如何防止名称重复但大小写不同的列重复的行数据?

时间:2019-02-27 13:33:43

标签: c# xml datatable linq-to-xml

我有一个像这样的xml文件:

<?xml version="1.0" encoding="UTF-8"?>
<Logging S="T006" version="2" >
<Log Date="2018-11-21" Severity="Error" id="22" ID="Opened" Msg="some text"/>
<Log Date="2018-11-21" Severity="Info" id="76" ID="Auth"/>
<Log Date="2018-11-21" Severity="Info" id="60" ID="Up"/>
<Log Date="2018-11-21" Severity="Warning" id="22" ID="Opened" Msg="some text"/>
<Log Date="2018-11-21" Severity="Info" id="96" ID="Locked"/>
<Log Date="2018-11-21" Severity="Info" id="84" ID="Done"/>
<Log Date="2018-11-21" Severity="Info" id="57" ID="Idle"/>
<Log Date="2018-11-21" Severity="Error" id="10" ID="Inspected" Pos="12"/>
<Log Date="2018-11-21" Severity="Info" id="148" ID="Started"/>
</Logging>

在datagridview上显示此文件中的数据。我使用xml属性名称的名称创建表列:日期,严重性,ID,ID。

DataTable dt = new DataTable();
dt.Columns.Add("Date", typeof(DateTime));
dt.Columns.Add("Severity", typeof(string));
dt.Columns.Add("id", typeof(string));
dt.Columns.Add("ID", typeof(string));
dt.Columns.Add("Msg", typeof(string));

XDocument docLog = XDocument.Load(XML_FILEPATH);

foreach (XElement log in docLog.Descendants("Log"))
{

   DateTime date = (DateTime)log.Attribute("Date");
   string severity = (string)log.Attribute("Severity");
   string id = (string)log.Attribute("id");
   string ID = (string)log.Attribute("ID");
   string msg = (string)log.Attribute("Msg");

   dt.Rows.Add(new object[] { date, severity, id, ID, msg });
}

但是显示的数据显示“ id”列的行,并重复显示“ ID”列的行。因此它会跳过/忽略“ ID”属性的实际数据/值。

the output table

如何防止这种重复并获取“ ID”属性的值并将其显示为“ ID”列下的行?

尝试通过StreamReader在表上输出结果:

 OpenFileDialog openFileDialog1 = new OpenFileDialog();

        openFileDialog1.Filter = @"All files (*.*)|*.*";
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            myLogfile = openFileDialog1.FileName;
            StreamReader sr = new StreamReader(openFileDialog1.FileName);
            var dataset = new DataSet();
            dataset.ReadXml(sr);

            var bindingSource = new BindingSource
            {
                DataSource = dataset,
                DataMember = "Log"
            };
            dataGridView1.DataSource = bindingSource;

但这还会将“ id”的值复制到“ ID”中!!

1 个答案:

答案 0 :(得分:0)

我测试了您的代码,它似乎正常工作。我在代码中发现的唯一问题是在XML文档的DateTime解析过程中,而不是:

DateTime date = (DateTime)log.Attribute("DT");

将其更改为:

DateTime date = (DateTime)log.Attribute("Date");

我在控制台应用程序上对此进行了测试。代码在这里:

using System;
using System.Data;
using System.Xml.Linq;

namespace Example
{
    class Program
    {
        static void Main(string[] args)
        {
            DataTable dt = new DataTable();
            dt.Columns.Add("Date", typeof(DateTime));
            dt.Columns.Add("Severity", typeof(string));
            dt.Columns.Add("id", typeof(string));
            dt.Columns.Add("ID", typeof(string));
            dt.Columns.Add("Msg", typeof(string));

            XDocument docLog = XDocument.Load("test.xml");

            foreach (XElement log in docLog.Descendants("Log"))
            {

                DateTime date = (DateTime)log.Attribute("Date");
                string severity = (string)log.Attribute("Severity");
                string id = (string)log.Attribute("id");
                string ID = (string)log.Attribute("ID");
                string msg = (string)log.Attribute("Msg");

                dt.Rows.Add(new object[] { date, severity, id, ID, msg });
            }
            foreach (DataRow dataRow in dt.Rows)
            {
                foreach (var item in dataRow.ItemArray)
                {
                    Console.WriteLine(item);

                }
            }
            Console.ReadLine();
        }
    }
}

进行此更改后的输出:

enter image description here

请注意,上图中此代码的输出是在调试过程中。我通过运行代码向您展示了要返回的数据表。另外,请根据您的环境更改FILEPATH。

        foreach (DataRow row in dt.Rows)
        {
            string date = row["Date"].ToString();
            string severity = row["Severity"].ToString();
            string id = row["id"].ToString();
            string ID = row["ID"].ToString();
            string msg = row["Msg"].ToString();
        }