how to select attribute value of a node in XQuery?

Question

In below XML:

<company>
    <customers>
    <customer cno="2222">
            <cname>Charles</cname>
            <street>123 Main St.</street>
            <city>Wichita</city>
            <zip>67226</zip>
            <phone>316-636-5555</phone>
        </customer>
        <customer cno="1000">
            <cname>Bismita</cname>
            <street>Ashford Dunwoody</street>
            <city>Wichita</city>
            <zip>67226-1555</zip>
            <phone>000-000-0000</phone>
        </customer>     
    </customers>
</company>

I need to get the customer's no which is an attribute. In XPath I know it is /company/customers/customer/@cno , in XQuery I've tried below expression but didn't work for me:

for $c in /company/customers/customer
return $c/@cno

Answer 1

You should use data to pick attribute value:-

for $c in /company/customers/customer
return data($c/@cno)

Answer 2

You can also use string to get attribute value:

for $c in /company/customers/customer
    return $c/@cno/string()