Question

请查看代码并告知需要进行哪些更改才能获得预期的输出。选择代码是伪的。

的xml：

<?xml version='1.0'  encoding='Cp1252' ?>
<RESULTS>
    <ROW>
        <COLUMN NAME="ID"><![CDATA[1234]]></COLUMN>
        <COLUMN NAME="FName"><![CDATA[Joe]]></COLUMN>
        <COLUMN NAME="LName"><![CDATA[Doe]]></COLUMN>
    </ROW>
    <ROW>
        <COLUMN NAME="ID"><![CDATA[5678]]></COLUMN>
        <COLUMN NAME="FName"><![CDATA[Jill]]></COLUMN>
        <COLUMN NAME="LName"><![CDATA[Martin]]></COLUMN>
    </ROW>
</RESULTS>

代码：

static void Main(string[] args)
{    
    XDocument doc = XDocument.Load("C:\\test.xml");
    var query = from headrow in doc.Descendants("ROW")
                from COLUMN in headrow.Descendants("COLUMN")
                select new 
                { 
                    ID = COLUMN.Element where (string)COLUMN.Attribute== "ID",
                    FName = COLUMN.Element where (string)COLUMN.Attribute== "FName",
                    LName = COLUMN.Element where (string)COLUMN.Attribute== "LName"
                };


    foreach (var result in query)
    {    
        Console.WriteLine(result);               
    }    
}

Exprected output
1234 Joe Doe
5678吉尔马丁

提前致谢

Answer 1

查看以下内容是否有帮助：

var query = from headrow in doc.Descendants("ROW") 
            select new 
            {
                ID = headrow.Elements("COLUMN").First(c => c.Attribute("NAME").Value == "ID").Value,
                FName = headrow.Elements("COLUMN").First (c => c.Attribute("NAME").Value == "FName").Value,
                LName = headrow.Elements("COLUMN").First (c => c.Attribute("NAME").Value == "LName").Value
            };

foreach (var result in query) 
{     
    Console.WriteLine("{0} {1} {2}", result.ID, result.FName, result.LName);                
}

使用Linq to XML读取xml

1 个答案: