Question

我有一个像

这样的xml文件

<?xml version="1.0"?>
<catalog>
<book id="bk101">
<author>Gambardella, Matthew</author>
<title>XML Developer's Guide [49-o]</title>
<genre>Computer</genre>
<price>44.95</price>
<publish_date>2000-10-01</publish_date>
<description>An in-depth look at [41-p] creating applications with XML.</description>
</book>
<book id="bk102">
<author>Ralls, Kim</author>
<title>Midnight Rain</title>
<genre>Fantasy</genre>
<price>5.95</price>
<publish_date>2000-12-16</publish_date>
<description>A former architect [100-x] battles corporate zombies, an evil sorceress, and her own childhood to become queen of the world.</description>
</book>
<book id="bk103">
<author>Corets, Eva</author>
<title>Maeve Ascendant</title>
<genre>Fantasy</genre>
<price>5.95</price>
<publish_date>2000-11-17</publish_date>
<description>After the collapse of a nanotechnology society in England, the [01-i] young survivors lay the foundation for a new society.</description>
</book>
</catalog>

如何使用 linq2xml 来提取值＆＃34; [（\ d +） - （[a-z]）]＆＃34;来自每个节点<description>并将其存储在一个变量中，或者可以使用它将这些提取的值添加到各个节点的新属性中，如<description val="41-p">等？

Answer 1

您可以使用Descendants

Regex regex = new Regex(@"(\d+)-([a-z])");
var xdoc = XDocument.Parse(xml);
var descriptions = xdoc.Descendants("description")
    .Where(x => regex.Match(x.Value).Success)
    .Select(x => regex.Match(x.Value).Value).ToList();

Output:
41-p
100-x
01-i

如果要将提取的值设置为属性;

Regex regex = new Regex(@"(\d+)-([a-z])");
var xdoc = XDocument.Parse(xml);
var descriptions = xdoc.Descendants("description")
                  .Where(x => regex.Match(x.Value).Success);
foreach (var description in descriptions)
{
    var regexResult = regex.Match(description.Value).Value;
    var attribute = new XAttribute("id", regexResult);
    description.Add(attribute);
}
xdoc.Save("sample.xml");

Answer 2

我不熟悉linq2xml，所以我使用XmlDocument和XPath表达式找到我感兴趣的节点。这样的事情：

XmlDocument doc = new XmlDocument();
doc.LoadXml(xmlString);

var books = doc.SelectNodes("//catalog/book");
foreach (XmlNode book in books)
{
     var description = book.SelectSingleNode("description");
     Regex regex = new Regex(@"(\[.*\])");
     var match = regex.Match(description.InnerText);
     if (match.Success)
     {
          var val = match.Groups[0].Value;
          var attribute = doc.CreateAttribute("val");
          attribute.Value = val;
          description.Attributes.SetNamedItem(attribute);
     }
}

// Convert XmlDocument back to string
var stringWriter = new StringWriter();
var xmlTextWriter = XmlWriter.Create(stringWriter);
doc.WriteTo(xmlTextWriter);
xmlTextWriter.Flush();
xmlString = stringWriter.GetStringBuilder().ToString();

如何使用c＃从节点内部获取字符串的一部分

2 个答案: