CODEWITHSUNDEEP
c#asp.netarraysdataset
MrCarder
MrCarder

Reputation: 438

Creating an array from scratch while code is running

I am trying to create a list via array like this:

private Application[] GetApps()
{
DataSet ds = new Dataset();
string query = "query";
ds = GetData(query);

var appList = new Application[ds.Tables[0].Rows.Count];

for(var i = 0; i <= ds.Tables[0].Rows.Count - 1; i++)
{
                DataRow item = ds.Tables[0].Rows[i];

                appList[i].Name = item["Name"].ToString();
                appList[i].Confidentiality = int.Parse(item["Conf"].ToString());
                appList[i].Id = item["ID"].ToString();
}
return appList;
}

I keep getting an object null error and I know I have to be missing something completely obvious that I'm just not seeing. Do I need to declare the new array in some other way?

Upvotes: 0

Views: 239

Answers (1)

Jason Watkins
Jason Watkins

Reputation: 3785

When you create appList, you are only creating the array itself. .NET does not automatically populate the array with Application objects for you to manipulate. You need to create a new Application object, and set the properties on that object, then you can assign the object to the array.

There are multiple Application classes withing the .NET framework, none of which seem to match your code, so the below example will simply assume that Application is a custom type of your own design.

for(var i = 0; i < ds.Tables[0].Rows.Count; i++)
{
    DataRow item = ds.Tables[0].Rows[i];

    Appliction app = new Application();
    app.Name = item["Name"].ToString();
    app.Confidentiality = int.Parse(item["Conf"].ToString());
    app.Id = item["ID"].ToString();
    appList[i] = app
}

As an aside, note that you can replace i <= x - 1 with i < x and the behavior is exactly the same.

Finally, you should introduce checks for all of your accessors if there is a chance that they could return null. For example, if item["Name"] returns null, then calling item["Name"].ToString() is equivelant to calling null.ToString(), which will also result in a NullReferenceException.

Upvotes: 6

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