Convert varying xml to another xml

Question

I have an input XML file which looks like this:

<Root>
 <Monday>Monday<Monday>
 <Indicator>true<Indicator>
 <Value>1<Value>
 <Tuesday>Tuesday<Tuesday>
 <Indicator>true<Indicator>
 <Value>2<Value>
 <Wednesday>Wednesday<Wednesday>
 <Indicator>true<Indicator>
 <Value>3<Value>
</Root>

It must be converted to the output XML file which is:

<Root>
 <Monday>Monday<Monday>
 <Value>1<Value>
 <Tuesday>Tuesday<Tuesday>
 <Value>2<Value>
 <Wednesday>Wednesday<Wednesday>
 <Value>3<Value>
</Root>

The problem is that the input XML can vary. Sometimes it might be

<Root>
 <Monday>Monday<Monday>
 <Indicator>true<Indicator>
 <Value>1<Value>
 <Thursday>Thursday<Thursday>
 <Indicator>true<Indicator>
 <Value>4<Value>
</Root>

Now the output must be

<Root>
 <Monday>Monday<Monday>
 <Value>1<Value>
 <Thursday>Thursday<Thursday>
 <Value>4<Value>
</Root>

I also have the list of valid tags like Monday, Tuesday etc, which can come in the input XML in an ArrayList in Java. Any ideas on how to accomplish this?

The question is not clear. Where do you get you input from? Where do you want to output? The only thing you are doing, as far as I can understand is that you want to remove the `Indicator` tag — Alkis Kalogeris, Aug 24 '14 at 11:49
The order of day, indicator, value is fixed? Also, have you tried anything yourself? If not, why not? If yes, show your attempt. — Tomalak, Aug 24 '14 at 11:49

score 1 · Answer 1 · edited May 23 '17 at 12:33

1

From this answer: How to remove elements from xml using xslt with stylesheet and xsltproc? :

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

    <xsl:output omit-xml-declaration="yes"/>

    <xsl:template match="node()|@*">
       <xsl:copy>
           <xsl:apply-templates select="node()|@*"/>
       </xsl:copy>
    </xsl:template>

    <xsl:template match="Indicator"/>
</xsl:stylesheet>

edited May 23 '17 at 12:33

Community

1
1

answered Aug 24 '14 at 12:23

Marcin Szawurski

1,313
12
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If you are saying this question is a duplicate, vote to close rather than copying another answer. – Raedwald Aug 24 '14 at 12:55
@Raedwald: this is not _exactly_ duplicate. Whether it is simple to find my answer from the original one is the question of OP's XSLT knowledge. – Marcin Szawurski Aug 24 '14 at 13:01
The question is trivial, so is the answer, and this answer says all that needs to be said. – Michael Kay Aug 24 '14 at 14:01

Convert varying xml to another xml

1 Answers1