desktops - that would be the solution to answer your question 5 that you could use a workstation as a terminal Wyse to allow a user to connect via AD and connect to their virtual office.

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