如何读取XMLFILE数据，发现号和日期时间

如何将XML文件中读取数和日期时间所有目录和子目录directories.i有txtnumber和txtdatetime，并搜索所有目录，并让我search.i的XMLDATA代表希望得到890001000011717.wav如何读取XMLFILE数据，发现号和日期时间

var allfiles = Directory.GetFiles(path, "*.*", System.IO.SearchOption.AllDirectories); 

foreach (var item in allfiles) 
{ 
    DateTime lastModified = System.IO.File.GetLastWriteTime(item); 

    string extension; 

    extension = Path.GetExtension(item); 

    if (lastModified.ToShortTimeString() == user_time2.ToShortTimeString() && extension == ".xml") 
    { 
     XmlReader xmlFile; 
     xmlFile = XmlReader.Create(item, new XmlReaderSettings()); 
     DataSet dss = new DataSet(); 
     DataView dv; 
     dss.ReadXml(xmlFile); 

     string ss = dss.Tables[0].Rows[0]["dataformat"].ToString(); 

     string number = dss.Tables["Party"].Rows[1]["number"].ToString(); 
    } 
}

的XML：

<?xml version="1.0" encoding="UTF-8" standalone="yes" ?> 
<recording> 
    <starttime>2016-05-13 15:03:14:000 +0100</starttime> 
    <endtime>2016-05-13 15:04:59:000 +0100</endtime> 
    <calldirection>Incoming</calldirection> 
    <filename>890001000011717.wav</filename> 
    <recordingowners> 
     <recordingowner>202</recordingowner> 
    </recordingowners> 
    <parties> 
     <party id="1"> 
      <number>0711111111</number> 
      <pstarttime>2016-05-13 15:05:00:703 +0100</pstarttime> 
      <pendtime>2016-05-13 15:05:00:703 +0100</pendtime> 
     </party> 
    </parties> 
</recording>

来源

2016-05-26 Adeel Khan

递归的文件夹，并使用数据夹http://stackoverflow.com/questions/ 929276 /如何做递归-list-所有的档案-IN-A-目录中-C和每个https://msdn.microsoft.com/en-us/library/hcebdtae(v=vs.110).aspx –

使用XPath我有txtnumber和txtdatetime，并搜索所有目录，并代表我的搜索xmldata。 –

你可以通过将XML文件反序列化为C＃对象，然后使用它来轻松完成此操作。

首先，您需要创建您的XML将反序列化为类（你可以做到这一点很容易地使用Visual Studio的编辑菜单 - >选择性粘贴 - >粘贴XML作为类）：

[XmlType(AnonymousType = true)] 
[XmlRoot(Namespace = "", IsNullable = false)] 
public class recording 
{ 
    public string starttime { get; set; } 

    public string endtime { get; set; } 

    public string calldirection { get; set; } 

    public string filename { get; set; } 

    [XmlArray] 
    [XmlArrayItem("recordingowner", IsNullable = false)] 
    public byte[] recordingowners { get; set; } 

    [XmlArrayItem("party", IsNullable = false)] 
    public recordingParty[] parties { get; set; } 
} 


[XmlType(AnonymousType = true)] 
public class recordingParty 
{ 
    public string number { get; set; } 

    public string pstarttime { get; set; } 

    public string pendtime { get; set; } 

    [XmlAttribute] 
    public int id { get; set; } 
}

然后只列出使用*.xml代替*.*的XML文件和反序列化他们然后访问你只需要像任何其他C＃对象的属性：

var allfiles = Directory.GetFiles(path, "*.xml", SearchOption.AllDirectories); 

var serializer = new XmlSerializer(typeof(recording)); 
foreach (var item in allfiles) 
{ 
    var lastModified = File.GetLastWriteTime(item); 

    if (lastModified.ToShortTimeString() != user_time2.ToShortTimeString()) continue; 

    recording recording; 
    using (var reader = new StreamReader(item)) 
    { 
     recording = (recording) serializer.Deserialize(reader); 
    } 

    string number = recording.parties[0].number; 
}

来源

2016-05-26 10:34:14 Nasreddine

感谢您的回复......先生，我有编辑自己的帖子请检查 –

如何读取XMLFILE数据，发现号和日期时间

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