XPath 3.0 compare dates diffrence with duration literal

Question

Given XML:

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <c cid="0"> 
        <d did="c0d0" d1="2015-02-11" d2="2015-06-15" />
        <d did="c0d1" d1="2015-04-01" d2="2015-04-14" />
    </c>
    <c cid="1"> 
        <d did="c1d0" d1="2014-11-15" d2="2015-07-21" />
        <d did="c1d1" d1="2016-02-10" d2="2016-02-25" />
    </c>    
</root>

Using XPath 3.0 I have to find all c nodes, which has at least one d child node, which d2-d1 attributes values difference is less or equal 30 days. d1 and d2 values are dates in YYYY-MM-DD format. I've tried XPath:

/root/c[d/xs:date(@d2)-d/xs:date(@d1)<=30]

but I've got and error:

Cannot compare xs:dayTimeDuration to xs:integer

I use XPath Builder in XPath 3.0 mode in Oxygen XML Editor v18. I suppose that this error is due to incorrect time duration setting. Please advice how to specify time duration as literal in XPath 3.0. Thanks!

score 2 · Accepted Answer · answered Oct 07 '16 at 15:47

2

You can substract dates but then you get a dayTimeDuration and you need to compare to such a value: /root/c[d[xs:date(@d2)-xs:date(@d1) <= xs:dayTimeDuration('P30D')]].

answered Oct 07 '16 at 15:47

Martin Honnen

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And for the benefit of others, there's nothing here that needs XPath 3.0 - everything you need is in 2.0. – Michael Kay Oct 07 '16 at 19:03

XPath 3.0 compare dates diffrence with duration literal

1 Answers1