Question

有人可以帮助我使用正确的LINQ语法从这个XML中提取状态吗？

理想情况下，我希望打印出以下内容：

状态：打开= 1

状态：待定= 2

状态：忽略= 3

状态：已关闭= 4

<?xml version="1.0" encoding="UTF-8"?>
<configuration>
    <scope name="com">
    <scope name="company">
        <scope name="app">
        <scope name="app_monitor">
            <scope name="statuses">
            <entry name="Open">1</entry>
            <entry name="Pending">2</entry>
            <entry name="Ignored">3</entry>
            <entry name="Closed">4</entry>
            </scope>
            <scope name="urgencies">
            <entry name="Critical">1</entry>
            <entry name="Alarm">2</entry>
            <entry name="Info">3</entry>
            </scope>                       
        </scope>
        </scope>
    </scope>
    </scope>
</configuration>

我尝试了不同的变化，但这里有多远：

    XDocument Xdocument = new XDocument();
    var doc = XDocument.Load(@"c:\temp\app_sett.xml");
    var returnedvalues = from app_sett in doc.Descendants("scope")
                where app_sett.Attribute("name").Value == "statuses"
                select new                                  
                {
                   blah = app_sett.Attribute("name").Value,
                };

Answer 1

听起来你需要获得相关范围元素的子元素。例如：

var query = doc.Descendants("scope")
               .Where(x => (string) x.Attribute("name") == "statuses")
               .Elements("entry")
               .Select(entry => new { Name = (string) entry.Attribute("name"),
                                      Value = (int) entry });

关于属性的C＃XML LINQ语法

1 个答案: