将每个子标记转换为具有多个分隔符的单列-SQL Server (2) [英] Converting Every Child Tags in to a Single Column with multiple Delimiters -SQL Server (2)

问题描述

declare @xml xml='<plan>
    <prescriptions id="1">
        <prescription>
            <name>ABC</name>
            <frequency>Daily</frequency>
            <dailyfrequency>
               <morning>2</morning>
               <afternoon></afternoon>
               <night>1</night>
            </dailyfrequency>
            <dayfrequency></dayfrequency>
        </prescription>
        <prescription>
            <name>EDF</name>
            <frequency>Daily</frequency>
            <dailyfrequency>
              <morning>5</morning>
              <afternoon>5</afternoon>
              <night>1</night>
           </dailyfrequency>
           <dayfrequency></dayfrequency>
        </prescription>
        <prescription>
            <name>YTER</name>
            <frequency>Weekly</frequency>
            <dailyfrequency>
              <morning>5</morning>
              <afternoon>5</afternoon>
              <night>1</night>
            </dailyfrequency>
            <dayfrequency>Monday,Tuesday,Wednesday</dayfrequency>
        </prescription>
    </prescriptions>
    <prescriptions id="2">
        <prescription>
            <name>YTRE</name>
            <frequency>Daily</frequency>
            <dailyfrequency>
               <morning>2</morning>
               <afternoon></afternoon>
               <night>1</night>
            </dailyfrequency>
            <dayfrequency></dayfrequency>
          </prescription>
    </prescriptions>
</plan>'

我们可以像下面这样查询，用多个分隔符分隔每个元素，以导致同一列.

We can query like below to segregate each elements with multiple delimiter to be resulted in same column.

SELECT STUFF(
(
    SELECT '!' + STUFF(p.query(N'for $n in .//*[local-name()!="dailyfrequency"]
                           return <a>{concat("$",($n/text())[1])}</a>'
                        ).value(N'.',N'nvarchar(max)'),1,1,'')
    FROM @xml.nodes(N'/plan/prescriptions/prescription') AS A(p)
    FOR XML PATH(''),TYPE).value(N'.',N'nvarchar(max)'),1,1,'')

结果:

ABC$Daily$2$$1$!EDF$Daily$5$5$1$!YTER$Weekly$5$5$1$Monday,Tuesday,Wednesday!YTRE$Daily$2$$1$

但这里的问题是这会将所有标签组合在一个列下.当我们查看xml时，它有2个主要部门Prescription id=1和prescritpion id=2.所以我们的最终结果将是这样的

But the problem here is this will combine all the tags under a single column.When we look the xml it has 2 major division Prescription id=1 and prescritpion id=2.So our final result will be like this

结果:

ABC$Daily$2$$1$!EDF$Daily$5$5$1$!YTER$Weekly$5$5$1$Monday,Tuesday,Wednesday
YTRE$Daily$2$$1$

我认为我们必须为每个 <Prescriptions> 粉碎 Xml 然后必须计算这个

I think we have to Shred the Xml for each <Prescriptions> then have to calculate this

有人可以在这里解决这个问题

Could someone please here to solve this

提前致谢，Jayendran

Thanks in advance,Jayendran

推荐答案

像这样改变你的查询:

SELECT STUFF(
(
    SELECT '!' + STUFF(p.query(N'for $n in .//*[local-name()!="dailyfrequency"]
                           return <a>{concat("$",($n/text())[1])}</a>'
                        ).value(N'.',N'nvarchar(max)'),1,1,'')
    FROM p.nodes(N'prescription') AS A(p)
    FOR XML PATH(''),TYPE).value(N'.',N'nvarchar(max)'),1,1,'')
FROM @xml.nodes(N'/plan/prescriptions') AS A(p);

首先我们为不同的处方创建一个派生表，其次我们对每个处方分别使用以前的代码.

First we create a derived table for the different prescriptions, Secondly we use the former code for each prescription separately.

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