how to structure availability relationships












0















I am building a recruitment database and need to structure 'current availability' data against candidates which can be one of three options - available now, a specific date (e.g. 5/3/2019) or a relative date (e.g. 4 weeks). When querying the database in my application the relative date will be calculated on the fly.



I am trying to work out the best way to store this data in a relational database. In my project I am using MySQL 5.7. I have tested and can see that storing this as JSON is possible e.g. the availability field in candidates table can be JSON with the following



{
'immediate': true
}


or



{
'relative': {
'period': '4',
'unit': 'Weeks'
}
}


or



{
'date': '2019-03-05'
}


the query can be something like this



select `candidates`.*, 
CASE
WHEN availability->>'$.immediate' = 'true' THEN date(now())
WHEN availability->>'$.date' THEN date(availability->>'$.date')
WHEN availability->>'$.relative.unit' = 'week' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' WEEK))
WHEN availability->>'$.relative.unit' = 'month' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' MONTH))
END as available from `candidates`
order by `available` desc


Although this works I am adverse to using JSON in a relational database so wondering how the same result could be achieved in other more relational ways.









share



























    0















    I am building a recruitment database and need to structure 'current availability' data against candidates which can be one of three options - available now, a specific date (e.g. 5/3/2019) or a relative date (e.g. 4 weeks). When querying the database in my application the relative date will be calculated on the fly.



    I am trying to work out the best way to store this data in a relational database. In my project I am using MySQL 5.7. I have tested and can see that storing this as JSON is possible e.g. the availability field in candidates table can be JSON with the following



    {
    'immediate': true
    }


    or



    {
    'relative': {
    'period': '4',
    'unit': 'Weeks'
    }
    }


    or



    {
    'date': '2019-03-05'
    }


    the query can be something like this



    select `candidates`.*, 
    CASE
    WHEN availability->>'$.immediate' = 'true' THEN date(now())
    WHEN availability->>'$.date' THEN date(availability->>'$.date')
    WHEN availability->>'$.relative.unit' = 'week' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' WEEK))
    WHEN availability->>'$.relative.unit' = 'month' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' MONTH))
    END as available from `candidates`
    order by `available` desc


    Although this works I am adverse to using JSON in a relational database so wondering how the same result could be achieved in other more relational ways.









    share

























      0












      0








      0








      I am building a recruitment database and need to structure 'current availability' data against candidates which can be one of three options - available now, a specific date (e.g. 5/3/2019) or a relative date (e.g. 4 weeks). When querying the database in my application the relative date will be calculated on the fly.



      I am trying to work out the best way to store this data in a relational database. In my project I am using MySQL 5.7. I have tested and can see that storing this as JSON is possible e.g. the availability field in candidates table can be JSON with the following



      {
      'immediate': true
      }


      or



      {
      'relative': {
      'period': '4',
      'unit': 'Weeks'
      }
      }


      or



      {
      'date': '2019-03-05'
      }


      the query can be something like this



      select `candidates`.*, 
      CASE
      WHEN availability->>'$.immediate' = 'true' THEN date(now())
      WHEN availability->>'$.date' THEN date(availability->>'$.date')
      WHEN availability->>'$.relative.unit' = 'week' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' WEEK))
      WHEN availability->>'$.relative.unit' = 'month' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' MONTH))
      END as available from `candidates`
      order by `available` desc


      Although this works I am adverse to using JSON in a relational database so wondering how the same result could be achieved in other more relational ways.









      share














      I am building a recruitment database and need to structure 'current availability' data against candidates which can be one of three options - available now, a specific date (e.g. 5/3/2019) or a relative date (e.g. 4 weeks). When querying the database in my application the relative date will be calculated on the fly.



      I am trying to work out the best way to store this data in a relational database. In my project I am using MySQL 5.7. I have tested and can see that storing this as JSON is possible e.g. the availability field in candidates table can be JSON with the following



      {
      'immediate': true
      }


      or



      {
      'relative': {
      'period': '4',
      'unit': 'Weeks'
      }
      }


      or



      {
      'date': '2019-03-05'
      }


      the query can be something like this



      select `candidates`.*, 
      CASE
      WHEN availability->>'$.immediate' = 'true' THEN date(now())
      WHEN availability->>'$.date' THEN date(availability->>'$.date')
      WHEN availability->>'$.relative.unit' = 'week' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' WEEK))
      WHEN availability->>'$.relative.unit' = 'month' THEN date(DATE_ADD(now(), INTERVAL availability->>'$.relative.period' MONTH))
      END as available from `candidates`
      order by `available` desc


      Although this works I am adverse to using JSON in a relational database so wondering how the same result could be achieved in other more relational ways.







      mysql database-design





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      asked 4 mins ago









      the-a-trainthe-a-train

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