Sql 每个ID有多少个“；第1列”“；第2列”；和“；总计”；并按“；总计”；

我在PostgreSQL 9.2中有下表：

CREATE TABLE leave
(
  "ID" text NOT NULL,
  "Leave class" text NOT NULL,
  "Leave Date" date NOT NULL,
  CONSTRAINT leave_pkey PRIMARY KEY (payroll, "Leave class", "Leave Date")
)

我想得到每个ID上有多少“紧急”、“病假”和“总假期”，并按“总假期”（降序）排序。大概是这样的：

类似的内容（不清楚是否使用ID或工资单）

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请参见

这也将显示零计数：如果要从另一个表中获取工资值，这一点很重要

查看更多信息

更简单的代码在这里

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显示的输入不清楚期望的结果：23266只有一次病假，不是吗？23234有一张病假和一张紧急病假。@RaphaëlAlthaus，我在最后一张桌子上搞错了对不起。

select Payroll,
       sum(case when "Leave class" = 'Emergency' then 1 else 0 end) as "emergency leave",
       sum(case when "Leave class" = 'Sick' then 1 else 0 end) as "Sick leave",
       count(*) as "Total"
FROM leave
group by Payroll
order by count(*) desc;

SELECT
  base.payroll,
  leaveCounts.Emergency,
  leaveCounts.Sick,
  leaveCounts.Total
FROM
 (SELECT DISTINCT payroll FROM leave) AS base
  LEFT JOIN
 (SELECT
     COUNT(CASE WHEN "Leave class" = 'Emergency' THEN 1 END) AS Emergency, 
     COUNT(CASE WHEN "Leave class" = 'Sick' THEN 1 END) AS Sick,
     COUNT(*) AS Total,
     payroll
  FROM
      leave
  GROUP BY
     payroll
  ) AS leaveCounts
      ON base.payroll = leaveCOunts.payroll
ORDER BY
  leaveCOunts.Total DESC;

SELECT
     payroll,
     COUNT(CASE WHEN "Leave class" = 'Emergency' THEN 1 END) AS Emergency, 
     COUNT(CASE WHEN "Leave class" = 'Sick' THEN 1 END) AS Sick,
     COUNT(*) AS Total
  FROM
      leave
  GROUP BY
      payroll
   ORDER BY
     COUNT(*) DESC;