Question

我目前有以下查询：

SELECT  name, id, 
  CASE 
    WHEN( status = 'Missing' AND severity = 'Optional' ) AND ( id=123 ) 
    THEN COALESCE(count(patchid),0)
    ELSE 0
  END AS missingoptional,

  CASE 
    WHEN( status = 'Missing' AND severity = 'Important' ) AND ( id=123 ) 
    THEN COALESCE(count(patchid),0)
    ELSE 0
  END  as missingimportant

FROM tablename
GROUP BY  name, id, status, severity
ORDER BY id

目前的结果：

name            id          missingoptional     missingimportant
abc             123         10                  0
abc             123         0                   20

预期结果：

name            id          missingoptional     missingimportant
abc             123         10                  20

请告诉我在上述查询中我必须做哪些更改才能获得预期的输出。

Answer 1

也许这就是你想要的：

SELECT name, id, 
  SUM(CASE 
      WHEN( status = 'Missing' AND severity = 'Optional' ) AND ( id=123 ) 
          THEN 1
      ELSE 0
      END) AS missingoptional,
  SUM(CASE 
      WHEN( status = 'Missing' AND severity = 'Important' ) AND ( id=123 ) 
          THEN 1
      ELSE 0
      END) as missingimportant
    FROM tablename
GROUP BY name, id
ORDER BY id

我从GROUP BY删除了状态和严重性，而是将SUM()放入选择列表进行聚合。

Answer 2

使用MAX（）聚合函数

interface IAnimal
{
    pubilc void Hunt();
}

interface IWeapon
{
}

Animal : IAnimal
{
    IWeapon _weapon;

    public void Animal(IWeapon weapon)
    {
       _weapon = weapon;
    }

    public void Hunt()
    {
        UseWeapon(_weapon);
    }

    protected abstract void UseWeapon(IWeapon weapon);
}

如何保存开关案例会导致不同的列

2 个答案: