我有一个有两列问题的表格。
我要做的是比较几周，看看EmpID重复了多少次。例如：
第1周是我的基础（开始日期范围，例如BETWEEN＆＃39; 2017-07-22＆＃39; AND＆＃39; 2017-07-29 23：59：59.993＆＃39;）。我现在想比较第2周和第1周。如果EmpID在第2周重复，我会看到2的计数，如果它在第2周第一次出现，那么计数为1.
继续第3周。如果EmpID出现在第1周，第2周和第3周，那么我想看到3的计数，如果它只出现在第2周和第3周然后计数为2，如果它只出现在第3周那么计数为1.
最后是第4周。如果EmpID出现在第1周，第2周，第3周和第4周，那么我希望看到计数为4.如果EmpID出现在第2周，第3周和第4周，则计数为3，如果它出现在第3周和第4周，则计数为2如果它仅在第4周出现，则计数为1.
非常感谢任何帮助。
添加了我迄今为止尝试但未获得预期效果的内容。
select t.emp_id,
(select count(emp_id)
from [Vacation Audit Care 2017]
WHERE ((UPLOAD_DATE BETWEEN '2017-07-22' AND '2017-07-29 23:59:59.993') or(UPLOAD_DATE BETWEEN '2017-07-29' AND '2017-08-05 23:59:59.993') or (UPLOAD_DATE BETWEEN '2017-08-12' AND '2017-08-19 23:59:59.993'))
and emp_id=t.emp_id) as counts
from [Vacation Audit Care 2017] t
group by t.emp_id
order by counts desc
示例数据
╔═════════════════╦═════════╗
║ Time_Stamp ║ Emp_ID ║
║ 7/20/2017 19:40 ║ 3140340 ║
║ 7/20/2017 19:40 ║ 2000950 ║
║ 7/20/2017 19:40 ║ 3118410 ║
║ 7/20/2017 19:40 ║ 311840 ║
║ 7/23/2017 21:19 ║ 3140340 ║
║ 7/23/2017 21:19 ║ 2000950 ║
║ 7/23/2017 21:19 ║ 3118410 ║
║ 7/23/2017 21:19 ║ 3124160 ║
║ 7/30/2017 7:00 ║ 3140340 ║
║ 7/30/2017 7:00 ║ 2000950 ║
║ 7/30/2017 7:00 ║ 3118410 ║
║ 7/30/2017 7:00 ║ 311840 ║
║ 8/6/2017 12:00 ║ 3140340 ║
║ 8/6/2017 12:00 ║ 3118410 ║
║ 8/6/2017 12:00 ║ 3124160 ║
║ 8/6/2017 12:00 ║ 311840 ║
║ 8/13/2017 12:00 ║ 3140340 ║
║ 8/13/2017 12:00 ║ 3118410 ║
║ 8/13/2017 12:00 ║ 3124160 ║
║ 8/13/2017 12:00 ║ 311840 ║
╚═════════════════╩═════════╝
预期输出
╔═════════╦════════╦═══════╦════════╗
║ ║ 30-Jul ║ 6-Aug ║ 13-Aug ║
║ emp_id ║ wk2 ║ wk3 ║ wk4 ║
║ 3140340 ║ 2 ║ 3 ║ 4 ║
║ 2000950 ║ 2 ║ ║ ║
║ 3118410 ║ 2 ║ 3 ║ 4 ║
║ 311840 ║ 1 ║ 2 ║ 3 ║
║ 3124160 ║ ║ 1 ║ 2 ║
╚═════════╩════════╩═══════╩════════╝
根据预期输出使用7/23作为我的基础（第1周） 该表显示了预期的数据结果。 在第2周，311840第一次出现，所以我希望计数为1。
接下来的一周（8/6）311840在wk2中出现一次，在wk3中出现一次，所以我预计计数为2，而3124160首次出现，所以我希望计数为1等。
答案 0 :(得分：2)
这将以您在样本中显示的格式显示结果。我在输出中前后加了一周。要删除那些额外的周，只需从枢轴中删除[29]和[33]。注释掉枢轴线以获取基础数据。
;WITH
weekcounts AS (
SELECT Time_Stamp, Emp_ID, DATEPART(week, Time_Stamp) AS int_week FROM sampleData
)
,counts AS (
SELECT Emp_ID, int_week, 1 AS int_count
FROM weekcounts
UNION ALL
SELECT weekcounts.Emp_ID, weekcounts.int_week, 1 AS int_count
FROM weekcounts
INNER JOIN counts
ON weekcounts.Emp_ID = counts.Emp_ID
AND (weekcounts.int_week - 1) = counts.int_week
)
,countsagg AS (
SELECT Emp_ID, int_week, SUM(int_count) AS int_count
FROM counts
GROUP BY Emp_ID, int_week
)
SELECT * FROM countsagg
PIVOT (MAX(int_count) FOR int_week IN ([29],[30],[31],[32],[33])) piv
答案 1 :(得分：1)
这样的事情应该有效
SET DATEFIRST 1; --this will set Monday as first day of week.
SELECT EmpID, count(1)
from table
group by DATEPART(wk, timestamp), EmpID
答案 2 :(得分：1)
试试这个..它应该有效。我相信你想把这个月的每周分组。
<html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
<title>Daily Tracker Settings</title>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<!-- Loading Bootstrap -->
<link href="../dist/css/vendor/bootstrap/css/bootstrap.min.css" rel="stylesheet">
<!-- Loading Flat UI -->
<link href="../dist/css/flat-ui.min.css" rel="stylesheet">
<link rel="shortcut icon" href="img/favicon.ico">
</head>
<body>
<form id="form1" runat="server" role="form">
<div class="container">
<div class="row mbl">
<div class="col-md-12">
<h4>Plugin methods</h4>
</div>
<div class="col-md-12">
<form role="form">
<div class="form-group">
<label class="checkbox" for="checkbox100">
<input type="checkbox" data-toggle="radio" value="" id="checkbox100" required>
Checkbox
</label>
</div>
<button type="submit" class="btn btn-sm btn-primary">Submit</button>
<button class="btn btn-sm btn-default" data-radiocheck-check="#checkbox100">Check</button>
<button class="btn btn-sm btn-default" data-radiocheck-uncheck="#checkbox100">Uncheck</button>
<button class="btn btn-sm btn-default" data-radiocheck-toggle="#checkbox100">Toggle</button>
<button class="btn btn-sm btn-default" data-radiocheck-indeterminate="#checkbox100">Indeterminate</button>
<button class="btn btn-sm btn-default" data-radiocheck-determinate="#checkbox100">Determinate</button>
<button class="btn btn-sm btn-default" data-radiocheck-disable="#checkbox100">Disable</button>
<button class="btn btn-sm btn-default" data-radiocheck-enable="#checkbox100">Enable</button>
<button class="btn btn-sm btn-default" data-radiocheck-destroy="#checkbox100">Destroy</button>
<button class="btn btn-sm btn-default" data-radiocheck-init="#checkbox100">Init</button><br>
<br>
</form>
</div>
</div>
<!-- /.row -->
</div>
<!-- jQuery (necessary for Flat UI's JavaScript plugins) -->
<script src="../dist/js/vendor/jquery.min.js"></script>
<!-- Include all compiled plugins (below), or include individual files as needed -->
<script src="../dist/js/vendor/video.js"></script>
<script src="../dist/js/flat-ui.min.js"></script>
<script>
$('button:not([type="submit"])').on('click', function (e) {
var $this = $(this);
if (!!$this.attr('data-radiocheck-check')) {
var el = $this.attr('data-radiocheck-check');
$(el).radiocheck('check');
} else if (!!$this.attr('data-radiocheck-uncheck')) {
var el = $this.attr('data-radiocheck-uncheck');
$(el).radiocheck('uncheck');
} else if (!!$this.attr('data-radiocheck-toggle')) {
var el = $this.attr('data-radiocheck-toggle');
$(el).radiocheck('toggle');
} else if (!!$this.attr('data-radiocheck-indeterminate')) {
var el = $this.attr('data-radiocheck-indeterminate');
$(el).radiocheck('indeterminate');
} else if (!!$this.attr('data-radiocheck-determinate')) {
var el = $this.attr('data-radiocheck-determinate');
$(el).radiocheck('determinate');
} else if (!!$this.attr('data-radiocheck-disable')) {
var el = $this.attr('data-radiocheck-disable');
$(el).radiocheck('disable');
} else if (!!$this.attr('data-radiocheck-enable')) {
var el = $this.attr('data-radiocheck-enable');
$(el).radiocheck('enable');
} else if (!!$this.attr('data-radiocheck-destroy')) {
var el = $this.attr('data-radiocheck-destroy');
$(el).radiocheck('destroy');
} else if (!!$this.attr('data-radiocheck-init')) {
var el = $this.attr('data-radiocheck-init');
$(el).radiocheck();
}
e.preventDefault();
});
</script>
</form>
</body>
</html>
答案 3 :(得分：1)
感谢您提供示例数据，让问题更清晰，对未来访问此问题的其他人来说肯定更有用。
示例数据设置：
为了方便起见，我将您的示例数据弹出到table variable中定义并填充如下。
declare @emp_ts table
(
time_stamp datetime not null
, emp_id int not null
)
insert into @emp_ts
values ('7/20/2017 19:40', 3140340),
('7/20/2017 19:40', 2000950),
('7/20/2017 19:40', 3118410),
('7/20/2017 19:40', 311840 ),
('7/23/2017 21:19', 3140340),
('7/23/2017 21:19', 2000950),
('7/23/2017 21:19', 3118410),
('7/23/2017 21:19', 3124160),
('7/30/2017 7:00 ', 3140340),
('7/30/2017 7:00 ', 2000950),
('7/30/2017 7:00 ', 3118410),
('7/30/2017 7:00 ', 311840 ),
('8/6/2017 12:00 ', 3140340),
('8/6/2017 12:00 ', 3118410),
('8/6/2017 12:00 ', 3124160),
('8/6/2017 12:00 ', 311840 ),
('8/13/2017 12:00', 3140340),
('8/13/2017 12:00', 3118410),
('8/13/2017 12:00', 3124160),
('8/13/2017 12:00', 311840 )
<强>答案：
我使用Recursive CTE生成周数，然后使用window function获取每周emp_id的运行总计数。得到了结果并pivoted它以预期的格式获得预期的周数。您可以将日期连接到列名称，但我认为提供的列名称将起作用。
; with wk_nbrs as
(
--recursive CTE that generates the week numbers.
-- 7/23 thru 7/29 is Week 1
select cast('2017-07-23' as date) as wk_bgn
, cast('2017-07-29' as date) as wk_end
, 1 as wk_nbr
union all
select dateadd(d,7,fw.wk_bgn) as wk_bgn
, dateadd(d,7,fw.wk_end) as wk_end
, fw.wk_nbr + 1 as wk_nbr
from wk_nbrs as fw
where 1=1
and fw.wk_nbr < 100
)
, emp_wk_cnt as
(
--Getting the running total count of emp_id by week
select a.emp_id
, b.wk_nbr
, count(*) over (partition by a.emp_id order by b.wk_nbr asc) as emp_wk_cnt
from @emp_ts as a
inner join wk_nbrs as b on cast(a.time_stamp as date) between b.wk_bgn and b.wk_end
group by a.emp_id
, b.wk_nbr
)
--pivoting the results out to final expected output
select post.emp_id
, post.wk2
, post.wk3
, post.wk4
from (
select a.emp_id
, 'wk' + cast(a.wk_nbr as varchar(10)) as wk_nbr
, a.emp_wk_cnt
from emp_wk_cnt as a
) as pre
pivot (sum(pre.emp_wk_cnt)
for pre.wk_nbr in
([wk2],[wk3],[wk4])
) post
order by post.emp_id
<强>结果：
查询结果如下。你的预期输出和打击之间的唯一区别是emp_id 3124160，我认为wk3应该是2，wk 4应该是3，因为有问题的emp_id有以下{{1}的记录} S：
time_stamp如果情况不是这样，请告诉我。
7/23/2017 21:19
8/6/2017 12:00
8/13/2017 12:00
答案 4 :(得分：1)
SELECT *
,ROW_NUMBER() OVER(PARTITION BY EMPID ORDER BY [TIMESTAMP]) AS [ROWCOUNT]
,DATENAME(WEEK, timestamp) weeknum
INTO #TMPweek
FROM [Vacation Audit Care 2017]
where timestamp>'2017-07-20 19:40:00.000'
select distinct timestamp, cast(day(timestamp) as nvarchar)+
cast('-' as nvarchar)+cast(datename(m,[timestamp]) as nvarchar)+
cast(' wk' as nvarchar) +cast(DATEPART(wk, timestamp) as nvarchar) weekname
into #tw from #TMPweek
select empid,[ROWCOUNT], weekname into #pvt
from
#TMPweek a
inner join #tw b
on a.timestamp=b.timestamp
SELECT empid, [23-July wk30], [30-July wk31], [6-August wk32], [13-August
wk33]
FROM
(select empid,[ROWCOUNT], Weekname from #pvt
) AS ST
PIVOT
(
SUM([ROWCOUNT])
FOR Weekname IN ([23-July wk30], [30-July wk31], [6-August wk32], [13-August
wk33])
) AS PivotTable;
OUTPUT -
empid 23-July wk30 30-July wk31 6-August wk32 13-August wk33
311840 NULL 1 2 3
2000950 1 2 NULL NULL
3118410 1 2 3 4
3124160 1 NULL 2 3
3140340 1 2 3 4