如何转换List< string>到C#中的xml文档 [英] How to convert List<string> to xml document in c#

问题描述

我想在C#中将列表转换为xml文档.

I would like to convert the List to a xml document in c#.

我的代码如下:

List<string> BugWSResponseList1 = new List<string>();
Logger.Write("\n\n" + DateTime.Now + " : " + " : START : Creation of a set of Bugs via bug.Add API");
BugWSResponseList1 = CreateBugs(FilePath_EXPRESS_API_BugAdd_CreateBugs_DataFile);
Logger.Write("\n\n" + DateTime.Now + " : " + " : END : Creation of a set of Bugs via bug.Add API");

我一直试图将列表响应转换为字符串，然后尝试将其转换为xml文档，但是当我解析它时，它表明xml中存在多个根元素.

I have been trying to convert the list response to string and then tried to convert it in xml document but when i parse it then it shows that there are more than one root elements in the xml.

xml的格式如下:

<Bug>
      <family>ESG</family>
      <product>Dr.Watson</product>
      <version>Xpress API</version>
      <productarea>1</productarea>
      <subarea></subarea>
      <qe>sdawar</qe>
      <duplicateId></duplicateId>
</Bug>
<Bug>
      <family>TEST22</family>
      <product>Dr.Watson</product>
      <version>Xpress API</version>
      <productarea>1</productarea>
      <subarea></subarea>
      <qe>sdawar</qe>
      <duplicateId></duplicateId>
</Bug>

请注意，它在xml中有两个不同的节点. 到目前为止，我正在做的是将List BugWSResponseList1转换为字符串，然后将其加载为xml.这是我正在执行的代码:

Please note that it has two different nodes in the xml. What i am doing so far is converting List BugWSResponseList1 to a string and then loading it as xml.here is the code i am doing:

XmlDocument xd = new XmlDocument();
                    string s = string.Join(",", BugWSResponseList2);                    
                    xd.LoadXml(s);

但是当我这样做时，它表示xml中有多个根元素.

But when i am doing it it says that there are more than one root element in the xml.

我想将BugWSResponseList1列表转换为xml文档，因为我需要对其进行进一步解析以执行代码. 任何帮助，将不胜感激. 谢谢

I would like to convert the List BugWSResponseList1 to an xml document as i need to parse it further for my code execution. Any help would be appreciated. Thanks

推荐答案

您可以使用XElement或XDocument和LINQ.提供XML外观的示例，我们可以提供更多信息.

You may use XElement or XDocument and LINQ. Providing a sample of what the XML should look like we could provide more info.

例如:

BugWSResponseList1.Add("<Bug><family>TEST22</family><product>Dr.Watson</product><version>Xpress API</version><productarea>1</productarea><subarea></subarea><qe>sdawar</qe><duplicateId></duplicateId></Bug>");

BugWSResponseList1.Add("<Bug><family>ESG</family><product>Dr.Watson</product><version>Xpress API</version><productarea>1</productarea><subarea></subarea><qe>sdawar</qe><duplicateId></duplicateId></Bug>");

XElement xe = new XElement
              (
                  "Bugs",
                  BugWSResponseList1
                  .Select 
                  (
                      x=>
                      XElement.Parse(x)
                  )
              );

因此，在我用两段xml加载列表之后，您提供了以下操作:

So after i have loaded the list with the two pieces of xml you have provided i do the following:

我创建一个新的XElement来保存XML的根.我将该根命名为Bugs.然后，作为该根的子级，我将您的List的内容解析为XElement对象后放入其中.

I create a new XElement that will hold the root of the XML. I name that root as Bugs. Then as children of that root i put the contents of your List after parsing them into XElement objects.

上面的代码将导致:

<Bugs>
  <Bug>
    <family>TEST22</family>
    <product>Dr.Watson</product>
    <version>Xpress API</version>
    <productarea>1</productarea>
    <subarea></subarea>
    <qe>sdawar</qe>
    <duplicateId></duplicateId>
  </Bug>
  <Bug>
    <family>ESG</family>
    <product>Dr.Watson</product>
    <version>Xpress API</version>
    <productarea>1</productarea>
    <subarea></subarea>
    <qe>sdawar</qe>
    <duplicateId></duplicateId>
  </Bug>
</Bugs>

您所缺少的是根元素. XmlDocument需要定义一个根元素.

What you were missing was the root element. XmlDocument needs to have a root element defined.

在您的情况下，只需在连接的字符串的开头和结尾加上如下代码，就可以避免在代码中使用它，但是我认为我的解决方案更可靠.

In your case you could get away with it in you code by just adding in the start of your joined string and in the end, like the following, but i think my solution is more robust.

XmlDocument xd = new XmlDocument();
string s = "<Bugs>" + string.Join(",", BugWSResponseList2) + "</Bugs>";                    
xd.LoadXml(s);

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