Re: [PHP] anyway to return more than one value?

[Boyan Nedkov]

One possible solution could be to write a simple class that will hold 
all data you need at one single place:

class MyData {
  var $_resultArray;
  var $_rows;
  var $_fields;
  function MyData($resultArray, $rows, $fields) {
    $this->data = $data;
    $this->rowcounr = $rowcount;
    $this->fildcount = $fildcount;
  }
  function getResultArray() {
    return $this->_resultArray;
  }
  function getRows() {
    return $this->_rows;
  }
  function getFilds () {
    return $this->_fields;
  }
}
To put some data in that object write:

$data = new MyData($theResultArray, $rows, $fields);

To get back your stuff in your code use something like:

$theResultArray = MyData->getResultArray();
$rows =  MyData->getRows();
$fields = MyData->getFields();
This is only an example code, you should adjust it according your needs

hth

Boyan
--
Chris W. Parker wrote:

Hi list.

Ok I know it's not possible to "return" more than one value. But I'm
going to explain what I'd like to do so maybe there's an easy way to do
it.
I've got some functions that query a database and turn the result into
an array and return that array. What I'd like to do is not only return
the array of results but ALSO return a row count and field count of that
result.
Here is some pseudo code:

function get_results()
{
// query database
        $result = query($sql);
        
// turn result set into a useable array
        $theResultArray = get_results($result);
        $rows = mysql_num_rows($result);
        $fields = mysql_num_fields($result);
        return $theResultArray;
        return $rows;
        return $fields;
}
Ok I know that won't work but that's just basically what I want to do.

The only way around this I've come up with is to stick all the values
into ANOTHER array and return then and then dissect that array in the
calling function, but that just seems messy.
All of a sudden the word "reference" came to mind. Is that what I want
to use?
Any help would be appreciated.

Thanks,
Chris.
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