如何在xquery中选择节点的属性值？ [英] how to select attribute value of a node in xquery ?

问题描述

以下xml：

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>

我需要获取客户的否属性。在XPath中，我知道它是 / company / customers / customer / @ cno ，在XQuery中，我尝试过以下表达但是对我没有用。

< b> for $ c in / company / customers / customer

返回$ c / @ cno



任何人都可以提供帮助我在这方面得到了这个。

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

Any one to help me to get this in XQuery in this regard.

推荐答案

c in / company / customers / customer

return

c in /company/customers/customer
return

c / @ cno



任何人都可以帮助我在这方面得到这个。

c/@cno

Any one to help me to get this in XQuery in this regard.

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