PHP mySQL resource query?

[ziycon]

I have this first piece of code which works fine but I'm trying to break it up into two parts like in the second code snippet below.

$tables = mysql_query("SHOW tables;");
while($row = mysql_fetch_array($tables)) {
    $tableNames .= "<a href=\"".$_SERVER['PHP_SELF'].">".$row[0]."</a><br/>\n";
    $tableCount++;
}

So it passes the sql query to the function and the function runs the query and should return the resource and store it in $tables variable and then the while loop should run through the $tables variable the same as in the first code snippet but it doesn't seem too work, its only outputting the first table name twice and there is over 100 tables!?

$tables = $this->dbQuery("SHOW tables;");
while($tables) {
    $tableNames .= "<a href=\"".$_SERVER['PHP_SELF'].">".$table."</a><br/>\n";
    $tableCount++;
}
/******************************************/
function dbQuery($queryIn) {
    $resource;
 
    switch ($_SESSION['dbtype']) {
        case 0:
            $sql = mysql_query($queryIn);
            $resource = mysql_fetch_array($sql);
            break;
    }
 
    return $resource;
}

[Dormilich]

personally, I’d use foreach() like

// use CSS to make it a block element!
$format = '<a href="'.$_SERVER['SCRIPT_NAME'].">%s</a>" . PHP_EOL;
$tables = $this->dbQuery("SHOW tables;");
foreach ($tables as $table) 
{
    $tableNames .= sprintf($format, $table);
}

for which you need PDO (MySQLi as of PHP 5.4)

why your code (snippet) doeasn’t work is
- $table is undefined
- $tables doesn’t change
- $tableCount does not influence anything