如何编写包含序列元素的 xquery？

Question

我有一个非常大的 xml 数据集，其结构如下：

<root>
    <person>
        <personid>HH3269732</personid>
        <firstname>John</firstname>
        <lastname>Smith</lastname>
        <entertime>01/02/2008 10:15</entertime>
        <leavetime>01/02/2008 11:45</leavetime>
        <entertime>03/01/2008 08:00</entertime>
        <leavetime>03/01/2008 10:00</leavetime>
        ... 
        // number of enter times and leave times vary from person to person
        // there may not be a final leave time (ie, they haven't left yet)
    </person>
    ...
</root>

数据的结构不在我的控制之下。此数据当前位于 MS SQL Server 2005 中单行中的单个 xml 列中。我正在尝试构建一个查询，该查询会产生以下输出：

HH3269732   John   Smith   01/02/2008 10:15   01/02/2008 11:45
HH3269732   John   Smith   03/01/2008 08:00   01/02/2008 10:00
HH3269735   Mark   Pines   02/01/2008 09:00   NULL
HH3263562   James  Frank   NULL               NULL
HH3264237   Harold White   04/18/2008 03:00   04/18/2008 05:00
...

我的查询目前如下所示：

DECLARE @xml xml
SELECT @xml = XmlCol FROM Data

SELECT
    [PersonId] = Persons.PersonCollection.value('(personid)[1]', 'NVARCHAR(50)')
    ,[First Name] = Persons.PersonCollection.value('(firstname)[1]', 'NVARCHAR(50)')
    ,[Last Name] = Persons.PersonCollection.value('(lastname)[1]', 'NVARCHAR(50)')
    ??????
FROM @xml.nodes('root\person') Persons(PersonCollection)

当我从内存中提取它时，该查询可能不是 100% 正确，但我遇到的问题是我不知道如何包含 entertime leavetime 序列元素以获取我在上面指出的所需行集。

谢谢。

更新： 我想补充一点，给定的人员记录可能根本没有 no entertime/leavetime 序列元素，但仍需要在行集中返回。我已经更新了所需输出的示例以反映这一点。

Answer 1

with cte_entertime as (
SELECT
    [PersonId] = t.c.value('(../personid)[1]', 'NVARCHAR(50)')
    ,[First Name] = t.c.value('(../firstname)[1]', 'NVARCHAR(50)')
    ,[Last Name] = t.c.value('(../lastname)[1]', 'NVARCHAR(50)')
    ,[Entertime] = t.c.value('.', 'NVARCHAR(50)')
    ,[entry_number] = ROW_NUMBER() OVER (ORDER BY t.c)
FROM @x.nodes('root/person/entertime') t(c))
, cte_leavetime as (
    SELECT
    [Leavetime] = t.c.value('.', 'NVARCHAR(50)')
    ,[entry_number] = ROW_NUMBER() OVER (ORDER BY t.c)
FROM @x.nodes('root/person/leavetime') t(c))
SELECT PersonID
    , [First Name]
    , [Last Name] 
    , [Entertime]
    , [Leavetime]
    FROM cte_entertime e 
    LEFT OUTER JOIN cte_leavetime l on e.entry_number = l.entry_number

Answer 2

我接受了 Remus 的回答，因为它让我 95% 得到了解决方案。仅供参考，这是最终的查询结构：

with cte_maindata as (
SELECT
    [PersonId] = t.c.value('(personid)[1]', 'NVARCHAR(50)')
    ,[First Name] = t.c.value('(firstname)[1]', 'NVARCHAR(50)')
    ,[Last Name] = t.c.value('(lastname)[1]', 'NVARCHAR(50)')
FROM @x.nodes('root/person') t(c))
, cte_entertime as (
    SELECT
    [PersonId] = t.c.value('(../personid)[1]', 'NVARCHAR(50)')
    ,[Entertime] = t.c.value('.', 'NVARCHAR(50)')
FROM @x.nodes('root/person/entertime') t(c))
, cte_leavetime as (
    SELECT
    [PersonId] = t.c.value('(../personid)[1]', 'NVARCHAR(50)')
    ,[Leavetime] = t.c.value('.', 'NVARCHAR(50)')
FROM @x.nodes('root/person/leavetime') t(c))
SELECT 
    m.PersonID
    ,[First Name]
    ,[Last Name] 
    ,[Entertime]
    ,[Leavetime]
FROM cte_maindata m
    LEFT OUTER JOIN cte_entertime e on m.PersonId = e.PersonId
    LEFT OUTER JOIN cte_leavetime l on m.PersonId = l.PersonId

Answer 3

尚未意识到文档中可能有多个人。在这种情况下，我的查询无论如何都是不正确的。我想如果你先把每个人分解成自己的 XML 片段，然后提取十个进入/离开时间可能会更好。我没有 215k 人 XML 可以尝试，但这里有一个想法：

declare @x xml;
select @x = N'<root>
    <person>
        <personid>HH3269732</personid>
        <firstname>John</firstname>
        <lastname>Smith</lastname>
        <entertime>01/02/2008 10:15</entertime>
        <leavetime>01/02/2008 11:45</leavetime>
        <entertime>03/01/2008 08:00</entertime>
        <leavetime>03/01/2008 10:00</leavetime>
        <entertime>04/01/2008 08:00</entertime>
    </person>
    <person>
        <personid>HH3269733</personid>
        <firstname>Jane</firstname>
        <lastname>Doe</lastname>
        <entertime>01/03/2008 10:15</entertime>
        <leavetime>01/03/2008 11:45</leavetime>
        <entertime>03/04/2008 08:00</entertime>
        <leavetime>03/04/2008 10:00</leavetime>
        <entertime>04/04/2008 08:00</entertime>
    </person>
</root>';


with cte_person as (
    select
        t.c.value('(personid)[1]', 'NVARCHAR(50)') as personid
        , t.c.value('(firstname)[1]', 'NVARCHAR(50)') as firstname
        , t.c.value('(lastname)[1]', 'NVARCHAR(50)') as lastname
        , t.c.query('entertime') as entertime
        , t.c.query('leavetime') as leavetime
    FROM @x.nodes('root/person') t(c))
, cte_cross_enter as (
    select
        p.personid
        , p.firstname
        , p.lastname
        , x.c.value('.', 'datetime') as entertime
        , row_number() over (partition by personid order by x.c) as row_enter
        from cte_person p
        cross apply p.entertime.nodes('/entertime') x(c))
, cte_cross_leave as (
    select
        p.personid 
        , x.c.value('.', 'datetime') as leavetime
        , row_number() over (partition by personid order by x.c) as row_leave
        from cte_person p
        cross apply p.leavetime.nodes('/leavetime') x(c))
select e.personid
    , e.firstname
    , e.lastname
    , e.entertime
    , l.leavetime
    from cte_cross_enter e
    left outer join cte_cross_leave l 
            on e.personid = l.personid and 
            e.row_enter = l.row_leave