如何读取 xmlfile 数据并查找数字和日期时间

Question

如何 xml 文件读取编号和日期时间所有目录和子 directories.i 有 txtnumber 和 txtdatetime，并搜索所有目录并代表我的 [=] 获取 xml 数据21=] 想得到 890001000011717.wav

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();
    }
}

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>

Answer 1

您可以通过将 XML 文件反序列化为 C# 对象然后使用它来轻松做到这一点。

首先，您需要创建类，您的 XML 将反序列化为该文件（您可以使用 Visual Studio 的编辑菜单 -> 选择性粘贴轻松完成此操作 - > 将 XML 粘贴为类):

[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; }
}

然后使用 *.xml 而不是 *.* 仅列出 xml 文件并反序列化它们，然后像访问任何其他 C# 对象一样访问您需要的属性：

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;
}

如何读取 xmlfile 数据并查找数字和日期时间

how to read xmlfile data and find number and datetime

c#

c#-4.0

linq-to-xml

c#-3.0