获取字符串中唯一元素的位置[]

Question

我有一个 xml 文件，我正在访问该文件以创建项目所用时间的报告。我将唯一日期返回到在 winform 上动态创建的标签，并希望编译每个唯一日期在项目上花费的时间。我已经能够退回每个日期下的所有项目或仅一个项目。目前我坚持只返回一个项目。谁能帮帮我吗？？如果数据正确，这就是数据的样子。

04/11/15
    26820   2.25
    27111   8.00
04/12/15
    26820   8.00
04/13/15
    01det   4.33
    26820   1.33
    27225   4.25

etc.

这就是我检索数据的方式

         string[] weekDateString = elementDateWeekstring();
         string[] uniqueDates = null;
         string[] weeklyJobNumber = elementJobNumWeek();
         string[] weeklyTicks = elementTicksWeek();

这就是我获取唯一日期的方式。

         IEnumerable<string> distinctWeekDateIE = weekDateString.Distinct();

         foreach (string d in distinctWeekDateIE)
         {
             uniqueDates = distinctWeekDateIE.ToArray();
         }

这就是我创建标签的方式。

        try
         {
              int dateCount;
              dateCount = uniqueDates.Length;

              Label[] lblDate = new Label[dateCount];
              int htDate = 1;
              int padDate = 10;

              for (int i = 0; i < dateCount; i++ )
              {
                   lblDate[i] = new Label();

                   lblDate[i].Name = uniqueDates[i].Trim('\r');

                   lblDate[i].Text = uniqueDates[i];

                   lblDate[i].TabIndex = i;

                   lblDate[i].Bounds = new Rectangle(18, 275 + padDate + htDate, 75, 22);

                   targetForm.Controls.Add(lblDate[i]);

                   htDate += 22;

                   foreach (string x in uniqueDates)
                   {
                        int[] posJobNumber;

                       posJobNumber = weekDateString.Select((b, a) => b == uniqueDates[i].ToString() ? a : -1).Where(a => a != -1).ToArray();

                        for (int pjn = 0; pjn < posJobNumber.Length; pjn++)
                        {

                             if (x.Equals(lblDate[i].Text))
                             {

                                  Label lblJobNum = new Label();

                                  int htJobNum = 1;
                                  int padJobNum = 10;

                                  lblJobNum.Name = weeklyJobNumber[i];

                                  lblJobNum.Text = weeklyJobNumber[i];

                                  lblJobNum.Bounds = new Rectangle(100, 295 + padJobNum + htJobNum, 75, 22);

                                  targetForm.Controls.Add(lblJobNum);

                                  htJobNum += 22;
                                  htDate += 22;
                                  padJobNum += 22;
                             }
                        }

                     }
              }
         }

我已经坚持了大约 3 个月。有没有人可以向我描述为什么我无法正确检索与特定日期相关的工作编号。我不相信这些是专门作为日期返回的。只是一个看起来像日期的字符串。

我非常感谢我能得到的任何帮助。我完全感到困惑。感谢您提前回复。我非常感谢您的帮助。

编辑：@Sayka - 这是 xml 示例。

<?xml version="1.0" encoding="utf-8"?>
<Form1>
  <Name Key="4/21/2014 6:51:17 AM">
    <Date>4/21/2014</Date>
    <JobNum>26820</JobNum>
    <RevNum>00000</RevNum>
    <Task>Modeling Secondary</Task>
    <Start>06:51 AM</Start>
    <End>04:27 PM</End>
    <TotalTime>345945089017</TotalTime>
  </Name>
  <Name Key="4/22/2014 5:44:22 AM">
    <Date>4/22/2014</Date>
    <JobNum>26820</JobNum>
    <RevNum>00000</RevNum>
    <Task>Modeling Secondary</Task>
    <Start>05:44 AM</Start>
    <End>06:56 AM</End>
    <TotalTime>43514201221</TotalTime>
  </Name>
  <Name Key="4/22/2014 6:57:02 AM">
    <Date>4/22/2014</Date>
    <JobNum>02e-n-g</JobNum>
    <RevNum>00000</RevNum>
    <Task>NET Eng</Task>
    <Start>06:57 AM</Start>
    <End>07:16 AM</End>
    <TotalTime>11706118875</TotalTime>
  </Name>
....
</Form1>

这就是我从 xml 文件中获取信息并返回字符串 [] 的方式。

    public static string[] elementDateWeekstring()
    {
        //string datetxtWeek = "";
        XmlDocument xmldoc = new XmlDocument();

        fileExistsWeek(xmldoc);

        XmlNodeList nodeDate = xmldoc.GetElementsByTagName("Date");

        int countTicks = 0;
        string[] dateTxtWeek = new string[nodeDate.Count];


        for (int i = 0; i < nodeDate.Count; i++)
        {
            dateTxtWeek[i] = nodeDate[i].InnerText;
            countTicks++;

        }
        return dateTxtWeek;
    }

Job number 和 Ticks 以类似的方式返回。我已经能够在整个代码中重用这些 sn-ps。这是一个一维的xml文件？？它将始终返回与日期或刻度相等的工作编号的位置。我永远不会有更多或更少的任何一种元素。

Answer 1

您可以使用 Linq-to-XML 解析 XML 文件，然后使用 Linq-to-objects 按作业日期对数据进行分组（和排序），并按作业名称对每个组进行排序。

解析XML文件的代码如下：

var doc = XDocument.Load(filename);
var jobs = doc.Descendants("Name");

// Extract the date, job number, and total time from each "Name" element.:

var data = jobs.Select(job => new
{
    Date = (DateTime)job.Element("Date"),
    Number = (string)job.Element("JobNum"),
    Duration = TimeSpan.FromTicks((long)job.Element("TotalTime"))
});

按日期对作业进行分组和排序并按作业名称对组进行排序的代码是：

var result = 
    data.GroupBy(job => job.Date).OrderBy(g => g.Key)
    .Select(g => new
    {
        Date = g.Key, 
        Jobs = g.OrderBy(item => item.Number)
    });

然后您可以通过遍历result 中的每个组来访问数据，然后遍历组中的每个作业，如下所示：

foreach (var jobsOnDate in result)
{
    Console.WriteLine("{0:d}", jobsOnDate.Date);

    foreach (var job in jobsOnDate.Jobs)
        Console.WriteLine("    {0}  {1:hh\\:mm}", job.Number, job.Duration);
}

将所有这些放在一个示例可编译控制台应用程序中（根据需要替换 XML 文件的文件名）：

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

namespace ConsoleApplication2
{
    class Program
    {
        private static void Main()
        {
            string filename = @"d:\test\test.xml"; // Substitute your own filename here.

            // Open XML file and get a collection of each "Name" element.            

            var doc = XDocument.Load(filename);
            var jobs = doc.Descendants("Name");

            // Extract the date, job number, and total time from each "Name" element.:

            var data = jobs.Select(job => new
            {
                Date = (DateTime)job.Element("Date"),
                Number = (string)job.Element("JobNum"),
                Duration = TimeSpan.FromTicks((long)job.Element("TotalTime"))
            });

            // Group the jobs by date, and order the groups by job name:

            var result = 
                data.GroupBy(job => job.Date).OrderBy(g => g.Key)
                .Select(g => new
                {
                    Date = g.Key, 
                    Jobs = g.OrderBy(item => item.Number)
                });

            // Print out the results:

            foreach (var jobsOnDate in result)
            {
                Console.WriteLine("{0:d}", jobsOnDate.Date);

                foreach (var job in jobsOnDate.Jobs)
                    Console.WriteLine("    {0}  {1:hh\\:mm}", job.Number, job.Duration);
            }
        }
    }
}

Answer 2

输出是这样的

创建一个新项目

将表单尺寸设置得更大。

应用这些代码。

为您的 XML 文件设置位置。

命名空间

using System.Xml;
using System.IO;

表单代码

public partial class Form1 : Form
{
    const string XML_FILE_NAME = "D:\\emps.txt";
    public Form1()
    {
        InitializeComponent();
    }

    private void Form1_Load(object sender, EventArgs e)
    {
        prepareDataGrid();
        List<JOBS> jobsList = prepareXML(XML_FILE_NAME);
        for (int i = 0; i < jobsList.Count; i++)
        {
            addDateRow(jobsList[i].jobDate.ToString("M'/'d'/'yyyy"));
            for (int j = 0; j < jobsList[i].jobDetailsList.Count; j++)
                dgv.Rows.Add(new string[] { 
                    jobsList[i].jobDetailsList[j].JobNumber,
                    jobsList[i].jobDetailsList[j].JobHours
                });
        }
    }

    DataGridView dgv;
    void prepareDataGrid()
    {
        dgv = new DataGridView();
        dgv.BackgroundColor = Color.White;
        dgv.GridColor = Color.White;
        dgv.DefaultCellStyle.SelectionBackColor = Color.White;
        dgv.DefaultCellStyle.SelectionForeColor = Color.Black;
        dgv.DefaultCellStyle.ForeColor = Color.Black;
        dgv.DefaultCellStyle.BackColor = Color.White;
        dgv.DefaultCellStyle.Alignment = DataGridViewContentAlignment.MiddleRight;
        dgv.Width = 600;
        dgv.Dock = DockStyle.Left;
        this.BackColor = Color.White;
        dgv.Columns.Add("Col1", "Col1");
        dgv.Columns.Add("Col2", "Col2");
        dgv.Columns[0].Width = 110;
        dgv.Columns[1].Width = 40;
        dgv.DefaultCellStyle.Font = new System.Drawing.Font("Segoe UI", 10);
        dgv.RowHeadersVisible = dgv.ColumnHeadersVisible = false;
        dgv.AllowUserToAddRows =
        dgv.AllowUserToDeleteRows =
        dgv.AllowUserToOrderColumns =
        dgv.AllowUserToResizeColumns =
        dgv.AllowUserToResizeRows =
        !(dgv.ReadOnly = true);
        Controls.Add(dgv);
    }

    void addJobRow(string jobNum, string jobHours)
    {
        dgv.Rows.Add(new string[] {jobNum, jobHours });
    }

    void addDateRow(string date)
    {
        dgv.Rows.Add(new string[] { date, ""});
        dgv.Rows[dgv.Rows.Count - 1].DefaultCellStyle.SelectionForeColor =
        dgv.Rows[dgv.Rows.Count - 1].DefaultCellStyle.ForeColor = Color.Firebrick;
        dgv.Rows[dgv.Rows.Count - 1].DefaultCellStyle.Font = new Font("Segoe UI Light", 13.5F);
        dgv.Rows[dgv.Rows.Count - 1].DefaultCellStyle.Alignment = DataGridViewContentAlignment.MiddleLeft;
        dgv.Rows[dgv.Rows.Count - 1].Height = 25;
    }

    List<JOBS> prepareXML(string fileName)
    {
        string xmlContent = "";
        using (FileStream fs = new FileStream(fileName, FileMode.Open, FileAccess.Read))
        using (StreamReader sr = new StreamReader(fs)) xmlContent = sr.ReadToEnd();
        XmlDocument doc = new XmlDocument();
        doc.LoadXml(xmlContent);

        List<JOBS> jobsList = new List<JOBS>();
        XmlNode form1Node = doc.ChildNodes[1];
        for (int i = 0; i < form1Node.ChildNodes.Count; i++)
        {
            XmlNode dateNode = form1Node.ChildNodes[i].ChildNodes[0].ChildNodes[0],
                jobNumNode = form1Node.ChildNodes[i].ChildNodes[1].ChildNodes[0],
                timeTicksNode = form1Node.ChildNodes[i].ChildNodes[6].ChildNodes[0];

            bool foundDate = false;
            for (int j = 0; j < jobsList.Count; j++) if (jobsList[j].compareDate(dateNode.Value))
                {
                    jobsList[j].addJob(jobNumNode.Value, Math.Round(TimeSpan.FromTicks(
                        (long)Convert.ToDouble(timeTicksNode.Value)).TotalHours, 2).ToString());
                    foundDate = true;
                    break;
                }
            if (!foundDate)
            {
                JOBS job = new JOBS(dateNode.Value);
                string jbnum = jobNumNode.Value;
                string tbtck = timeTicksNode.Value;
                long tktk = Convert.ToInt64(tbtck);
                double tkdb = TimeSpan.FromTicks(tktk).TotalHours;
                job.addJob(jobNumNode.Value, Math.Round(TimeSpan.FromTicks(
                        Convert.ToInt64(timeTicksNode.Value)).TotalHours, 2).ToString());
                jobsList.Add(job);
            }
        }
        jobsList.OrderByDescending(x => x.jobDate);
        return jobsList;
    }

    class JOBS
    {
        public DateTime jobDate;
        public List<JobDetails> jobDetailsList = new List<JobDetails>();

        public void addJob(string jobNumber, string jobHours)
        {
            jobDetailsList.Add(new JobDetails() { JobHours = jobHours, JobNumber = jobNumber });
        }
        public JOBS(string dateString)
        {
            jobDate = getDateFromString(dateString);
        }

        public JOBS() { }

        public bool compareDate(string dateString)
        {
            return getDateFromString(dateString) == jobDate;
        }

        private DateTime getDateFromString(string dateString)
        {
            string[] vals = dateString.Split('/');
            return new DateTime(Convert.ToInt32(vals[2]), Convert.ToInt32(vals[0]), Convert.ToInt32(vals[1]));
        }
    }

    class JobDetails
    {
        public string JobNumber { get; set; }
        public string JobHours { get; set; }
    }
}