本文介绍了通过XElement的属性值获取的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有以下XML:
<rootNode>
... some stuff
<ReportCellRef>
<dang n="DVCompany" h="0" u="0" o="0" fmt="0">
... some stuff
</dang>
</ReportCellRef>
</rootNode>
我想得到< dang ...>...</dang>
节点作为XElement,因此只要具有 n
属性的值,我就可以将其替换为另一个节点.
And I want get the <dang ...> ... </dang>
node as XElement, so I can replace it with another node, provided I have the value of the n
attribute.
我有以下代码:
Dim nameToSearch = importNode.Attribute("n").Value
Dim replaceable = From dangToTake In xdoc.Elements("ReportCellRef") _
Where CStr(dangToTake.Element("dang").Attribute("n")) = nameToSearch
Select dangToTake
For Each nodeToReplace As XElement In replaceable
nodeToReplace.ReplaceWith(importNode)
Next nodeToReplace
但是LINQ查询不会产生任何结果...有什么想法吗?
But the LINQ query does not yield any results... Any ideas?
推荐答案
在其中抛出"Descendants()"调用:
Throw a "Descendants()" call in there:
dim xdoc as XDocument = XDocument.Parse("<rootNode><ReportCellRef><dang n=""DVCompany"" h=""0"" u=""0"" o=""0"" fmt=""0""></dang></ReportCellRef></rootNode>")
Dim replaceable = From dangToTake In xdoc.Descendants().Elements("ReportCellRef") _
Where dangToTake.Element("dang").Attribute("n").Value = "DVCompany"
Select dangToTake
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