How To Get Certain Parent Element Node When Child Condition Found

Question

I'm having problems writing a C# script that'll get me a certain parent element of a given child element. To clear things up, this is an example of the XML tree.

<parent attribute='X'>
  <child_element1>A</child_element1>
  <child_element2 attribute='Y'>
    <grandchild_element1>B</grandchild_element1>
    <grandchild_element2>
      <key>C</key>
    </grandchild_element2>
    <grandchild_element3>D</grandchild_element3>
  </child_element2>
  <child_element3>E</child_element3>
</parent>

Traversing the tree, my script has found the <key> node. What I'm trying to produce is the following tree.

<parent attribute='X'>
  <child_element2 attribute='Y'>
    <grandchild_element1>B</grandchild_element1>
    <grandchild_element2>
      <key>C</key>
    </grandchild_element2>
    <grandchild_element3>D</grandchild_element3>
  </child_element2>
</parent>

As you can see, I'm trying to remove child_element1 and child_element3 in original tree.

I've written a pseudo-code for the script as follows, but so far have failed to figure out how to script the real thing in C#.

Show all the first and second parent elements above of the key node. grandchild_element2 is first parent of key and child_element2 is second parent of key. So all their elements are shown.
Show only node/element that has immediate relation with key for next parent level. parent is third parent of key. The parent element that has direct relation with key is child_element2. So only child_element2 is shown. child_Element1 and child_element2 are not shown because they do not have direct relation with the key

I've taken a look at the following solutions on stackoverflow, but unfortunately they all only show how to get all parent elements (instead of showing the ones which are ancestors of the <key> node.

`` and `` never have the same name? This is what makes this one tricky... You would have to carry a list, which describes until which node we should go up... — Jan P., Jan 29 '13 at 10:37
Have you got any code you've already setup that we can work with? — Jon Egerton, Jan 29 '13 at 10:37
Its easy enough to isolate the `key` but eliminating the siblings of parents without using brute force might be interesting. — Jon Egerton, Jan 29 '13 at 10:38
So why would `grandchild_element1` and `grandchild_element3` be included? — JLRishe, Jan 30 '13 at 04:42
@JLRishe Perhaps I was not being clear. The first XML is the original/source, the second one is the result I want — user2020692, Jan 30 '13 at 04:58
The second one has `grandchild_element1` and `grandchild_element3` in it. — JLRishe, Jan 30 '13 at 05:21
Actually grandchild_element2 contains the key, and other grandchild_elements contain the data. And this requirement/format comes from third party. Sometimes the XML data need to be corrected. The third party will give the keys, so I can correct the data based by the keys. In the example, third party give me the C key. So i have to correct the data of grandchild_element1 or grandchild_element2 or both, and send the XML back like second XML — user2020692, Jan 30 '13 at 07:17
Ok, well that seems to contradict your described requirements about only taking the nodes that are parents of the key, or parents of those parents and so on. Is there something in your described requirements that I'm misunderstanding, or are your requirements incomplete? Either way, could you please clarify why `grandchild_element1` and `grandchild_element3` should be included, and not `child_element1` or `child_element3`, because I don't think that is clear. — JLRishe, Jan 30 '13 at 12:25

JLRishe · Answer 1 · 2013-01-31T17:11:17.033

1

This is quite easy if you use an XSLT:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:param name="valueToFind" select="'C'" omit-xml-declaration="yes" />

  <xsl:template match="/*">
    <xsl:apply-templates select="node()[descendant-or-self::node() = $valueToFind]" />
  </xsl:template>

  <xsl:template match="*">
    <xsl:copy>
      <xsl:copy-of select="@*"/>
      <xsl:apply-templates 
            select="node()[descendant-or-self::node() = $valueToFind]" />
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

When the example XML is given a single root element and fed into this XSLT, the result is:

<a>
  <a.1 name="First">
    <a.1.2>
      <a.1.2.2>
        <key>C</key>
      </a.1.2.2>
    </a.1.2>
  </a.1>
</a>

In case you're unfamiliar with XSLT, an XSLT parameter:

<xsl:param name="valueToFind" select="'C'" />

is essentially a variable whose value you can pass in from the executing code. Here, I've just given it it the default value C, so it can be run in an XSLT debugger and produce a demonstrable result.

edited Jan 31 '13 at 17:11

answered Jan 29 '13 at 11:08

JLRishe

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LINQ query will return parents with all other elements – Sergey Berezovskiy Jan 29 '13 at 11:45
@lazyberezovsky Yes, it certainly will. That LINQ will obtain a list of the sought value's parent elements, and those will possess all of the children they have in the source XML. But if all the asker wants is a list of the parent elements of the sought value, then that may be sufficient. If the asker's true objective (since it's not 100% clear from the question) is to obtain an XML structure with all of the other stuff omitted, I suggest that s/he use my XSLT solution since it's a lot more efficient and clean than (as far as I know) anything that could be written in C#. – JLRishe Jan 29 '13 at 11:48
hi JLRishe and lazyberezovsky, i change the example. maybe it can make 100% clear :) – user2020692 Jan 30 '13 at 04:26

How To Get Certain Parent Element Node When Child Condition Found

1 Answers1