C# ListView Problem Adding Items

Question

So, here's my question: Why won't the code in the first snippet work when the second one works fine. Also, I have set the view property to details. I've read all over how to add lvi's to the listview, and it fails every time... except for then I do it manually.

So, this doesn't work...

// Iterating through the rows...
for (int x = 0; x < numRows; x++) {
    row = new List<string>();
    // Iterating through the cols...
    for (int y = 0; y < numCols; y++) {
        row.Add(data[y][x]);
    }
    lv.Items.Add(new ListViewItem(row.ToArray()));
}

But this will work:

lv.Items.Add(new ListViewItem("foo"));

Can you define "fails"? Also - perhaps check you have the array indexes the right way around... — Marc Gravell
– Marc Gravell, Commented May 21, 2009 at 20:44
In what way does it not work? I just tried it here and it works fine. — Fredrik Mörk
– Fredrik Mörk, Commented May 21, 2009 at 20:45

John Feminella · Accepted Answer · 2009-05-21 20:45:21Z

1

row.Add(data[y][x]) seems suspicious. Why do you access the data in column-first order but iterate in row-first order? Also, make sure the type of row (you didn't tell us this) is actually List<string>.

answered May 21, 2009 at 20:45

John Feminella

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Larry Hipp · Accepted Answer · 2009-05-21 20:44:48Z

0

The ListViewItem is looking for a String[] try casting row.ToArray() to a String[].

answered May 21, 2009 at 20:44

Larry Hipp

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5 Comments

Marc Gravell Over a year ago

Assuming that row is typed as List<string> (per the init), then row.ToArray() is already a string[]

Fredrik Mörk Over a year ago

Since row is declared as List<string> ToArray is already coming as string[]; no need to cast.

Marc Gravell Over a year ago

@Fredrik - actually, the row declaration isn't shown... we might assume it is List<string>...

Fredrik Mörk Over a year ago

I assumed based on this code in the sample: row = new List<string>();

Larry Hipp Over a year ago

I just ran the code from above with some quick test data and it worked just fine. What error are you seeing?

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