Question

我在c＃中有一个三个列表，变量名称为l_lstData1, l_lstData2, l_lstData3。

文件结构

<FileDetails>  
  <Date FileModified="29/04/2010 12:34:02" />   
  <Data Name="Data_1" DataList="India" Level="2" />   
  <Data Name="Data_2" DataList="chennai" Level="2" />   
  <Data Name="Data_3" DataList="hyderabad" Level="2" />   
  <Data Name="Data_4" DataList="calcutta" Level="2" />  
  <Data Name="Data_5" DataList="vijayawada" Level="1" /> 
  <Data Name="Data_6" DataList="cochin" Level="1" /> 
  <Data Name="Data_7" DataList="madurai" Level="0" />  
  <Data Name="Data_8" DataList="trichy" Level="0" />   
</FileDetails>

3个列表的值如下：

 l_lstData1[0] = "India";
 l_lstData1[1] = "chennai";
 l_lstData1[2] = "hyderabad";
 l_lstData1[3] = "calcutta";

所以上面的XML（element：Data）的level属性值=“2”。

 l_lstData2[0] = "vijayawada";
 l_lstData2[1] = "cochin";

所以上面的XML（element：Data）的level属性值=“1”。

 l_lstData3[0] = "madurai";
 l_lstData3[1] = "trichy";

所以上面的XML（element：Data）的level属性值=“0”。

Answer 1

目前尚不清楚为什么“Level”属性是如此指定的，但这会为您创建相关的XML：

// Used for side-effects in the XElement constructor. This is a bit icky. It's
// not clear what the "Name" part is really for...
int count = 1;

var doc = new XDocument(
    new XElement("FileDetails",
        new XElement("Date", new XAttribute("FileModified", DateTime.UtcNow)),
        l_lstData1.Select(x => new XElement("Data",
            new XAttribute("Name", "Data_" + count++),
            new XAttribute("DataList", x),
            new XAttribute("Level", 2))),
        l_lstData2.Select(x => new XElement("Data",
            new XAttribute("Name", "Data_" + count++),
            new XAttribute("DataList", x),
            new XAttribute("Level", 1))),
        l_lstData3.Select(x => new XElement("Data",
            new XAttribute("Name", "Data_" + count++),
            new XAttribute("DataList", x),
            new XAttribute("Level", 0)))));

如果您可以将列表项中的投影提取到其元素，那么可能会更整洁，但"Data_" + count位使得这很棘手。目前尚不清楚为什么你需要这样的事情才能说实话......如果没有它就可以逃脱，代码可能更清晰。

我想一种替代方法是创建文档而不用 Name属性，然后再填充它们。例如：

private static IEnumerable<XElement> ProjectList(IEnumerable<string> list,
    int level)
{
    return list.Select(x => new XElement("Data",
        new XAttribute("DataList", x),
        new XAttribute("Level", level)));
}

然后：

var doc = new XDocument(
    new XElement("FileDetails",
        new XElement("Date", new XAttribute("FileModified", DateTime.UtcNow)),
        ProjectList(l_lstData1, 2),
        ProjectList(l_lstData2, 1),
        ProjectList(l_lstData3, 0)));

int count = 1;
foreach (var element in doc.Descendants("Data"))
{
    element.SetAttributeValue("Name", "Data_" + count++);
}

Answer 2

怎么样：

XDocument doc = new XDocument();

var total = (from a in list1 select new { Name = a, Level = 2 }).Concat(
             from b in list2 select new { Name = b, Level = 1 }).Concat(
             from c in list3 select new { Name = c, Level = 0 });

XElement root = new XElement("FileDetails", from i in Enumerable.Range(0, total.Count())
                                            let element = total.ElementAt(i)
                                            let name = new XAttribute("Name", String.Format("Data_{0}", i + 1))
                                            let level = new XAttribute("Level", element.Level)
                                            let datalist = new XAttribute("DataList", element.Name)
                                            select new XElement("Data", name, datalist, level),
                                            new XElement("Date", new XAttribute("FileModified", DateTime.Now)));

LINQ和XDocument：如何创建XML文件？

2 个答案: