sql

编码相同取id最大的

SELECT * from project_financial_status p

  WHERE NOT EXISTS
  (SELECT * FROM project_financial_status WHERE id >p.id
  AND p.project_code = project_code)

根据时间和code排序取时间最大的

SELECT * FROM
  (SELECT * FROM project_historical_cost
  WHERE project_code =
  'C20190101034' AND project_date <= '2019-05-01'
  ORDER BY
  project_date DESC ) a GROUP BY project_code