[xsl] Find the number of elements that are prior to the series of elemen

Hi Folks,

I have an XML document containing a long list of Byte elements, e.g.,

<Bytes>
    <Byte hexAddress="0">4D</Byte>
    <Byte hexAddress="1">5A</Byte>
    <Byte hexAddress="2">90</Byte>
    <Byte hexAddress="3">00</Byte>
    <Byte hexAddress="4">03</Byte>
    <Byte hexAddress="5">00</Byte>
    <Byte hexAddress="6">00</Byte>
    <Byte hexAddress="7">00</Byte>
    <Byte hexAddress="8">04</Byte>
    <Byte hexAddress="9">00</Byte>
    <Byte hexAddress="A">00</Byte>
    <Byte hexAddress="B">00</Byte>
    <Byte hexAddress="C">FF</Byte>
    <Byte hexAddress="D">FF</Byte>
    ...
</Bytes>

I have a string of hex values, e.g.,

04000000FF

You can see that the string is contained within this series of Byte elements:

    <Byte hexAddress="8">04</Byte>
    <Byte hexAddress="9">00</Byte>
    <Byte hexAddress="A">00</Byte>
    <Byte hexAddress="B">00</Byte>
    <Byte hexAddress="C">FF</Byte>

I want to find the number of <Byte> elements that occur prior to the series of 
Byte elements.

I figured one approach is to convert all the Byte text node values into a 
string and then perform string manipulations:

string-length(substring-before(string-join(/Bytes/Byte/text(), ''), 
'04000000FF')) idiv 2

Note: I divided by two because each Byte has 2 digits.

That seems to work, but I suspect it is horribly inefficient because it creates 
a string containing all the Byte text node values and there may be tens or 
hundreds of thousands of Byte elements. 

What is an efficient way to solve this problem?

/Roger



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