Re: [Phplib-users] DB_Sql question

[Andrew C. <ph...@ev...>]

Howdy,

I haven't spent a lot of time dissecting the code in DB_Sql but, my
understanding is that DB_Sql's connect() is an *internal* instance
method, not intended for use like that.  I would probably do the same
thing by declaring a separate DB_Sql extension class for each database.
 That would be something like this:

    class DB_Example extends DB_Sql {
      var $classname = "db_example";
      var $Host     = "localhost";
      var $Database = "DB1";
      var $User     = "example";
      var $Password = "password";
      var $Auto_Free = "true";
      var $Halt_On_Error = "report";
    }

    class DB_Example2 extends DB_Sql {
      var $classname = "db_example2";
      var $Host     = "localhost";
      var $Database = "DB2";
      var $User     = "example2";
      var $Password = "password2";
      var $Auto_Free = "true";
      var $Halt_On_Error = "report";
    }

Then, you could do basically the same thing this way:

    $exampledb = new DB_Example;
    $query = "SELECT name,id FROM FILE WHERE datatype='folder'";
    $exampledb->query($query);
    while($exampledb->next_record()) {
        echo $exampledb->f(0) . "[" . $exampledb->f(1) . "]<br />\n";
    }


    $exampledb2 = new DB_Example2;
    $query2 = "SELECT name,id FROM FILE WHERE datatype='folder'";
    $exampledb2->query($query2);
    while($exampledb2->next_record()) {
        echo $exampledb->f(0) . "[" . $exampledb->f(1) . "]<br />\n";
    }

    $exampledb->query($query);
    while($exampledb->next_record()) {
        echo $exampledb->f(0) . "[" . $exampledb->f(1) . "]<br />\n";
    }

This definitely works.

Andrew Crawford

Kelby Zorgdrager wrote:
> Folks:
> 
> Gotta question for you. I am experiencing a unexpected problem when 
> using the DB_Sql class. It appears that if I create a DB_Sql object  to
> one database (and perform a query) then create a second DB_Sql  object
> to another database (and perform a query), when I return to  the first
> object, it's connection state is invalid. Due to the  architecture of
> our app, I can not consolidate the two objects into  one (that would be
> reused).
> 
> I have attached a sample as an example to show the problem.
> 
> Assume for the example there are:
> --> two databases, DB1 and DB2
> --> a table in each db (DB1.FILE and DB2.USER)
> 
> I have attached two SQL files for the tables along with the example. 
> Any insight into why I am seeing this would be greatly appreciated.
> 
> Thanks.
> 
> KZ
> 
> 
> ------------------------------------------------------------------------
> 
> <?php
>   include_once ("db_mysql.inc");
> 
>   $dbConnection = new DB_Sql;
>   $dbConnection->Database = $dbName;
>   if (!$dbConnection->connect("DB1")) {
>       $error = "Failed to connect to $dbName";
>       echo "Failed to initialize: $error<br>";
>       exit;
>   }
> 
>   $dbConnection->query("SELECT name,id FROM FILE WHERE datatype='folder'");
>   while($dbConnection->next_record()) {
>     echo "<BR>".$dbConnection->f(0)."[".$dbConnection->f(1)."]";
>   }
> 
>   $dbConnection2 = new DB_Sql;
>   $dbConnection2->connect("DB2");
>   $dbConnection->query("SELECT name,id FROM USER");
>   while($dbConnection->next_record()) {
>     echo "<BR>".$dbConnection->f(0)."[".$dbConnection->f(1)."]";
>   }
> 
>   /*
>   * when this is queried, it returns nothing.. even though it should
>   * return the same 'set' as the first time it was queried above.
>   */
>   $dbConnection->query("SELECT name,id FROM FILE WHERE datatype='folder'");
>   while($dbConnection->next_record()) {
>     echo "<BR>".$dbConnection->f(0)."[".$dbConnection->f(1)."]";
>   }
> ?>