how to count same names in textfile

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I wanna create text file containing one name on each line. Compute the number of times any name occurs. Output one line for each name in file and on each line print the number of occurrences followed by name.

and i would like to diplay the answer in messageBox

this is what i have done so far

C#

var nameCount = new Dictionary<string, int>();

          foreach (String s in File.ReadAllLines(openFileDialog1.FileName))
          {
              if (nameCount.ContainsKey(s))
              {
                  nameCount[s] = nameCount[s] + 1;
              }
              else
              {
                  nameCount.Add(s, 1);
              }
          }
          var output = new StringBuilder();
          foreach (var pair in nameCount(openFileDialog1.FileName))
          {
              MessageBox.Show("{0} {1}\n", pair.Key, pair.Value);
          }

but i still have some error

Posted 26-Jul-12 4:22am

ahmedalialshehri

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[no name] 26-Jul-12 10:25am

"but i still have some error"... does not tell us anything about whatever problem you are having,

Kaushik Saha from Kolkata,India 26-Jul-12 10:25am

Where And which error occur

Kaushik Saha from Kolkata,India 26-Jul-12 10:37am

what you need explain with some content of your files

ahmedalialshehri 26-Jul-12 10:55am

Could not find file 'C:\Users\acer\AppData\Local\Temporary Projects\FourthProgramm\bin\Debug\openFileDialog1'.

this is the error

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OriginalGriff · Answer 1 · 2012-07-26T04:38:00

The problem is in the last part:

C#

foreach (var pair in nameCount(openFileDialog1.FileName))
{
    MessageBox.Show("{0} {1}\n", pair.Key, pair.Value);
}

1) You don't need the (openFileDialog1.FileName) bit.
2) You need to format your string before you can hand it to a messgae box:

C#

foreach (var pair in nameCount)
    {
    MessageBox.Show(String.Format("{0} {1}\n", pair.Key, pair.Value));
    }

But I would do it slightly differently:

C#

foreach (string key in nameCount.Keys)
    {
    MessageBox.Show(string.Format("{0} {1}\n", key, nameCount[key]));
    }

And I really hope you don't have a lot of names, or you will spend all afternoon pressing the "OK" button!

"Could not find file 'C:\Users\acer\AppData\Local\Temporary Projects\FourthProgramm\bin\Debug\openFileDialog1'."

Did you call ShowDialog on the OpenFileDialog at any time?
Normally it would be:

C#

private void button1_Click(object sender, EventArgs e)
    {
    OpenFileDialog openFileDialog1 = new OpenFileDialog();
    if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
        var nameCount = new Dictionary<string, int>();

        foreach (String s in File.ReadAllLines(openFileDialog1.FileName))
            if (nameCount.ContainsKey(s))
                {
                nameCount[s] = nameCount[s] + 1;
                }
            else
                {
                nameCount.Add(s, 1);
                }
        var output = new StringBuilder();
        foreach (var pair in nameCount)
            {
            MessageBox.Show(String.Format("{0} {1}\n", pair.Key, pair.Value));
            }
        }
    }