列表不增长或无法向asp.net中的列表添加新元素

Question

I have created a custom type Food which has 4 attributes name , weight , fat , and calorie . The web application will get name and weight from users and calculate the fat and calorie .

I have an Add button to click in order to add a choice to a list.

My problem is when I display the message box to check the list, it always show 1 , no matter how many you have added. Can anyone point out where I am wrong?

P/S: I tried to display the message box before I add an item to a list and it shows 0 <-- this is true.

    List<Food> foodList = new List<Food>();
    Food temp = new Food();

    public void addButton_Click(object sender, EventArgs e)
    {
      //Get weight value from a textbox
        temp.weight = Convert.ToDouble(weightTextBox.Text);
      //Get name from a provided dropdownlist
        temp.name = DropDownList1.Text;
      //Add to list
        foodList.Add(temp);
      //Check the number of elements in the list
        ClientScript.RegisterStartupScript(this.GetType(), "myalert", "alert('" + foodList.Count + "');", true);
    }

Answer 1

Every time your page is instantiated, it has an empty foodList to which you add exactly one item in addButton_Click . And a new instance of the page is created to handle each request from the browser, such as request caused by addButton click. You have to implement server-side storage of added items and load them into foodList every time.

Answer 2

You should move your temp object creation into addButton_Click . Like this:

List<Food> foodList = new List<Food>();

public void addButton_Click(object sender, EventArgs e)
{
    Food temp = new Food();
  //Get weight value from a textbox
    temp.weight = Convert.ToDouble(weightTextBox.Text);
  //Get name from a provided dropdownlist
    temp.name = DropDownList1.Text;
  //Add to list
    foodList.Add(temp);
  //Check the number of elements in the list
    ClientScript.RegisterStartupScript(this.GetType(), "myalert", "alert('" + foodList.Count + "');", true);
}

Else you're adding the same object all the time.