Why don't you just use Vector?
--------
Vector v_queryResult = New Vector();
...
if(tmpString.indexOf('C')!=-1)
{
        v_queryResult.addElement((Object)tmpString);
}
--------
Then there's no need for Array2.


> -----Original Message-----
> From: Shirley Chen [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, December 20, 2000 11:39 AM
> To: [EMAIL PROTECTED]
> Subject: arrays
>
>
> Hi,
>
> I have a question, somebody please help me:
>
> I have a vector which will keep all the names the user typed
> in.  Then the
> user supposes can do a query like "give me all the name which
> starts with
> 'C'".  In order to do that, I do something like this:
>
> ...
> String[] array1 = new String[myVector.size()];
> String tempString = "";
> int j=0;
>
> for (int i=0; i<myVector.size(); i++)
> {
>    tmpString = (String)myVector.get(i);
>    if (tmpString.indexOf('C') != -1)
>       {
>         array1[j] = tmpString;
>         j++;
>       }
> }
>
> //since array1 may not be filled up with data, I transfer it
> to another
> String array so I know how many element in the array
>
> String array2 = new String[j];
> for (int i=0; i<j; i++)
> {
>    array2[i] = array1[i];
> }
>
> // then I will do display. Supposedly, say, array2.length is
> 20, if the user
> wants to display first 5, I will display only the first 5
> records, if the
> user wants to display 30, and I will display 20 instead.
> ...
>
> I can display all the situations ok except when myVector
> contains the names
> that all start with "C" (which means array2.length should be
> myVector.size()). then the program will crash.  I think it
> might be array
> out of bound problem, but I don't know how to fix it. Please help me.
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