用于解析XML属性的XPath

Question

Can someone please tell me the "correct / most-efficient way" of getting the "Status" XML attributes (ID, CssClass, Description, and IsActive for the XML below:

Implementation will be Java - (but I am more interested in the XPath):

<ArrayOfLineStatus>
<LineStatus ID="0" StatusDetails="">
    <BranchDisruptions/><Line ID="1" Name="Bakerloo"/>
    <Status ID="GS" CssClass="GoodService" Description="Good Service" IsActive="true">          
    <StatusType ID="1" Description="Line"/></Status></LineStatus>
[snip]
 </ArrayOfLineStatus>

Thank,

Miles.

Answer 1

@* selects all attributes of the context node. Use:

/*/*/Status/@*

Or, more specifically:

/ArrayOfLineStatus/LineStatus/Status/@*

Or, for Status elements appearing anywhere in the document:

//Status/@*