XPath 1 query and attributes name - xpath

First question: is there any way to get the name of a node's attributes?
<node attribute1="value1" attribute2="value2" />
Second question: is there a way to get attributes and values as value pairs? The situation is the following:
<node attribute1="10" attribute2="0" />
I want to get all attributes where value>0 and this way: "attribute1=10".

First question: is there any way to
get the name of a node's attributes?
<node attribute1="value1"
attribute2="value2" />
Yes:
This XPath expression (when node is the context (current) node)):
name(#*[1])
produces the name of the first attribute (the ordering may be implementation - dependent)
and this XPath expression (when node is the context (current) node)):
name(#*[2])
produces the name of the second attribute (the ordering may be implementation - dependent).
Second question: is there a way to get
attributes and values as value pairs?
The situation is the following:
<node attribute1="10" attribute2="0"
/>
I want to get all attributes where
value>0 and this way: "attribute1=10".
This XPath expression (when the attribute named "attribute1" is the context (current) node)):
concat(name(), '=', .)
produces the string:
attribute1=value1
and this XPath expression (when the node node is the context (current) node)):
#*[. > 0]
selects all attributes of the context node, whose value is a number, greater than 0.
In XPath 2.0 one can combine them in a single XPath expression:
#*[number(.) > 0]/concat(name(.),'=',.)
to get (in this particular case) this result:
attribute1=10
If you are using XPath 1.0, which is less powerful, you'll need to embed the XPath expression in a hosting language, such as XSLT. The following XSLT 1.0 thransformation :
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:template match="/*">
<xsl:for-each select="#*[number(.) > 0]">
<xsl:value-of select="concat(name(.),'=',.)"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
when applied on this XML document:
<node attribute1="10" attribute2="0" />
Produces exactly the same result:
attribute1=10

It depends a little bit on the context, I believe. In most cases, I expect you'd have to query "#*", enumerate over the items, and call "name()" - but it may work in some tests.
Re the edit - you can do:
#*[number(.)>0]
to find attributes matching your criteria, and:
concat(name(),'=',.)
to display the output. I don't think you can do both at once, though. What is the context here? xslt? what?

Related

Split methods on XPath 1.0

I use 'XPath', how I can simulate split method?
I read documentation and I know that XPath version 1.0 not have this method.
I have document contains this tags:
<TestCategoryModule>
<ItemCategories>
<![CDATA[Birthday Travel,Travel]]>
</ItemCategories>
</TestCategoryModule>
<TestCategoryModule2>
<ItemCategories>
<![CDATA[Travel]]>
</ItemCategories>
</TestCategoryModule2>
I want filter item by 'ItemCategories', but when I filtered by world 'Travel', return 2 item. I use this filter "ItemCategories[contains(text(), 'Travel')]".
I want that I filter by "Travel" return only second item. How can do it?
Use:
/*/*/*[contains(concat(',', ., ','), ',Travel,')]
Here is XSLT-based verification:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/">
<xsl:copy-of select=
"/*/*/*[contains(concat(',', ., ','), ',Travel,')]"/>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on this XML document (essentially the provided XML fragment, extended with one more test case and made a well-formed XML document:
<t>
<TestCategoryModule>
<ItemCategories>Birthday Travel,Travel</ItemCategories>
</TestCategoryModule>
<TestCategoryModule2>
<ItemCategories>Birthday Travel</ItemCategories>
</TestCategoryModule2>
<TestCategoryModule2>
<ItemCategories>Travel</ItemCategories>
</TestCategoryModule2>
</t>
The wanted, correct result is produced:
<ItemCategories>Birthday Travel,Travel</ItemCategories>
<ItemCategories>Travel</ItemCategories>
I was a little wrong, or poorly described problumu. The problem is that the categories are stored as a string. I have three items, the first one contains categories: (Birthday Travel,Travel), second: (Birthday Travel), third: (Travel). When I request filtering for the word "Travel", I need to get the first and third items, but I get all three items, because all items contain world "Travel".
You actually don't need split() for the problem that you've described. If you want to match Travel but not Travel,Travel you want = instead of contains(). To deal with the whitespace around your CDATA sections, wrap it in normalize-space().
All put together, try ItemCategories[normalize-space(text()) = 'Travel'].

Self axis in xslt

<element>
<bye>do not delete me</bye>
<hello>do not delete me</hello>
<hello>delete me</hello>
<hello>delete me</hello>
</element>
Applied to the above xml, this deletes all the nodes except the first hello child of /element:
<xsl:template match="hello[not(current() = parent::element/hello[1])]" />
Why these ones doesn't work? (assuming the first node is not a text node)
<xsl:template match="hello[not(self::hello/position() = 1)]" />
<xsl:template match="hello[not(./position() = 1)]" />
Or this one?
<xsl:template match="hello[not(self::hello[1])]" />
What is the self axis selecting? Why isn't this last example equivalent to not(hello[1])?
First, you are wrong when you say that:
This deletes all the nodes except the first hello child of /element
The truth is that it deletes (if that's the correct word) any hello child of /element whose value is not the same as the value of the first one of these. For example, given:
XML
<element>
<hello>a</hello>
<hello>b</hello>
<hello>c</hello>
<hello>a</hello>
</element>
the template:
<xsl:template match="hello[not(current() = parent::element/hello[1])]" />
will match the second and the third hello nodes - but not the first or the fourth.
Now, with regard to your question: in XSLT 1.0, position() is not a valid location step - so this:
<xsl:template match="hello[not(self::hello/position() = 1)]" />
should return an error.
In XSLT 2.0, the pattern hello[not(self::hello/position() = 1)] will not match any hello element - because there is only one node on the self axis, and therefore its position is always 1.
Similarly:
<xsl:template match="hello[not(./position() = 1)]" />
is invalid in XSLT 1.0.
In XSLT 2.0, ./position() will always return 1 for the same reason as before: . is short for self::node() and there is only one such node.
Finally, this template:
<xsl:template match="hello[not(self::hello[1])]" />
is looking for a node that doesn't have (the first instance of) itself. Of course, no such node can exist.
Using position() on the RHS of the "/" operator is never useful -- and in XSLT 1.0, which is the tag on your question, it's not actually permitted.
In XSLT 2.0, the result of the expression X/position() is a sequence of integers 1..count(X). If the LHS is a singleton, like self::E, then count(X) is one so the result is a single integer 1.

Is there any method to get any type of sibling of a particular node in Xpath 2.0

Is there any method to get any type of sibling of a particular node in Xpath 2.0
The axes "following-sibling" only supports for the same type of siblings.
Ex:
<node>
<b name="bold">abc</b>
<div>gef</div>
</node>
I want to select all the sibling of the <b name="bold">.
Is there any method to get any type of sibling of a particular node in Xpath 2.0
The axes following-sibling only supports for the same type of siblings.
Use:
following-sibling::node()
this select all siblings nodes of any type -- elements, text-nodes, processing-instruction nodes and comment nodes.
Here is a complete XSLT - based verification:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/">
<xsl:for-each select="/*/b[#name='bold']/following-sibling::node()">
"<xsl:copy-of select="."/>"
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the provided XML document:
<node>
<b name="bold">abc</b>
<div>gef</div>
</node>
the XPath expression is applied (off the wanted element) and all selected three nodes are copied to the output:
"
"
"<div>gef</div>"
"
"
As we can see, all sibling nodes are selected -- a whitespace-only text node, a div element and another whitespace-only text node.
Do note: This is an XPath 1.0 expression and I don't believe XPath 2.0 adds any new feature for selecting siblings than what is already in XPath 1.0.
In case by "sibling" you mean something different than the meaning of "sibling" in XPath, then you must define precisely what you mean.
Not sure I understand the question, but how about:
//*[preceding-sibling::b]
That will get all previous siblings of the <b name="bold">abc</b> element. The * selects any type of element.
If you want all siblings:
//*[preceding-sibling::b or following-sibling::b]
And if you want to be more specific in how you select the b element:
//*[preceding-sibling::b[#name="bold"]]

Sorting XPath results in the same order as multiple select parameters

I have an XML document as follows:
<objects>
<object uid="0" />
<object uid="1" />
<object uid="2" />
</objects>
I can select multiple elements using the following query:
doc.xpath("//object[#uid=2 or #uid=0 or #uid=1]")
But this returns the elements in the same order they're declared in the XML document (uid=0, uid=1, uid=2) and I want the results in the same order as I perform the XPath query (uid=2, uid=0, uid=1).
I'm unsure if this is possible with XPath alone, and have looked into XSLT sorting, but I haven't found an example that explains how I could achieve this.
I'm working in Ruby with the Nokogiri library.
There is no way in XPath 1.0 to specify the order of the selected nodes.
XPath 2.0 allows a sequence of nodes with any specific order:
//object[#uid=2], //object[#uid=1]
evaluates to a sequence in which all object items with #uid=2 precede all object items with #uid=1
If one doesn't have anXPath 2.0 engine available, it is still possible to use XSLT in order to output nodes in any desired order.
In this specific case the sequence of the following XSLT instructions:
<xsl:copy-of select="//object[#uid=2]"/>
<xsl:copy-of select="//object[#uid=1]"/>
produces the desired output:
<object uid="2" /><object uid="1" />
I am assuming you are using XPath 1.0. The W3C spec says:
The primary syntactic construct in XPath is the expression. An expression matches the production Expr. An expression is evaluated to yield an object, which has one of the following four basic types:
* node-set (an unordered collection of nodes without duplicates)
* boolean (true or false)
* number (a floating-point number)
* string (a sequence of UCS characters)
So I don't think you can re-order simply using XPath. (The rest of the spec defines document order and reverse document order, so if the latter does what you want you can get it using the appropriate axis (e.g. preceding).
In XSLT you can use <xsl:sort> using the name() of the attribute. The XSLT FAQ is very good and you should find an answer there.
An XSLT example:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:param name="pSequence" select="'2 1'"/>
<xsl:template match="objects">
<xsl:for-each select="object[contains(concat(' ',$pSequence,' '),
concat(' ',#uid,' '))]">
<xsl:sort select="substring-before(concat(' ',$pSequence,' '),
concat(' ',#uid,' '))"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Output:
<object uid="2" /><object uid="1" />
I don't think there is a way to do it in xpath but if you wish to switch to XSLT you can use the xsl:sort tag:
<xsl:for-each select="//object[#uid=1 or #uid=2]">
<xsl:sort: select="#uid" data-type="number" />
{insert new logic here}
</xsl:for-each>
more complete info here:
http://www.w3schools.com/xsl/el_sort.asp
This is how I'd do it in Nokogiri:
require 'nokogiri'
xml = '<objects><object uid="0" /><object uid="1" /><object uid="2" /></objects>'
doc = Nokogiri::XML(xml)
objects_by_uid = doc.search('//object[#uid="2" or #uid="1"]').sort_by { |n| n['uid'].to_i }.reverse
puts objects_by_uid
Running that outputs:
<object uid="2"/>
<object uid="1"/>
An alternative to the search would be:
objects_by_uid = doc.search('//object[#uid="2" or #uid="1"]').sort { |a,b| b['uid'].to_i <=> a['uid'].to_i }
if you don't like using sort_by with the reverse.
XPath is useful for locating and retrieving the nodes but often the filtering we want to do gets too convoluted in the accessor so I let the language do it, whether it's Ruby, Perl or Python. Where I put the filtering logic is based on how big the XML data set is and whether there are a lot of different uid values I'll want to grab. Sometimes letting the XPath engine do the heavy lifting makes sense, other times its easier to let XPath grab all the object nodes and filter in the calling language.

Select top 10 events from wevtutil using xpath

I am currently working on a project that uses the Windows event log. I am using wevtutil to get the results from the event logs. I know that wevtutil supports xpath queries, but since I'm new to xpath I don't know that I can achieve what I'm trying to do.
In SQL, what I would be doing is something like this:
SELECT log.*, COUNT(1) numHits
FROM Application log
GROUP BY Source, Task, Level, Description
ORDER BY numHits DESC
LIMIT 10
Is it possible to do such a thing using xpath?
Edit: Here is a sample Event:
<Event xmlns='http://schemas.microsoft.com/win/2004/08/events/event'>
<System>
<Provider Name='MSSQL$SQLEXPRESS' />
<EventID Qualifiers='16384'>17403</EventID>
<Level>4</Level>
<Task>2</Task>
<Keywords>0x80000000000000</Keywords>
<TimeCreated SystemTime='2010-10-20T20:06:18.000Z' />
<EventRecordID>9448</EventRecordID>
<Channel>Application</Channel>
<Computer>SHAZTOP</Computer>
<Security />
</System>
<EventData>
<Data>73094</Data>
<Binary>
FB4300000A000000130000005300480041005A0054004F0050005C00530051004C004500580050005200450053005300000000000000</Binary>
</EventData>
</Event>
XPath 1.0 has four data types: string, number, boolean and node set.
The only XPath ordering criteria is document order (in the given axis direction). That is how you can limit any result node set as #Dimitre and #Welbog have sugested with fn:position().
But, there is no specification that an XPath engine must provide a node set result in any given order. So, you can't sort nor grouping in XPath 1.0. You can select the firsts of each group, but not efficiently. As example:
//Event[not(System/Level = preceding::Level) or
not(System/Task = preceding::Task)]
XPath 2.0 has the sequence data type. A sequence has the exclicit order of construction. So, you can group. As example:
for $event (//Event)[index-of(//Event/System/concat(Level,'++',Task),
System/concat(Level,'++',Task))[1]]
result //Event[System/Level = $event/System/Level]
[System/Task = $event/System/Task]
But, because XPath 2.0 has not built-in sorting nor recursion mechanism (you could provide an extension function...) you can't sort.
For that you need a language with built-in sorting or a way to express its algorithm. Both XSLT (1.0 or 2.0) and XQuery have these features.
In SQL, what I would be doing is
something like this:
SELECT log.*, COUNT(1) numHits
FROM Application log
GROUP BY Source, Task, Level, Description
ORDER BY numHits DESC
LIMIT 10
Is it possible to do such a thing
using xpath?
In case no sorting is necessary, one can get the first $n nodes selected by any XPath expression by:
(ExpressionSelectingNodeSet)[not(position() > $n)]
where $n can be substituted by a specific number
If there is a requirement that the nodes be sorted on one or more sort-keys, then this is not possible pure XPath, but one can easily perform such tasks with XSLT, using the <xsl:sort> instruction and the XPath position() function:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/*">
<nums>
<xsl:for-each select="num">
<xsl:sort data-type="number" order="descending"/>
<xsl:if test="not(position() > 5)">
<xsl:copy-of select="."/>
</xsl:if>
</xsl:for-each>
</nums>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the following XML document:
<nums>
<num>01</num>
<num>02</num>
<num>03</num>
<num>04</num>
<num>05</num>
<num>06</num>
<num>07</num>
<num>08</num>
<num>09</num>
<num>010</num>
</nums>
the correct result, containing only the top 5 numbers is produced:
<nums>
<num>010</num>
<num>09</num>
<num>08</num>
<num>07</num>
<num>06</num>
</nums>
You can use the position() function to limit the results you're getting:
/root/element[position()<=10]
For example, that would select the first ten element elements which are children of the root.
If your structure is more complicated, you can use the position element in different places. For example, if the element element can exist in more than one parent, but you want the first ten of them regardless of parent, you can do it this way:
(/root/parent1/element | /root/parent2/element)[position()<=10]

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