Question

过去一天我一直在尝试阅读和修改这个xml

<TimeSeriesDocument xmlns:xsd="http://www.w3.org/2001/XMLSchema"                xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="urn:Test">
<DocumentIdentification v="ZDAE_02417b6a-477a-a-4aa19e1c4712"/>
<TimeSeries>
<Period>
  <TimeInterval v="12" />
  <Resolution v="PT1H" />
  <Interval>
    <Pos v="17" />
    <Qty v="0" />
  </Interval>
  <Interval>
    <Pos v="19" />
    <Qty v="10" />
  </Interval>
</Period>
</TimeSeries>
</TimeSeriesDocument>

我想修改期间内每个时间间隔内的每个数量

我一直在使用后代，但我正在努力绕过每个区间并挑选和更改数量

var doc = XDocument.Load(test.xml);
var periods = doc.Descendants("TimeSeriesDocument/TimeSeries").Select(e => new { name = e.Element("Period").Value });

你能帮忙吗？

Answer 1

循环访问Qty节点并为v属性设置新值的示例控制台应用程序：

class Program
{
    static void Main(string[] args)
    {
        var document = new XmlDocument();
        document.Load("Sample.xml");

        XmlNamespaceManager nsmgr = new XmlNamespaceManager(document.NameTable);
        nsmgr.AddNamespace("ns", "urn:Test");

        var intervalNodes = document.SelectNodes("//ns:Interval", nsmgr);
        foreach(XmlNode node in intervalNodes)
        {
            var quantityNode = node.SelectSingleNode("./ns:Qty", nsmgr);
            quantityNode.Attributes["v"].Value = "New Value";
        }
    }
}

Answer 2

XNamespace ns = "urn:Test";

XElement doc = XElement.Load("file.xml");

var periods = doc.Descendants(ns + "Period");
var intervals = periods.Elements(ns + "Interval");

foreach (var qty in intervals.Elements(ns + "Qty"))
{
    qty.Attribute("v").Value = "42";
}

修改XML文件

2 个答案: