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    <blockquote> <p>When the feed comes in, I want to filter <strong>each xml file</strong> by removing the nodes that are <strong>not in my array of xpath's</strong></p> </blockquote> <p><strong>Step1. Select all nodes that aren't selected by the given XPath expressions</strong></p> <p><strong>I guess that by "nodes" you mean elements. If so, this XPath expression</strong>:</p> <pre><code>//*[count(. | yourExpr1 | yourExpr2 ... | yourExprN) &gt; count(yourExpr1 | yourExpr2 ... | yourExprN) ] </code></pre> <p><strong>selects all elements in the XML document that aren't selected by any of your N XPath expressions</strong> <code>yourExpr1</code>, <code>yourExpr2</code>, ... , <code>yourExprN</code></p> <p>If by "nodes" you mean elements, text-nodes, processing-instruction-nodes (PIs), comment-nodes and attribute nodes, use this XPath expression to select all nodes not selected by your N XPath expressions:</p> <pre><code>(//node() | //*/@*) [count(. | yourExpr1 | yourExpr2 ... | yourExprN) &gt; count(yourExpr1 | yourExpr2 ... | yourExprN) ] </code></pre> <p><strong>Step2. Delete all nodes selected in Step1.</strong></p> <p>For each of the nodes selected in Step1 above, use: </p> <pre><code> node.ParentNode.RemoveChild(node); </code></pre> <p><strong>Explanation</strong>:</p> <ol> <li><p>The XPath union operator <code>|</code> produces the union of two node-sets. Therefore the expression <code>yourExpr1 | yourExpr2 ... | yourExprN</code> when applied on the XML document produces the set of all nodes that are selected by any of the N given XPath expressions.</p></li> <li><p>A node <code>$n</code> doesn't belong to a set of nodes <code>$ns</code> exactly when ...</p> <p><code>count($n | $ns) &gt; count($ns)</code></p></li> </ol>
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    1. POHow to remove xml nodes that are not in an array of xpath strings?
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    1. USDimitre Novatchev
    1. USDimitre Novatchev
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    1. POHow to remove xml nodes that are not in an array of xpath strings?
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    1. VO
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    1. COThis is very helpful! thanks!
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      1. USLaguna
    2. CO@Lagoona: You are welcome.
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      1. USDimitre Novatchev
    3. COi wish i could upvote twice for this answer!
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