Question

我有如下的XML显示RoadPart和Vehicles元素：

<RoadPart>
    <Vehicles>
        <ConfigurationSummaryListPosition>
          <ConfigurationSummary>Test name</ConfigurationSummary>
        </ConfigurationSummaryListPosition>
    </Vehicles>
    <Vehicles>
        <ConfigurationSummaryListPosition>
          <ConfigurationSummary>Test name</ConfigurationSummary>
        </ConfigurationSummaryListPosition>
    </Vehicles>
    <Vehicles>
        <ConfigurationSummaryListPosition>
          <ConfigurationSummary>Test name</ConfigurationSummary>
        </ConfigurationSummaryListPosition>
    </Vehicles>
</RoadPart>

我有一个变量

var test = @"<DataType>
                <ConfigurationSummaryListPositionType>
                  <ConfigurationSummaryType> name </ConfigurationSummaryType>
                </ConfigurationSummaryListPositionType>
            </DataType>
            <DataType>
                <ConfigurationSummaryListPositionType>
                  <ConfigurationSummaryType> Test</ConfigurationSummaryType>
                </ConfigurationSummaryListPositionType>
            </DataType>";

想要更换具有可变输出的车辆元素应如下：

<RoadPart>
    <DataType>
        <ConfigurationSummaryListPositionType>
          <ConfigurationSummaryType> name </ConfigurationSummaryType>
        </ConfigurationSummaryListPositionType>
    </DataType>
    <DataType>
        <ConfigurationSummaryListPositionType>
          <ConfigurationSummaryType> Test</ConfigurationSummaryType>
        </ConfigurationSummaryListPositionType>
    </DataType>
</RoadPart>

帮我解决这个问题

Answer 1

为什么不用linq读取xml，重复每个节点并更改该节点值。

var nodes = (from node in document.Descendant("RoadPart") select node);

其中document是加载的xml文件

var document = XDocument.Load("your xml file path");

然后像你一样在文档中引用foreach节点

foreach (var node in nodes)
{
    // do something
}

Answer 2

这会将结果返回为System.Xml.Linq.XElement：

test = string.Concat("<root>", test, "</root>"); // set dummy root
XDocument xdoc = XDocument.Parse(test);

XElement replaced = new XElement("RoadPart", xdoc.Descendants("DataType").Select(e => e));

如果要将结果保存为文件，

using (FileStream fs = new FileStream("{filePath}", FileMode.Create))
{
    replaced.Save(fs);
    fs.Close();
}

替换XML中的元素

2 个答案: