Xquery 帮助排序父子关系 [英] Xquery help sorted parent child relationship
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问题描述
我有以下 xml(从实际输入中简化和匿名化)它基本上包含一个策略列表，这些策略具有策略开始日期、策略参考和父策略参考(0 表示没有父策略)
I have the below xml (simplified and anonymized from real input) it basically contains a list of policies which have a policy start date, policy reference, and a parent policy reference (with 0 indicating no parent)
我想要实现的是表单的输出.
What I am trying to achieve is the output of the form.
- 最旧的策略在顶部(最旧的开始日期)
- 如果它有孩子，它的孩子必须跟随(也按最早的开始日期排序)
- 如果它有孩子，它的孩子必须跟随(也按最早的开始日期排序)
它实际上让我难住了，我尝试了各种方法，但这是我最近的尝试.
It actually has me stumped, I've tried various things, but here is my latest attempt.
{ let $rows:= for $x in SOAP-ENV:Envelope/SOAP-ENV:Body/cus:GetCustPolicyResponse/cus:Policy/cus:PolicyRow order by $x/cus:DateStart return $x for$policy in distinct-values($rows/cus:PolRef) for $parentPolicy in distinct-values($rows/cus:parentPolRef) for $row in $rows where $row/cus:parentPolRef =$parentPolicy and $row/cus:PolRef =$policy return <tr> <td>{$row/cus:PolRef/text()}</td> <td>{$row/cus:parentPolRef/text()}</td> <td>{$row/cus:DateStart/text()}</td> </tr> }XML
<SOAP-ENV:Envelope xmlns:SOAP-ENV="hp://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="hp://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="hp://www.w3.org/2001/XMLSchema"> <SOAP-ENV:Body> <GetCustPolicyResponse xmlns="hp://www.client.com/services/customer"> <Policy> <PolicyRow> <PolRef>1</PolRef> <DateStart>2011-04-01</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>2</PolRef> <DateStart>2011-04-01</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>3</PolRef> <DateStart>2011-04-20</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>20</PolRef> <DateStart>2011-04-02</DateStart> <parentPolRef>1</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>21</PolRef> <DateStart>2011-04-01</DateStart> <parentPolRef>1</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>26</PolRef> <DateStart>2011-04-22</DateStart> <parentPolRef>3</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>4</PolRef> <DateStart>2011-04-03</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>25</PolRef> <DateStart>2011-04-21</DateStart> <parentPolRef>3</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>24</PolRef> <DateStart>2011-04-16</DateStart> <parentPolRef>2</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>23</PolRef> <DateStart>2011-04-17</DateStart> <parentPolRef>2</parentPolRef> </PolicyRow> </Policy> </GetCustPolicyResponse> </SOAP-ENV:Body> </SOAP-ENV:Envelope>想要的输出
<table> <tr> <td>Policy Reference</td> <td>Policy start date</td> </tr> <tr> <td>1</td> <td>2011-04-01</td> </tr> <tr> <td>21</td> <td>2011-04-21</td> </tr> <tr> <td>20</td> <td>2011-04-02</td> </tr> <tr> <td>2</td> <td>2011-04-01</td> </tr> <tr> <td>24</td> <td>2011-04-16</td> </tr> <tr> <td>23</td> <td>2011-04-17</td> </tr> <tr> <td>4</td> <td>2011-04-03</td> </tr> <tr> <td>3</td> <td>2011-04-20</td> </tr> <tr> <td>25/td> <td>2011-04-21</td> </tr> <tr> <td>26</td> <td>2011-04-22</td> </tr>推荐答案
I.此 XQuery 代码:
declare namespace x = "hp://www.client.com/services/customer"; declare function x:PolicyByParentRef($pNodes as element()*, $pRef as xs:string) as element()* { $pNodes[x:parentPolRef eq $pRef] }; declare function x:ProcessPolicy($pNodes as element()*, $pPol as element()) as element()* { if(not(empty($pPol))) then (<tr> <td>{$pPol/x:PolRef/text()}</td>, <td>{$pPol/x:DateStart/text()}</td> </tr>, for $child-policy in x:PolicyByParentRef($pNodes, $pPol/x:PolRef) order by $child-policy/x:DateStart descending return x:ProcessPolicy($pNodes, $child-policy) ) else () }; <table> {for $topPolicy in x:PolicyByParentRef(/*/*/*/*/x:PolicyRow, '0') order by $topPolicy/x:DateStart descending return x:ProcessPolicy(/*/*/*/*/x:PolicyRow, $topPolicy) } </table>应用于提供的 XML 文档时:
<SOAP-ENV:Envelope xmlns:SOAP-ENV="hp://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="hp://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="hp://www.w3.org/2001/XMLSchema"> <SOAP-ENV:Body> <GetCustPolicyResponse xmlns="hp://www.client.com/services/customer"> <Policy> <PolicyRow> <PolRef>1</PolRef> <DateStart>2011-04-01</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>2</PolRef> <DateStart>2011-04-01</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>3</PolRef> <DateStart>2011-04-20</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>20</PolRef> <DateStart>2011-04-02</DateStart> <parentPolRef>1</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>21</PolRef> <DateStart>2011-04-01</DateStart> <parentPolRef>1</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>26</PolRef> <DateStart>2011-04-22</DateStart> <parentPolRef>3</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>4</PolRef> <DateStart>2011-04-03</DateStart> <parentPolRef>0</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>25</PolRef> <DateStart>2011-04-21</DateStart> <parentPolRef>3</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>24</PolRef> <DateStart>2011-04-16</DateStart> <parentPolRef>2</parentPolRef> </PolicyRow> <PolicyRow> <PolRef>23</PolRef> <DateStart>2011-04-17</DateStart> <parentPolRef>2</parentPolRef> </PolicyRow> </Policy> </GetCustPolicyResponse> </SOAP-ENV:Body> </SOAP-ENV:Envelope>产生想要的结果:
<?xml version="1.0" encoding="UTF-8"?> <table> <tr> <td>3</td>, <td>2011-04-20</td> </tr> <tr> <td>26</td>, <td>2011-04-22</td> </tr> <tr> <td>25</td>, <td>2011-04-21</td> </tr> <tr> <td>4</td>, <td>2011-04-03</td> </tr> <tr> <td>1</td>, <td>2011-04-01</td> </tr> <tr> <td>20</td>, <td>2011-04-02</td> </tr> <tr> <td>21</td>, <td>2011-04-01</td> </tr> <tr> <td>2</td>, <td>2011-04-01</td> </tr> <tr> <td>23</td>, <td>2011-04-17</td> </tr> <tr> <td>24</td>, <td>2011-04-16</td> </tr> </table>二.仅供比较 - XSLT 解决方案:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:x="hp://www.client.com/services/customer" exclude-result-prefixes="x"> <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:strip-space elements="*"/> <xsl:key name="kPolicyByRef" match="x:PolicyRow" use="x:parentPolRef"/> <xsl:template match="/"> <table> <xsl:apply-templates select= "key('kPolicyByRef', '0')"> <xsl:sort select="x:DateStart" order="descending"/> </xsl:apply-templates> </table> </xsl:template> <xsl:template match="x:PolicyRow"> <tr> <xsl:apply-templates/> </tr> <xsl:apply-templates select= "key('kPolicyByRef', x:PolRef)"> <xsl:sort select="x:DateStart" order="descending"/> </xsl:apply-templates> </xsl:template> <xsl:template match="x:PolicyRow/*"> <td><xsl:value-of select="."/></td> </xsl:template> <xsl:template match="x:parentPolRef" priority="2"/> </xsl:stylesheet>这篇关于Xquery 帮助排序父子关系的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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