Question

您好我有以下xml文件：

<?xml version="1.0" encoding="utf-8"?>
<Resources>
    <Category Cause="ONE">
        <Resource Component="AAA" Name="cBackgroundTextA" />
        <Resource Component="BBB" Name="cBackgroundTextB" />
        <Resource Component="CCC" Name="cBackgroundTextC" />
    </Category>
    <Category Cause="TWO">
        <Resource Component="DDD" Name="cBackgroundTextD" />
        <Resource Component="EEE" Name="cBackgroundTextE" />
        <Resource Component="FFF" Name="cBackgroundTextF" />
    </Category>
    <Category Cause="THREE">
        <Resource Component="GGG" Name="cBackgroundTextG" />
        <Resource Component="HHH" Name="cBackgroundTextH" />
        <Resource Component="III" Name="cBackgroundTextI" />
    </Category>
</Resources>

我如何得到以下内容：

ONE，AAA，cBackgroundTextA

ONE，BBB，cBackgroundTextB

ONE，CCC，cBackgroundTextC

TWO，DDD，cBackgroundTextD

TWO，EEE，cBackgroundTextE

...

Answer 1

尝试一下这一行：

XElement root;

foreach(var folder in root.Children()) { //For each folder
    var folderType = folder.Attribute("type").Value;
    foreach(var file in folder.Children()) { //For each  file

        Console.WriteLine($"{folderType} AND {file}");
    }
}

Answer 2

我终于通过以下方式获得了价值：

xmlDocument = new XDocument();
xmlDocument = XDocument.Load(filePath, LoadOptions.None);

var categoryNodeList = XmlDocument.Descendants("Category");

    foreach (XElement categoryItem in categoryNodeList)
    {
        foreach (XElement resourceItem in categoryItem.Elements())
        {
            var XmlObject = new ResultDetail
            {
                Cause = categoryItem.Attribute("Cause").Value,
                Component = resourceItem.Attribute("Component").Value,
                Name = resourceItem.Attribute("Name").Value
            };
        }
    }

Answer 3

可以使用单个LINQ查询收集所有必需的信息：

var query = from cat in XmlDocument.Descendants("Category");
            from res in cat.Elements()
            select new ResultDetail
                    {
                        Cause = cat.Attribute("Cause").Value,
                        Component = res.Attribute("Component").Value,
                        Name = res.Attribute("Name").Value
                    };

C＃使用重复的节点模式解析XML

3 个答案: