How to find the max attribute from an XML document using Xpath 1.0

In XPath 2.0, use the max function. To find the book with the highest id, do

/library/book[@id = max(/library/book/@id)]

The following XPath selects the book with highest id:

/library/book[not(@id <= preceding-sibling::book/@id) and not(@id <=following-sibling::book/@id)]

Note: The following information assumes use of XPath 1.0.

The following expression returns the element(s) with the largest id value:

/*/book[not(@id < preceding-sibling::book/@id) and 
        not(@id < following-sibling::book/@id)]

Note that this is slightly different than @timbooo's answer in that this will return more than one element when there are duplicates with the same max value (@timbooo's would return none). If you want only one element in this case, then you need a resolution strategy. To choose the first such element in document order, use this:

/*/book[not(@id < preceding-sibling::book/@id) and 
        not(@id < following-sibling::book/@id)][1]

To choose the last one, use this:

/*/book[not(@id < preceding-sibling::book/@id) and 
        not(@id < following-sibling::book/@id)][last()]

This approach is very inefficient (O(n^2)) because it requires you to compare each element to every other potential max. For this reason, it's probably best to use your host programming language to select the maximum element. Simply select all of the book elements first and then choose the max from that list. This is (most likely) a linear operation (O(n)), which would be noticeably faster on very large documents. For example, in Java (JAXP) you might do it like this:

XPath xpath = XPathFactory.newInstance().newXPath();
NodeList nodes = (NodeList) xpath.evaluate("/*/book", doc,
        XPathConstants.NODESET);
Node max = nodes.item(0);
for (int i = 0; i < nodes.getLength(); i++) {
    int maxval = Integer.parseInt(max.getAttributes()
            .getNamedItem("id").getNodeValue());
    int curval = Integer.parseInt(nodes.item(i).getAttributes()
            .getNamedItem("id").getNodeValue());
    if (curval >= maxval)
        max = nodes.item(i);
}
System.out.println(max.getAttributes().getNamedItem("name"));

Note that this is just a demonstration; be sure to include null-checks where appropriate.

I've found that answers like the lwburk's or timbooo's work fine for attributes representing numbers having just one digit. However, if the attribute is a number having more than one digit, extrange things seem to happen when comparing between attributes' values. For example, try changing the original XML data with something like this:

<?xml version="1.0" encoding="utf-8"?>
<library>
        <book id="250" name="Dragon Tatoo"/>
        <book id="700123" name="Ender's Game"/>
        <book id="305" name="Catch 22"/>
        <book id="1070" name="Lord of the rings"/>
</library>

Running the suggested snippets won't work. I got a solution using the casting operator xs:int() applied on id attribute, like in:

/library/book[not(xs:int(@id) <= preceding-sibling::book/@id) and not(xs:int(@id) <=following-sibling::book/@id)]

That will give the correct answer!

If you're willing to use external tooling - which depends on your implementation featuring implementations of these tools - try the EXSLT:Math function highest().

The fact that EXSLT implements this implies that such a feature isn't directly available in plain xpath, of course. If you're not using Transforms, or want to stick purely with standards-compliant markup, other posters' suggestions would be a better choice.

XPath 1.0

/library/book[not(@id < /library/book/@id)]

This query style is more generic and works even if books are grouped i.e.

<?xml version="1.0" encoding="utf-8"?>
<library>
    <genre id="1">
        <book id="2" name="Dragon Tatoo"/>
        <book id="7" name="Ender's Game"/>
    </genre>
    <genre id="2">
        <book id="3" name="Catch 22"/>
        <book id="1" name="Lord of the rings"/>
    </genre>
</library>

Same query still works (the path should be modified)

/library/genre/book[not(@id < /library/genre/book/@id)]

or even

//book[not(@id < //book/@id)]

To avoid performance troubles use XPath 2 max() instead

This example can be used to find the max.

XmlDocument doc = new XmlDocument();                    
doc.Load("../../Employees.xml");
XmlNode node = doc.SelectSingleNode("//Employees/Employee/@Id[not(. <=../preceding-sibling::Employee/@id) and not(. <=../following-sibling::Employee/@Id)]");
int maxId = Convert.ToInt32(node.Value);

For other similar topics on xpath and linq check out http://rmanimaran.wordpress.com/2011/03/20/xml-find-max-and-min-value-in-a-attribute-using-xpath-and-linq/