How to display/concatenate several lines into one line

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  I have three tables:
  1 project Pk = proj_id
  2 Resources_on_Project Pk = Resources_on_proj_id Foreign_key = proj_id
  3 cost Pk = cost_id Foreign_key = proj_id
  
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  Project: proj_id, proj_name, proj_description
  Resources_on_Project: Resources_on_proj_id, proj_id, id_ressource
  cost: cost_id, proj_id, cost_type, cost_incurred
  
  the table relationships:
  
  Project: Resources_on_project 1:m (project has several resources in Appendix)
  Cost of the project: 1:m (project has many costs in annex)
  
  Data:
  
  Project:
  PROJ_ID, proj_name, proj_description
  1, CDM, Advisory Services
  
  2, efficient lighting, efficient ESCO
  
  Resources_on_project
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  11.2, travel, 800
  12.2, travel, 1150
  13.2, food, 3000
  14.2, food, 1000
  15.2, food, 2400
  
  
  First project has 2 resources attached to it and 6 different costs.
  Second project includes 2 resources and 9 heads of different costs.
  
  I want a report that gives me:
  
  Proj_name, Proj_description, id_ressource, cost_incurred
  Advisory services CDM 10500 1/2
  
  ESCO effective effective lighting 14850 3/4
  
  
  I wrote a query:
  «Select «PROJECT»» PROJ_NAME' as 'PROJ_NAME. "
  ««PROJECT»» PROJ_DESCRIPTION' as 'PROJ_DESCRIPTION. "
  sum (COST. COST_INCURRED) as "COST_INCURRED."
  'RESOURCES_ON_PROJECTS '. "" RESOURCE_ID ' as 'RESOURCE_ID.
  'COST', 'cost ',.
  "RESOURCES_ON_PROJECTS" "RESOURCES_ON_PROJECTS"
  'PROJECT' 'PROJECT '.
  where "PROJECT. "" PROJ_ID "=" RESOURCES_ON_PROJECTS. " "" PROJ_ID.
  and 'PROJECT '. "" "PROJ_ID '=' COST. PROJ_ID.
  PROJECT group. PROJ_NAME, PROJECT. PROJ_DESCRIPTION, RESOURCES_ON_PROJECTS. ACCOUNTABLE_RESOURCE_ID
  
  the result is:
  
  Advisory services CDM 1 10500
  Advisory services CDM 2 10500
  ESCO effective effective lighting 3 14850
  ESCO effective effective lighting 4 14850
  
  It gives a new line for each different resource.
  
  
  I want to:
  Advisory services CDM 10500 1/2
  ESCO effective effective lighting 14850 3/4
  
  any help will be appreciated.
  
  Published by: Rahul Gupta on September 7, 2009 16:07

  Rahul Gupta says:
  Shadow of Blu,

  It's exactly what I wanted. But the measures specified by you are not clear for me.

  can you let me know how I can implement?

  Okay... Let's take the internal selection...

  SQL> select proj_name
    2        ,proj_desc
    3        ,resource_id
    4        ,cost
    5        ,row_number() over (partition by proj_name order by resource_id) as rn
    6  from output
    7  /
  
  PROJ_NAME PROJ_DESC               RESOURCE_ID       COST         RN
  --------- ----------------------- ----------- ---------- ----------
  CDM       Advisory Services                 1      10500          1
  CDM       Advisory Services                 2      10500          2
  Efficient ESCO Efficient lighting           3      14850          1
  Efficient ESCO Efficient lighting           4      14850          2
  
  SQL>
  

  This affects a line number for each line, each group proj_name (partition). Line numbers are allocated in the order of the id_ressource.

  Now, using a (hierarchical) query connection we create a hierarchy of folders as well as RN = 2 is considered to be a child of RN = 1, etc..

  SQL> ed
  Wrote file afiedt.buf
  
    1  select level
    2        ,lpad(' ',(level-1)*2,' ')||x.proj_name as proj_name -- indent for hierarchical illustration only
    3        ,x.proj_desc
    4        ,x.resource_id
    5        ,x.cost
    6        ,x.rn
    7  from (
    8        select proj_name
    9              ,proj_desc
   10              ,resource_id
   11              ,cost
   12              ,row_number() over (partition by proj_name order by resource_id) as rn
   13        from output
   14       ) x
   15  connect by proj_name = prior proj_name and rn = prior rn + 1
   16* start with rn = 1
  SQL> /
  
       LEVEL PROJ_NAME       PROJ_DESC               RESOURCE_ID       COST         RN
  ---------- --------------- ----------------------- ----------- ---------- ----------
           1 CDM             Advisory Services                 1      10500          1
           2   CDM           Advisory Services                 2      10500          2
           1 Efficient       ESCO Efficient lighting           3      14850          1
           2   Efficient     ESCO Efficient lighting           4      14850          2
  
  SQL>
  

  Now, using the sys_connect_by_path function we can get it to collect the data, because it passes through to the bottom of the hierarchy...

  SQL> ed
  Wrote file afiedt.buf
  
    1  select level
    2        ,lpad(' ',(level-1)*2,' ')||x.proj_name as proj_name
    3        ,x.proj_desc
    4        ,x.resource_id
    5        ,sys_connect_by_path(resource_id,'/') as resources
    6        ,x.cost
    7        ,x.rn
    8  from (
    9        select proj_name
   10              ,proj_desc
   11              ,resource_id
   12              ,cost
   13              ,row_number() over (partition by proj_name order by resource_id) as rn
   14        from output
   15       ) x
   16  connect by proj_name = prior proj_name and rn = prior rn + 1
   17* start with rn = 1
  SQL> /
  
       LEVEL PROJ_NAME       PROJ_DESC               RESOURCE_ID RESOURCES        COST         RN
  ---------- --------------- ----------------------- ----------- ---------- ---------- ----------
           1 CDM             Advisory Services                 1 /1              10500          1
           2   CDM           Advisory Services                 2 /1/2            10500          2
           1 Efficient       ESCO Efficient lighting           3 /3              14850          1
           2   Efficient     ESCO Efficient lighting           4 /3/4            14850          2
  
  SQL>
  

  Now, we can store up to the sys_connect_by_path to eliminate the left "/" simply by using the TRIM function.
  We also need to do however is to choose only the lines that have reached the bottom of the hierarchy (the nodes). We have a pseudo-device column we can refer to called CONNECT_BY_ISLEAF...

  SQL> ed
  Wrote file afiedt.buf
  
    1  select level
    2        ,lpad(' ',(level-1)*2,' ')||x.proj_name as proj_name
    3        ,x.proj_desc
    4        ,x.resource_id
    5        ,ltrim(sys_connect_by_path(resource_id,'/'),'/') as resources
    6        ,x.cost
    7        ,x.rn
    8        ,connect_by_isleaf
    9  from (
   10        select proj_name
   11              ,proj_desc
   12              ,resource_id
   13              ,cost
   14              ,row_number() over (partition by proj_name order by resource_id) as rn
   15        from output
   16       ) x
   17  connect by proj_name = prior proj_name and rn = prior rn + 1
   18* start with rn = 1
  SQL> /
  
       LEVEL PROJ_NAME       PROJ_DESC               RESOURCE_ID RESOURCES        COST         RN CONNECT_BY_ISLEAF
  ---------- --------------- ----------------------- ----------- ---------- ---------- ---------- -----------------
           1 CDM             Advisory Services                 1 1               10500          1             0
           2   CDM           Advisory Services                 2 1/2             10500          2             1
           1 Efficient       ESCO Efficient lighting           3 3               14850          1             0
           2   Efficient     ESCO Efficient lighting           4 3/4             14850          2             1
  
  SQL>
  

  So now we can filter only the lines where CONNECT_BY_ISLEAF = 1, also remove the output columns and calculates values etc, we don't need to see.

  SQL> ed
  Wrote file afiedt.buf
  
    1  select x.proj_name
    2        ,x.proj_desc
    3        ,ltrim(sys_connect_by_path(resource_id,'/'),'/') as resources
    4        ,x.cost
    5  from (
    6        select proj_name
    7              ,proj_desc
    8              ,resource_id
    9              ,cost
   10              ,row_number() over (partition by proj_name order by resource_id) as rn
   11        from output
   12       ) x
   13  where connect_by_isleaf = 1
   14  connect by proj_name = prior proj_name and rn = prior rn + 1
   15* start with rn = 1
  SQL> /
  
  PROJ_NAME       PROJ_DESC               RESOURCES        COST
  --------------- ----------------------- ---------- ----------
  CDM             Advisory Services       1/2             10500
  Efficient       ESCO Efficient lighting 3/4             14850
  

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