英文:
Reading xmlnode child tag value
问题
XmlDocument xmlDocNew = new XmlDocument();XmlElement CATLOG = xmlDocNew.CreateElement("mappings");XmlNode xmlNodeTab = xmltest.DocumentElement;XmlNodeList xmlNodeListCD = xmlNodeTab.SelectNodes("//mapping");foreach (XmlNode xmlNodeCD in xmlNodeListCD){string innerText = xmlNodeCD["a"].InnerText;string xmlNodeapp = xmlNodeCD["b/c/d"].InnerText;}
在上面的代码中，我能够遍历所有的 "a" 标签元素，但是无法读取 "d" 标签的值。要读取 "d" 标签的值，你可以使用 "b/c/d" 的路径来访问，就像上面的代码示例中所示。这样，在 foreach 循环中，你可以同时获取 "a" 标签和其对应的 "d" 标签的值。
英文:
My xml file looks like below.
<mappings><mapping><a>a1value</a><b><c><d>d1value</d><e>e1value</e></c></b></mapping><mapping><a>a2value</a><b><c><d>d2value</d><e>e2value</e></c></b></mapping></mappings>
C# code to read the value of a,d tags.
XmlDocument xmlDocNew = new XmlDocument();XmlElement CATLOG = xmlDocNew.CreateElement("mappings");XmlNode xmlNodeTab = xmltest.DocumentElement;XmlNodeList xmlNodeListCD = xmlNodeTab.SelectNodes("//mapping");foreach (XmlNode xmlNodeCD in xmlNodeListCD){string innerText = xmlNodeCD["a"].InnerText;string xmlNodeapp = xmlNodeCD["//b/c/d"].InnerText;}
With the above code i am able to iterate through all the "a" tag elements.
But I am not able to read "d" tag value.
How do I read the value of "d" tag?
When I iterate through each element in the foreach loop, I want to get the value of "a" tag and its corresponding "d" tag value.
答案1
得分: 0
这段代码查看每个映射，并在其下查找标签 a 和 d 的值。还要注意，如果您想查找“sub”标签并仅获取这些子项，您必须使用 .// 而不仅仅是 //。点符号提取位于当前节点下的节点。
XmlDocument doc = new XmlDocument();doc.Load($@"{Directory.GetParent(Environment.CurrentDirectory).Parent.Parent.FullName}\json.txt");var nodes = doc.SelectNodes("//mapping");foreach (XmlNode node in nodes){var aNodes = node.SelectNodes(".//a");foreach (XmlNode aNode in aNodes)Console.WriteLine(aNode.InnerText);// 如果您知道 .//b/c/d 下只有一个节点，请使用 SelectSingleNode。Console.WriteLine(node.SelectSingleNode(".//b/c/d").InnerText);}
输出
a1valued1valuea2valued2value
注意：
a 标签在某种程度上是 d 标签的兄弟，并且它们的单一父元素是 mapping 元素。如果您获取所有映射节点，然后逐个处理这些节点，您将获得最佳结果。
英文:
This code looks at each of the mapping and under it, looks up the value for tags a and d. Also note that if you want to look up 'sub' tags and only get these child items, you have to use .// instead of just //. Dot notation pulls node "under" the current node.
XmlDocument doc = new XmlDocument();doc.Load($@"{Directory.GetParent(Environment.CurrentDirectory).Parent.Parent.FullName}\json.txt");var nodes = doc.SelectNodes("//mapping");foreach (XmlNode node in nodes){var aNodes = node.SelectNodes(".//a");foreach (XmlNode aNode in aNodes)Console.WriteLine(aNode.InnerText);// If you know there will only be one node at .//b/c/d, use SelectSingleNode instead.Console.WriteLine(node.SelectSingleNode(".//b/c/d").InnerText);}
Output
a1valued1valuea2valued2value
Note:
a tag is sort of a sibling of d tag and their singular parent is the mapping element. You would get best results if you got all the mapping nodes and then worked through those.
答案2
得分: 0
自 .Net 3.5 开始，引入了 Linq 和 Linq To XML，使 XML 解析变得更加简便。使用 Linq To XML，您可以这样读取它：
void Main(){var s = @"<mappings><mapping><a>a1value</a><b><c><d>d1value</d><e>e1value</e></c></b></mapping><mapping><a>a2value</a><b><c><d>d2value</d><e>e2value</e></c></b></mapping></mappings>";var data = from x in XDocument.Parse(s).Descendants("mapping")select new {a = (string)x.Element("a"),d = (string)x.Element("b").Element("c").Element("d")};foreach (var e in data){Console.WriteLine($"{e.a}, {e.d}");}}
如果您有一个与此结构匹配的类，如下所示：
public class MyData{public string A { get; set; }public string D { get; set; }}
然后，您可以像这样使用这个类，而不是匿名类型：
var data = from x in XDocument.Parse(s).Descendants("mapping")select new MyData {A = (string)x.Element("a"),D = (string)x.Element("b").Element("c").Element("d")};foreach (var e in data){Console.WriteLine($"{e.A}, {e.D}");}
PS：由于您获得的是一个 IEnumerable，您可以直接将数据绑定到 DataGridView、ListBox 等（...DataSource = data.ToList()）。
英文:
Since .Net 3.5 there is Linq and Linq To XML which makes XML parsing easier. With Linq To XML you could read it as:
void Main(){var s = @"<mappings><mapping><a>a1value</a><b><c><d>d1value</d><e>e1value</e></c></b></mapping><mapping><a>a2value</a><b><c><d>d2value</d><e>e2value</e></c></b></mapping></mappings>";var data = from x in XDocument.Parse(s).Descendants("mapping")select new {a=(string)x.Element("a"),d=(string)x.Element("b").Element("c").Element("d")};foreach (var e in data){Console.WriteLine($"{e.a}, {e.d}");}}
With a file, instead of .Parse, you would use .Load( filename ).
Output:
a1value, d1valuea2value, d2value
If you have a class matching this structure, like:
public class MyData{public string A { get; set; }public string D { get; set; }}
Then instead of the anonymous type you could use this class like so:
var data = from x in XDocument.Parse(s).Descendants("mapping")select new MyData {A=(string)x.Element("a"),D=(string)x.Element("b").Element("c").Element("d")};foreach (var e in data){Console.WriteLine($"{e.A}, {e.D}");}
PS: Since you are getting an IEnumerable, you could directly bind data to a DataGridView, ListBox ... ( ...DataSource = data.ToList()).
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论