RE: [PHP] References
| From: | Neil Kimber | Date: | Tue, 07 Nov 2000 20:57:24 +0000 |
| Subject: | RE: [PHP] References | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-24187@lists.php.net to get a copy of this message | ||
Just had a thought that isn't so off topic. You may want to try the
following.
1. Create a class
2. Store your result set in the class
3. Provide a method to access the db
4. provide a method to return the resultset
5. instantiate your class from whereever you desire.
So, you'll get something like:
class myDataClass{
var $result;
function call_db(){
// Db connections etc goes here...
$sql = "select * from hugetable";
$this->result = mysql_query($sql);
}
function get_RecordSet(){
return $this->result;
}
} // End class myDataClass
/// Now in your normal code you can instantiate the object thus:
$DbObj = new myDataClass();
$DbObj->call_db();
$ref = $DbObj->get_RecordSet();
This should do you fine as the memory address will exist for as long as the
object exists.
-----Original Message-----
From: Vidyut Luther [mailto:vluther@phpcult.com]
Sent: 07 November 2000 19:44
To: php-general@lists.php.net
Subject: [PHP] References
I was wondering why this is not possible.
function get_a_lot_of_data() {
$sql = "select * from hugetable";
$result = mysql_query($sql);
return &result;
}
I have to do this all the time,
$ref = &$result;
return $ref;
why do i have to create the reference first, and then pass it ? i have no
formal education in programming.. so any of you computer sci majors wanna
clear this up for me ?
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