efficient xslt conditional increment

夙愿已清 提交于 2020-01-06 02:16:07

问题


In this question i asked how to perform a conditional increment. The provided answer worked, but does not scale well on huge data-sets.

The Input:

<Users>
    <User>
        <id>1</id>
        <username>jack</username>
    </User>
    <User>
        <id>2</id>
        <username>bob</username>
    </User>
    <User>
        <id>3</id>
        <username>bob</username>
    </User>
    <User>
        <id>4</id>
        <username>jack</username>
    </User>
</Users>

The desired output (in optimal time-complexity):

<Users>
   <User>
      <id>1</id>
      <username>jack01</username>
   </User>
   <User>
      <id>2</id>
      <username>bob01</username>
   </User>
   <User>
      <id>3</id>
      <username>bob02</username>
   </User>
   <User>
      <id>4</id>
      <username>jack02</username>
   </User>
</Users>

For this purpose it would be nice to

  • sort input by username
  • for each user
    • when previous username is equals current username
      • increment counter and
      • set username to '$username$counter'
    • otherwise
      • set counter to 1
  • (sort by id again - no requirement)

Any thoughts?


回答1:


This transformation produces exactly the specified wanted result and is efficient (O(N)):

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
 xmlns:ext="http://exslt.org/common" exclude-result-prefixes="ext">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:key name="kUserByName" match="User" use="username"/>
 <xsl:key name="kUByGid" match="u" use="@gid"/>

 <xsl:variable name="vOrderedByName">
  <xsl:for-each select=
  "/*/User[generate-id()=generate-id(key('kUserByName',username)[1])]">
     <xsl:for-each select="key('kUserByName',username)">
       <u gid="{generate-id()}" pos="{position()}"/>
     </xsl:for-each>
  </xsl:for-each>
 </xsl:variable>

  <xsl:template match="node()|@*">
     <xsl:copy>
       <xsl:apply-templates select="node()|@*"/>
     </xsl:copy>
 </xsl:template>

 <xsl:template match="username/text()">
     <xsl:value-of select="."/>
     <xsl:variable name="vGid" select="generate-id(../..)"/>

     <xsl:for-each select="ext:node-set($vOrderedByName)[1]">
      <xsl:value-of select="format-number(key('kUByGid', $vGid)/@pos, '00')"/>
     </xsl:for-each>
 </xsl:template>
</xsl:stylesheet>

When applied on the provided XML document:

<Users>
    <User>
        <id>1</id>
        <username>jack</username>
    </User>
    <User>
        <id>2</id>
        <username>bob</username>
    </User>
    <User>
        <id>3</id>
        <username>bob</username>
    </User>
    <User>
        <id>4</id>
        <username>jack</username>
    </User>
</Users>

the wanted, correct result is produced:

<Users>
   <User>
      <id>1</id>
      <username>jack01</username>
   </User>
   <User>
      <id>2</id>
      <username>bob01</username>
   </User>
   <User>
      <id>3</id>
      <username>bob02</username>
   </User>
   <User>
      <id>4</id>
      <username>jack02</username>
   </User>
</Users>



回答2:


This is kind of ugly and I'm not fond of using xsl:for-each, but it should be faster than using preceding-siblings, and doesn't need a 2-pass approach:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
  <xsl:key name="count" match="User" use="username" />

  <xsl:template match="Users">
    <Users>
      <xsl:for-each select="User[generate-id()=generate-id(key('count',username)[1])]">
        <xsl:for-each select="key('count',username)">
          <User>
            <xsl:copy-of select="id" />
            <username>
              <xsl:value-of select="username" />
              <xsl:number value="position()" format="01"/>
            </username>
          </User>
        </xsl:for-each>
      </xsl:for-each>
    </Users>
  </xsl:template>
</xsl:stylesheet>

If you really need it sorted by ID afterwards, you can wrap it into a two-pass template:

<xsl:stylesheet version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
   xmlns:msxsl="urn:schemas-microsoft-com:xslt">
  <xsl:key name="count" match="User" use="username" />

  <xsl:template match="Users">
    <xsl:variable name="pass1">
      <xsl:for-each select="User[generate-id()=generate-id(key('count',username)[1])]">
        <xsl:for-each select="key('count',username)">
          <User>
            <xsl:copy-of select="id" />
            <username>
              <xsl:value-of select="username" />
              <xsl:number value="position()" format="01"/>
            </username>
          </User>
        </xsl:for-each>
      </xsl:for-each>
    </xsl:variable>

    <xsl:variable name="pass1Nodes" select="msxsl:node-set($pass1)" />

    <Users>
      <xsl:for-each select="$pass1Nodes/*">
        <xsl:sort select="id" />
        <xsl:copy-of select="." />
      </xsl:for-each>
    </Users>
  </xsl:template>
</xsl:stylesheet>



回答3:


Here's a slight variation, but possible not a great increase in efficiency

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
   <xsl:output method="xml" indent="yes"/>
   <xsl:key name="User" match="User" use="username" />

   <xsl:template match="username/text()">
      <xsl:value-of select="." />
      <xsl:variable name="id" select="generate-id(..)" />
      <xsl:for-each select="key('User', .)">
         <xsl:if test="generate-id(username) = $id">
            <xsl:number value="position()" format="01"/>
         </xsl:if>
      </xsl:for-each>
   </xsl:template>

   <xsl:template match="@*|node()">
      <xsl:copy>
         <xsl:apply-templates select="@*|node()"/>
      </xsl:copy>
   </xsl:template>
</xsl:stylesheet>

What this is doing is defining a key to group Users by username. Then, for each username element, you look through the elements in the key for that username, and output the position when you find a match.

One slight advantage of this approach is that you are only looking at user records with the same name. This may be more efficient if you don't have huge numbers of the same name.



来源:https://stackoverflow.com/questions/13727201/efficient-xslt-conditional-increment

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!