Dennis Gaya, Nairobi, Kenya
The error shows that the resulting array values for $i are undefined. eg. 4=1,2,3,4.. ETC.
Here is a quick Solution
   function DataFetch(){
        //Suggestions: The Table names can be changed according to the User. But be carefull with the Query and other variables
        // So that the Script will function Properly
        $this->typeSql="select * from e_categories";    //Query to fetch the Data from the table "type"
        $this->typeResult=mysql_query($this->typeSql); //Execute the Query and Store the result in new variable
        $numrows=mysql_num_rows($this->typeResult);
        // Start fetching the Values from the Result Resource (typeResult)
        while ( $this->typeRow = mysql_fetch_array($this->typeResult,MYSQL_BOTH)){ // TYPE table Fetch While Loop begins Here
			for($i=0;isset($this->typeRow[$i]);$i++){
			//for($i=0; isset($sponsorheading[$i]); $i++){
                  $this->arrType[$this->IDx][$i]= $this->typeRow[$i];   // Store the Value of the Type table to the Array
            }
Regards
Dennis Gaya, Nairobi, Kenya
[email protected]