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using an Array in a Switch Statement

Posted on 2006-06-12
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Last Modified: 2010-04-16
I have this code:
        string strSessionPageID = Session["SessionPageID"].ToString();

        switch (strSessionPageID)
        {
            case "3":
                //do x
                break;
            case "5":
                //do y
                break;
        }

what I'd like to be able to do is set up arrays, and use the switch like this:

        string strSessionPageID = Session["SessionPageID"].ToString();

        array xArr = array("3","43","56","68");
        array yArr = array("34","42","51","63");

        switch (strSessionPageID)
        {
            case xArr:
                //do x
                break;
            case yArr:
                //do y
                break;
        }

where I use the ID in the switch, and check it against the arrays to determine the case to perform.

Is there a way to do this, and if so, might I have an example?
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Question by:Paul Kahl
2 Comments
 
LVL 11

Accepted Solution

by:
Expert1701 earned 500 total points
ID: 16887293
If your arrays are static (i.e. can be hard-coded):

  switch (strSessionPageID)
  {
    case "3":
    case "43":
    case "56":
    case "68":
      {
        //do x
        break;
      }
    case "34":
    case "42":
    case "51":
    case "63":
      {
        //do x
        break;
      }
    default:
      {
        break;
      }
  }

If your arrays are dynamic (i.e. determined at run-time), you can do something similar to the following.  You would likely use an ArrayList or List<string> or even a SortedList to store your reference x and y values.

  string[] xArr = new string[] { "3", "43", "56", "68" };
  string[] yArr = new string[] { "34", "42", "51", "63" };

  foreach (string x in xArr)
    if (strSessionPageID == x)
    {
      //do x
      break;
    }

  foreach (string y in yArr)
    if (strSessionPageID == y)
    {
      //do y
      break;
    }
0
 
LVL 3

Author Comment

by:Paul Kahl
ID: 16887448
That looks like it will work fantastically well. Thank you very much.
0

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