xml 如何在XQuery中选择节点的属性值？
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how to select attribute value of a node in XQuery?
提问by Itz.Irshad
In below XML:
在下面的 XML 中：
<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:
我需要得到客户的否，这是一个属性。在 XPath 中，我知道它是/company/customers/customer/@cno，在 XQuery 中，我尝试了以下表达式，但对我不起作用：
for $c in /company/customers/customer
return $c/@cno
回答by Navin Rawat
You should use data to pick attribute value:-
您应该使用数据来选择属性值：-
for $c in /company/customers/customer
return data($c/@cno)
回答by Nikunj Vekariya
You can also use stringto get attribute value:
您还可以使用string来获取属性值：
for $c in /company/customers/customer
return $c/@cno/string()
