C# structure and listview

Question

I supposed to create a program that will load salary data, consisting of name and salary, from a text file. The data will be stored in a List<> of structures, and displayed using a ListView. The contents of the List<> of structures will be sorted by either the name, or the salary.

My Load button is not working: Everytime i click the load button it gives me error message
"Empty path name is not legal"...

       struct SPatient
        {
            public string sName;
            public int iSalary;

            public SPatient(string n, int a)
            {
                sName = n;
                iSalary = a;
            }

            public override string ToString()
            {
                return string.Format("{0}, {1}", sName, iSalary);
            }
        }

        List<SPatient> m_List = new List<SPatient>();
        void SelectionSort()
        {
            int iMinimum = 0;    //index where min value found
            SPatient iTemp;    //temporary storage for swapping
            int iCurrent = 0;    //current location for selected value
            int iScan = 0;    //index to scan the unsorted array

            for (iCurrent = 0; iCurrent < m_List.Count - 1; iCurrent++)
            {
                iMinimum = iCurrent;

                for (iScan = iCurrent + 1; iScan < m_List.Count; iScan++)
                {
                    if (m_List[iScan].iSalary < m_List[iMinimum].iSalary)
                    {
                        iMinimum = iScan;
                    }
                }
                iTemp = m_List[iMinimum];
                m_List[iMinimum] = m_List[iCurrent];
                m_List[iCurrent] = iTemp;
            }
        }
        public Form1()
        {
            InitializeComponent();
        }

        private void btn_SORT_Click(object sender, EventArgs e)
        {
            if (rb_NAME.Checked)
            {
                SelectionSort();
                listView1.Items.Clear();
                foreach (SPatient p in m_List)
                    listView1.Items.Add(p.ToString());
            }
            if (rb_SALARY.Checked)
            {
                listView1.Items.Clear();
                foreach (SPatient p in m_List)
                    listView1.Items.Add(p.ToString());
            }
        }

        private void btn_LOAD_Click(object sender, EventArgs e)
        {
            try
            {
                FileStream fs = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);
                BinaryFormatter bf = new BinaryFormatter();
                m_List = (List<SPatient>)bf.Deserialize(fs);
                fs.Close();
                listView1.Items.Clear();
                foreach (SPatient p in m_List)
                    listView1.Items.Add(p.ToString());
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }

Vulpes · Answer

The error you're getting sounds to me like a serialization problem. How did you serialize the List to disk in the first place? Incidentally the instructions don't necessarily oblige you to use serialization to write the data to disk though I can see why you thought it would be convenient to do it this way as it's a technique which I often use myself.

Jonathan Crispe · Answer

i figure it out this assignment... thank you for your help....

Jonathan Crispe · Answer

This is original instruction... When the Load button is pressed the OpenFileDialog will be used to select a text file. Use a filter for text files (*.txt) or all files (*.*) with a default extension of txt. The test file will contain the salary information per person on each line of text with the name separated from the salary using a comma. Load the salary information for each person into a structure containing a string name and a double salary then add the structure to a List and display it using the ListView. If an error occurs during the file load display an error message using a MessageBox.

Vulpes · Answer

There are several possible reasons why nothing may be showing up in the ListView. However I'd first check that openFileDialog1.FileName does in fact contain the correct filepath. You can do that by placing a breakpoint on this line and then hovering your mouse over that particular expression: FileStream fs = new FileStream(openFileDialog1.FileName FileMode.Open FileAccess.Read); Alternatively just place this line immediately above the previous one : MessageBox.Show(openFileDialog1.FileName); If that seems to be working OK then the next questions are whether the file is deserializing properly to the List and whether the SPatient class has a suitable override of the ToString() method.

Jonathan Crispe · Answer

i tried using your code vulpes the openfiledialog is working but the problem is it won't show the txt file i chose in the listview....

Vulpes · Answer

It looks to me as though you haven't actually displayed the OpenFileDialog so the user can choose a file. If that's the case openFileDialog1.FileName will be null (it's default value) and you'll then get the error you described. Although the following code is unlikely to be appropriate it will give you an idea of what to do to display the OpenFileDialog: private void btn_LOAD_Click(object sender EventArgs e) { openFileDialog1.InitialDirectory = c:\ ; openFileDialog1.Filter = txt files (*.txt)|*.txt|All files (*.*)|*.* ; openFileDialog1.FilterIndex = 2 ; openFileDialog1.RestoreDirectory = true ; if(openFileDialog1.ShowDialog() == DialogResult.OK) { try { FileStream fs = new FileStream(openFileDialog1.FileName FileMode.Open FileAccess.Read); BinaryFormatter bf = new BinaryFormatter(); m_List = (List )bf.Deserialize(fs); fs.Close(); listView1.Items.Clear(); foreach (SPatient p in m_List) listView1.Items.Add(p.ToString()); } catch (Exception ex) { MessageBox.Show(ex.Message); } } }

Sam Hobbs · Answer

You say  it gives me error message  (emphasis added). What is it ? What line of code does the error occur on?