I have a requirement to specify wild card in the following xpath
Field[#name="/Root/Table[i]/FirstName"]
Basically the "i" would be a variable which can have either a GUID or a running number. I would like to pick up all elements that basically have the attribute pattern
"/Root/Table[*]/FirstName"
i.e. starting with "/Root/Table[" and ending with "]/FirstName". Any ideas as to how this can be done ?
Here is a sample payload:
<Package>
<Input>
<Data id="36e9f0fe3f8d4508ac20710e07cfddd4">
<Input>
<Field name="/Root/Table[1]/FirstName">Thomas</Field>
</Input>
</Data>
</Input>
</Package>
You should be able to do this using starts-with() and a makeshift ends-with() (since XPath 1.0 doesn't actually have an ends-with() function):
//*[starts-with(#name, '/Root/Table[') and
substring(#name, string-length(#name) - 11 + 1) = ']/FirstName']
Here, 11 is the length of ]/FirstName.
Related
I'm trying to get with xPath the position only of the first element which has the attribute value true.
<?xml version="1.0" encoding="UTF-8"?>
<elements>
<element attribute="false"/>
<element attribute="true"/>
<element attribute="true"/>
</elements>
What I have so fare is:
head(/elements/element[#attribute='true']/position())
Result:
1
But it should be:
2
What am I doing wrong?
position() returns the position of the element in the nodelist created by the predicate, i.e. with the false excluded. Instead of position, you can e.g. count the number of preceding elemements.
For example, this works even in XPath 1.0:
1+count(/elements/element[#attribute="true"][1]/preceding-sibling::element)
I think it's (with XPath 3):
head(index-of(/elements/element/#attribute, 'true'))
saxon-lint --xpath 'count(//element[#attribute="true"]/position())' file.xml
From Michael answer:
saxon-lint --xpath 'head(index-of(/elements/element/#attribute, "true"))' file.xml
Output
2
Assume the following XML:
<data>
<node id="1" />
<node id="2" />
<node id="12" />
<node id="16" />
</data>
This xpath expression should be valid:
count(//node)
.. and should produce the number 4
I'm new to robot frameworks. Is it possible to use this xpath in robot framework?
for example something like:
${value}= Get something something source=${xml} xpath=count(//node)
The one below works but I would like the xpath to produce the end value, not a list.
#{nodelist}= Get Elements ${xml} xpath=node
Length Should Be ${nodelist} 4
Edit
I know that I can count the nodes in a list of nodes. However, I would like to get the absolute value (integer or string) using xpath. Now I need to write different code depending on if the xpath result is a node, list or attribute when the xpath could theoretically produce the final value.
You can use the Get Element Count Keyword it returns the number of elements matching the locator
You can do something as simple as this
${count} = Get Element Count name:div_name
Should Be True ${count} > 2
For more info on Keywords Have a look at this Keyword Page
When working with XML it is generally best to use the XML library. In the below example you'll find a solution for counting the elements using the XML library Get Element Count.
data.xml
<data>
<node id="1" />
<node id="2" />
<node id="12" />
<node id="16" />
</data>
Testcase.robot
*** Settings ***
Library XML
Library OperatingSystem
*** Test Cases ***
TC
${xml} Get File ./data.xml
${count} Get Element Count ${xml} xpath=node
Should Be Equal As Integers ${count} ${4}
I have an xpath-expression like this:
element[#attr="a"] | element[#attr="b"] | element[#attr="c"] | … which is an »or« statement. So can I create an expression that guarantees the result to appear in the order as in the query, even if the elements appear in a different order in the document?
f.e. an document fragment in this order:
<doc>
<element attr="c" />
<element attr="b" />
<element attr="a" />
.
.
.
</doc>
and a result list ordered like this:
[0] <element attr="a" />
[1] <element attr="b" />
[2] <element attr="c" />
.
.
.
The | operator computes the union of its operands and with XPath 1.0 you simply get a set of nodes, the order is undefined, though most XPath APIs then return the result in document order or allow you to say which order you want or whether order matters (see for instance http://www.w3.org/TR/DOM-Level-3-XPath/xpath.html#XPathResult).
With XPath 2.0 you get a sequence of nodes ordered in document order, with XPath 2.0 if you want the order of your subexpressions you would need to use the comma operator, not the union operator i.e. element[#attr="a"] , element[#attr="b"] , element[#attr="c"].
can I create an expression that guarantees the result to appear in the
order as in the query, even if the elements appear in a different
order in the document?
Not with any XPath 1.0 engine -- they return the resulting XmlNodeList in document order.
With XPath 2.0 one can specify that a sequence is to be returned, using the comma , operator, like this:
element[#attr="a"] , element[#attr="b"] , element[#attr="c"]
Finally, If you are limited with an XPath 1.0 implementation, one way of getting the results in the desired order is to evaluate these three XPath expressions:
element[#attr="a"]
element[#attr="b"]
element[#attr="c"]
Then you can access the first result first, the second result -- second and the third result -- third.
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.
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?