Question

我有一个控制台应用程序，其中包含一些foreach循环，这些循环遍历路径并从csv文件中解析出电子邮件地址。但是，我需要我的循环从配置文件中读取而不是通过目录，然后最终为每个电子邮件地址调用现有的API端点。这就是我现在所拥有的：

 static void RunTask(string[] args)
        {
            foreach (string folders in Directory.EnumerateDirectories("C:\\tests"))//looks at every folder within this main folder called tests --this needs to read the list of paths
            {
                foreach (string path in Directory.EnumerateFiles(folders, "*.csv"))//looks at every file with the extension ".csv" in each folder
                {
                    Debug.Write("\n" + path + "\n"); //writes out file names
                    using (StreamReader sr = new StreamReader(path))
                    {
                        String line;

                        while ((line = sr.ReadLine()) != null)
                        {
                            string[] parts = line.Split(',');

                            string email = parts[1];
                            Debug.Write(email + "\n");//writes out email column

                        }
                    }

                }
            }
        }

这是我的Path.Config文件中的内容：

<pathsSettings>
  <paths>
    <add Path="C:\tests\first" 
         TemplateId="123456">
    </add>
    <add Path="C:\tests\second"
         TemplateId="tem_56hyNijCXxGP52ZrgdWziC ">
    </add>
    <add Path="C:\tests\third"
         TemplateId="tem_2wWT6YfGkDXSntEPKhHCWB ">
    </add>
  </paths>
</pathsSettings>

我不确定如何编写我的foreach来从这些路径列表中绘制。

Answer 1

 public class PathTemplate
 {
     public string path { get; set; }
     public string template { get; set; }
 }

  public static string getxml()
    {
        return "<pathsSettings>" +
               "<paths>" +
               "<add Path=\"C:\\tests\\first\" TemplateId=\"123456\"></add>" +
               "<add Path=\"C:\\tests\\second\" TemplateId=\"tem_56hyNijCXxGP52ZrgdWziC \"></add>" +
               "<add Path=\"C:\\tests\\third\" TemplateId=\"tem_2wWT6YfGkDXSntEPKhHCWB \"></add>" +
               "</paths>" +
               "</pathsSettings>";
    }


var xdoc = XDocument.Parse(getxml());

        Console.WriteLine(xdoc);

        var pts = xdoc.Elements("pathsSettings")
            .Elements("paths")
            .Elements("add").
            Select(p => new PathTemplate()
            {
                path = p.Attribute("Path").Value.ToString(),
                template = p.Attribute("TemplateId").Value.ToString()
            }).ToList();

Answer 2

如果对象正确，可以反序列化xml。我会说你的课程应该如下：

[Serializable()]
public class add
{
    [System.Xml.Serialization.XmlAttribute("Path")]
    public string Path { get; set; }

    [System.Xml.Serialization.XmlAttribute("TemplateId")]
    public string TemplateId { get; set; }
}


[Serializable()]
[System.Xml.Serialization.XmlRoot("pathsSettings")]
public class pathsSettings
{
    [XmlArray("paths")]
    [XmlArrayItem("add", typeof(add))]
    public add[] paths { get; set; }
}

你可以像这样迭代它

public void DoWhatever(){
        pathsSettings myObject;
        // Construct an instance of the XmlSerializer with the type
        // of object that is being deserialized.
        XmlSerializer mySerializer = new XmlSerializer(typeof(pathsSettings));
        // To read the file, create a FileStream.
        FileStream myFileStream = new FileStream(@"yourpath\Path.config", FileMode.Open);
        // Call the Deserialize method and cast to the object type.
        myObject = (pathsSettings)mySerializer.Deserialize(myFileStream);
        foreach(add addObj in myObject.paths)
        {
            Console.WriteLine(addObj.Path);
            Console.WriteLine(addObj.TemplateId);
        }
}

此方法要求您拥有正确的对象，但从长远来看，如果需要，您可以使用这些对象序列化您的值，从而编辑您的配置文件。如果您还需要TemplateId，那么您已经在“添加”对象

中

Answer 3

这对我有用：

var doc = XDocument.Parse(@"<pathsSettings>
  <paths>
    <add Path=""C:\tests\first"" 
         TemplateId=""123456"">
    </add>
    <add Path=""C:\tests\second""
         TemplateId=""tem_56hyNijCXxGP52ZrgdWziC "">
    </add>
    <add Path=""C:\tests\third""
         TemplateId=""tem_2wWT6YfGkDXSntEPKhHCWB "">
    </add>
  </paths>
</pathsSettings>");

var paths =
    doc
        .Root
        .Descendants("add")
        .Select(e => e.Attribute("Path").Value)
        .ToList();

这给了我这些路径：

C:\tests\first 
C:\tests\second 
C:\tests\third

迭代配置文件中的路径列表？

3 个答案: