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Village Idiot · 06-13-2010, 07:00 PM

Quote:

Originally Posted by pipesportugal

PHP Code:


			
// class name is record
// number of records is 1.
// vector_fields is an array with the fieldnames in the table 
// The field names are like: code_customer, name_customer, and so on.

//*************************************
while ($row2 = mysql_fetch_array($this->result_sql)) {
  for($j= 0; $j<count($this->vector_fields); $j++) {
     $firstpartstring = $this->vector_fields[$j];
     $stringfieldname = 'scr_'.$firstpartstring;
     $this->$stringfieldname = $row2['$firstpartstring'];
  } // END of for
} // END of while

//*************************************
// From the main program I want to retrieve the name of the customer with:
//  $customer = new record();
//  $customer_invoice = $customer->scr_name_customer;

I am geting the following error:
Catchable fatal error: Object of class registo could not be converted to string in /home/vnhdpwxq/public_html/classes/classe_ler_registos.php on line 68

I believe it is because of the sentence:

PHP Code:


			
$this->$stringfieldname

The program is not expecting the "$" on the stringfieldname variable.
What I really need is to make a $this->scr_name_customer.

Can someone help me overpass this problem?

I have completely run out of ideas.

Thanks in advance,
PP

It should be

PHP Code:


			
$this->stringfieldname

. I don't understand what you mean by "What I really need is to make a $this->scr_name_customer."