There is xslt code for version 2.0 which gives output shown below but I want to convert it to version 1.0. I referred following links.. How to compare two XML nodes and get compared result using XSLT? and xslt compare two different nodes and then combine
Input XML file:
<?xml version="1.0" encoding="utf-8"?>
<OperatorStationCollection xmlns="http://www.w3.org" >
<OperatorStation xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Name>OS001</Name>
<Nodes>
<DataNodeBase xsi:type="Adaptor">
<Family>NetworkSettings</Family>
<Name>Network A</Name>
<IPAddress>111.11.11.1</IPAddress>
</DataNodeBase>
<DataNodeBase xsi:type="Adaptor">
<Family>Network1111</Family>
<Name>Network B</Name>
<IPAddress>111.22.11.1</IPAddress>
</DataNodeBase>
<DataNodeBase xsi:type="Adaptor">
<Family>Network2222</Family>
<Name>Network C</Name>
<IPAddress>111.33.11.1</IPAddress>
</DataNodeBase>
</Nodes>
</OperatorStation>
<OperatorStation xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Name>OS002</Name>
<Nodes>
<DataNodeBase xsi:type="Adaptor">
<Family>NetworkSettings</Family>
<Name>Network A</Name>
<IPAddress>111.11.11.1</IPAddress>
</DataNodeBase>
<DataNodeBase xsi:type="Adaptor">
<Family>Network1111</Family>
<Name>Network B</Name>
<IPAddress>111.22.11.2</IPAddress>
</DataNodeBase>
<DataNodeBase xsi:type="Adaptor">
<Family>NetworkSettings</Family>
<Name>Network D</Name>
<IPAddress>111.33.11.2</IPAddress>
</DataNodeBase>
</Nodes>
</OperatorStation>
<OperatorStation xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<Name>OS003</Name>
<Nodes>
<DataNodeBase xsi:type="Adaptor">
<Family>NetworkSettings</Family>
<Name>Network A</Name>
<IPAddress>111.11.11.1</IPAddress>
</DataNodeBase>
<DataNodeBase xsi:type="Adaptor">
<Family>NetworkSettings</Family>
<Name>Network B</Name>
<IPAddress>111.22.11.3</IPAddress>
</DataNodeBase>
<DataNodeBase xsi:type="Adaptor">
<Family>NetworkSettings</Family>
<Name>Network E</Name>
<IPAddress>111.33.11.3</IPAddress>
</DataNodeBase>
</Nodes>
</OperatorStation>
</OperatorStationCollection>
XSLT code for version 2.0:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="/">
<xsl:variable name="allStations"
select="/*:OperatorStationCollection/*:OperatorStation" />
<table>
<!-- Header row - two fixed columns plus one per station name -->
<tr>
<td>Name</td><td>Status</td>
<xsl:for-each select="$allStations">
<td><xsl:value-of select="*:Name" /></td>
</xsl:for-each>
</tr>
<!-- main rows - one per "group" of DataNodeBase elements which share the
same Name -->
<xsl:for-each-group
select="$allStations/*:Nodes/*:DataNodeBase"
group-by="*:Name">
<!-- calculate the column values - the IPAddress if this network (i.e. the
current-group) has an entry for this station, and "None" if not -->
<xsl:variable name="addresses"
select="for $s in ($allStations)
return (current-group()[../.. is $s]/*:IPAddress, 'None')[1]" />
<tr>
<td><xsl:value-of select="current-grouping-key()" /></td>
<td>
<!-- equal if all the $addresses are the same, unequal otherwise -->
<xsl:value-of select="if (count(distinct-values($addresses)) = 1)
then 'Equal' else 'Unequal'" />
</td>
<xsl:for-each select="$addresses">
<td><xsl:value-of select="."/></td>
</xsl:for-each>
</tr>
</xsl:for-each-group>
</table>
</xsl:template>
</xsl:stylesheet>
Expected OUTPUT: As there is no provision to add table , I have made hatml code for result, please save this code to html file and see expected output.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title></title>
</head>
<body>
<table>
<tr>
<td>Name</td><td>Status</td><td>OS01</td><td>OS02</td><td>OS03</td>
</tr>
<tr>
<td>Network A</td><td>Equal</td><td>111.11.11.1</td><td>111.11.11.1</td><td>111.11.11.1</td>
</tr>
<tr>
<td>Network B</td><td>Unequal</td><td>111.22.11.1</td><td>111.22.11.2</td><td>111.22.11.2</td>
</tr>
<tr>
<td>Network C</td><td>Unequal</td><td>111.33.11.1</td><td>Not Exist</td><td>Not Exist</td>
</tr>
<tr>
<td>Network D</td><td>Unequal</td><td>Not Exist</td><td>111.33.11.2</td><td>Not Exist</td>
</tr>
<tr>
<td>Network E</td><td>Unequal</td><td>Not Exist</td><td>Not Exist</td><td>111.33.11.3</td>
</tr>
</table>
</body>
</html>
In the previous question it is mentioned about Muenchian Grouping, so you must be already aware this is the approach to take. In this case, you are grouping by the network name, which will form the basis of your table rows, so you define a key like this
(The w3: prefix is used here because of namespaces. All your nodes in your XML are in the namespace "http://www.w3.org" and in XSLT 1.0 you will need to explicitly declare this)
Also, you will need to check for distinct value of IP Address for each network later, so it is also worth defining a key to help doing this
Note the
|
symbol here. It can be any character you choose, just as long as it doesn't occur in either Name or IPAddress)For getting the columns, which are your OperatorStation elements, the code is very much as before (with the main difference being about the namespace prefix)
Outputting the columns is then much as before too
The first main difference though is when you want to get the distinct networks, for your columns. This is where you use Muenchian Grouping
Within the template that matches this, you can check for whether all IP addresses are the same by using the second key
Outputing the IP addresses for the row, is then simply a case of looping over the stations, and outputting the one with the matching network
Note that the following expression would also work in this case
Anyway, try this XSLT
EDIT: If you wish to restrict the entries to a certain family, in this case just add a condition to check the family to the two keys