HP Pavilion dv6626us needs special support to connect the HDD and I have no

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  Hello
  
  I have a query which can be simplified to the following data, if
  Assume that it is my dataset, with THE_VALUE of
  Sign (Anum) therefore,-1, 0 or 1.
  
  * (Sorry for crushed registration. "I can't understand this website's rich text formatting.) *
  
  THE_VALUE THE_DATE THE_MEMBER
  1 0 1 January 2007
  1 0 1 February 2007
  1 0 1 March 2007
  1. 1 April 1, 2007
  1 1 1 May 2007
  1 0 1 June 2007
  1. 1 July 1, 2007
  1. 1 August 1, 2007
  1 0 01 - sep - 2007
  2 0 1 January 2007
  2 1 February 1, 2007
  2 0 1 March 2007
  2. 1 April 1, 2007
  2 1 1 May 2007
  2 0 1 June 2007
  -2 1 1 July 2007
  2. 1 August 1, 2007
  2. 1 01-sep-2007
  3 0 1 January 2007
  3. 1 February 1, 2007
  3 0 1 March 2007
  3 0 1 April 2007
  3. 1 May 1, 2007
  3 0 1 June 2007
  3. 1 July 1, 2007
  3. 1 August 1, 2007
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  SELECT the_member
  the_value
  the_date
  first_value (the_date)
  in the first_date (PARTITION the_member, ORDER BY the_member the_value, the_date)
  last_value (the_date)
  during the last_date (PARTITION the_member, ORDER BY the_member the_value, the_date)
  OF togola
  ORDER BY the_member
  the_date;
  
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  THE_MEMBER THE_VALUE THE_DATE FIRST DATE LAST DATE
  1 0 January 1, 2007 January 1, 2007 January 1, 2007
  1 0 1 February 2007 1 January 2007 1 February 2007
  1 0 1 March 2007 1 January 2007 March 1, 2007
  1. 1 April 1, 2007 April 1, 2007 April 1, 2007
  1. 1 May 1, 2007 April 1, 2007 may 1, 2007
  1-0 June 1, 2007 January 1, 2007 June 1, 2007
  1. 1 July 1, 2007 April 1, 2007 July 1, 2007
  1. 1 August 1, 2007 April 1, 2007 August 1, 2007
  1 0 01 - sep - 2007 1 January 2007 01-Sep-2007
  
  2 0 January 1, 2007 January 1, 2007 January 1, 2007
  2 1 1 February 2007 February 1, 2007 February 1, 2007
  2 0 1 March 2007 1 January 2007 March 1, 2007
  2. 1 April 1, 2007 February 1, 2007 April 1, 2007
  2 1 1 May 2007 February 1, 2007 may 1, 2007
  2-0 June 1, 2007 January 1, 2007 June 1, 2007
  -2 1 1 July 2007 February 1, 2007 July 1, 2007
  2. 1 August 1, 2007 February 1, 2007 August 1, 2007
  2. 1 01-sep-2007 1 February 2007 01-Sep-2007
  
  3 0 January 1, 2007 January 1, 2007 January 1, 2007
  3. 1 February 1, 2007 February 1, 2007 February 1, 2007
  3 0 1 March 2007 1 January 2007 March 1, 2007
  3 0 April 1, 2007 January 1, 2007 April 1, 2007
  3. 1 May 1, 2007 February 1, 2007 may 1, 2007
  3 0 June 1, 2007 January 1, 2007 June 1, 2007
  3. 1 July 1, 2007 February 1, 2007 July 1, 2007
  3. 1 August 1, 2007 February 1, 2007 August 1, 2007
  3 0 01 - sep - 2007 1 January 2007 01-Sep-2007
  
  That's what I need to get in the end.
  
  THE_MEMBER FIRST DAY DATE OF LAST
  July 2 1 2007-01 - Sep - 2007
  
  Any ideas would be really appreciated.
  
  Published by: user10700981 on December 17, 2008 13:57

  Hello

  We'll call the lines whose the_value =-1 'good' lines and lines with the_value! =-1 ranks of 'bad '.

  For each value of the_value, we need the first and last values of the_date where
  the last row in the group is good
  (b) the current line is good
  (c) there is a lot of lines from the current line to the end of the Group
  Condition (a) is just a special case of (c).
  Conditions (b) and (c) are similar: (b) if applied to the current line, (c) applies to all lines after him. They can easily be combined into a single test that counts the number of lines after the current line or bad.

  Which produces the results you requested of the sample data that you provided:

  WITH     count_bad     AS
  (
       SELECT     the_member
       ,     the_date
       ,     COUNT     ( CASE
                      WHEN  the_value != -1 -- assuming the_value is never NULL
                      THEN  1
                   END
                 ) OVER     ( PARTITION BY     the_member
                        ORDER BY     the_date     DESC
                      )     AS bad_cnt
       FROM     table_x
  )
  SELECT     the_member
  ,     MIN (the_date)     AS first_date
  ,     MAX (the_date)     AS last_date
  FROM     count_bad
  WHERE     bad_cnt     = 0
  GROUP BY     the_member
  ORDER BY     the_member;
  

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