根据两表统计任务完成情况

存在两个表，任务表worktodo和完成任务表workdone，worktodo记录所有工作人员worker的任务，而workdone记录完成的部分，一条workorder可能分多次记录到workdone中。结构如下：

CREATE TABLE worktodo
(
    workorder VARCHAR(32),
    worker VARCHAR(64),
    duedate DATETIME,
    PRIMARY KEY workorder
);
CREATE TABLE workdone
(
    workorder VARCHAR(32),
    donedate DATETIME
);

示例数据：

worktodo:
workorder  worker  duedate
10001      JOE     2012-01-01
10002      JOE     2012-01-02
10003      FRED    2012-01-03
10004      BILL    2013-12-31

workdone:
workerorder     donedate
10001      2012-01-31
10002      2011-12-30
10002      2012-01-04

期望统计每个worker的任务总数以及超期的任务，超期任务指的是donedate > duedate或者没有workdone记录且duedate < 当前时间NOW。

就以上数据而言，JOE的总数为2,10001超期，10002由于在workdone中有在duedate之前的记录，不算超期；FRED总数为1,10003由于无workdone记录且duedate小于NOW，超期；BILL总数为1,10004的duedate未到，则不算超期。综上，预期结果如下：

Desired:
worker   num_total_workorders   num_late_workorders
BILL     1                      0
FRED     1                      1
JOE      2                      1

如何编写单条SQL实现该功能？

本题来源stackflow

select t1.worker,
	   sum(1) as num_total_workorders,
	   sum
	   (
	    case when t2.donedate is null and t1.duedate < getdate() then 1
	         when t2.donedate > t1.duedate then 1
	         else 0
	    end
	   ) as num_late_workorders
from worktodo t1
left outer join
	(
		select workorder,min(donedate) as donedate
		from workdone
		group by workorder
	) t2
  on t1.workorder = t2.workorder
group by t1.worker

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