Handling results from MySQLi query after MySql to MySqli conversion - php

Question

I am trying to change the following code to use MySqli instead of MySql. I have removed some methods that seem unimportant to what I'm addressing here.
class db {
var $hostname,
$database,
$username,
$password,
$connection,
$last_query,
$last_i,
$last_resource,
$last_error;
function db($hostname=DB_HOSTNAME,$database=DB_DATABASE,$username=DB_USERNAME,$password=DB_PASSWORD) {
$this->hostname = $hostname;
$this->database = $database;
$this->username = $username;
$this->password = $password;
$this->connection = mysql_connect($this->hostname,$this->username,$this->password) or $this->choke("Can't connect to database");
if($this->database) $this->database($this->database);
}
function database($database) {
$this->database = $database;
mysql_select_db($this->database,$this->connection);
}
function query($query,$flag = DB_DEFAULT_FLAG) {
$this->last_query = $query;
$resource = mysql_query($query,$this->connection) or $this->choke();
list($command,$other) = preg_split("|\s+|", $query, 2);
// Load return according to query type...
switch(strtolower($command)) {
case("select"):
case("describe"):
case("desc"):
case("show"):
$return = array();
while($data = $this->resource_get($resource,$flag)) $return[] = $data;
//print_r($return);
break;
case("replace"):
case("insert"):
if($return = mysql_insert_id($this->connection))
$this->last_i = $return;
break;
default:
$return = mysql_affected_rows($this->connection);
}
return $return;
}
function resource_get($resource = NULL,$flag = DB_DEFAULT_FLAG) {
if(!$resource) $resource = $this->last_resource;
return mysql_fetch_array($resource,$flag);
}
}
This is what I've got so far:
class db {
var $hostname = DB_HOSTNAME,
$database = DB_DATABASE,
$username = DB_USERNAME,
$password = DB_PASSWORD,
$connection,
$last_query,
$last_i,
$last_resource,
$last_error;
function db($hostname, $database, $username, $password) {
$this->hostname = $hostname;
$this->database = $database;
$this->username = $username;
$this->password = $password;
$this->connection = new mysqli($this->hostname, $this->username, $this->password, $this->database) or $this->choke("Can't connect to database");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if($this->database)
$this->database($this->database);
}
function database($database) {
$this->database = $database;
mysqli_select_db($this->connection, $this->database );
}
function query($query, $flag = DB_DEFAULT_FLAG) {
$this->last_query = $query;
//print_r($query);
$result = mysqli_query($this->connection, $query) or $this->choke("problem connecting to DB");
while($row=mysqli_fetch_assoc($result)) {
$resource[]=$row;
}
//print($command);
//print_r($resource);print("<br>");
list($command, $other) = preg_split("|\s+|", $query, 2);
// Load return according to query type...
switch(strtolower($command)) {
case("select"):
case("describe"):
case("desc"):
case("show"):
$return = array();
while($data = $this->resource_get($resource, $flag))
$return[] = $data;
//print_r($return);
break;
case("replace"):
case("insert"):
if($return = mysqli_insert_id($this->connection))
$this->last_i = $return;
break;
default:
$return = mysqli_affected_rows($this->connection);
}
return $return;
}
function resource_get($resource = NULL, $flag = DB_DEFAULT_FLAG) {
if(!$resource)
$resource = $this->last_resource;
return mysqli_fetch_array($resource, $flag);
}
So here's the problem: I've checked the results with a print_r() and the $resource array is loading correctly, but the value of $return when checked with print_r() just ends up being "Array()". Therefore, as near as I can figure, something isn't being handled correctly in this part of the code, which is why I included the resource_get() function call:
$return = array();
while($data = $this->resource_get($resource, $flag))
$return[] = $data;
//print_r($return);
break;
If I use mysqli_fetch_row($resource, $flag) instead of mysqli_fetch_array($resource, $flag) I still get the same result, i.e. print_r($return) yields simply "Array()".

Answer 1

The variable $resource does not represent a mysqli_result resource object at the time you pass it into $this->resource_get(). Instead, it is already a 2D array of results since you previously ran a mysqli_fetch_assoc() loop.
To make this work with your current code, you may either remove the earlier fetch loop:
$result = mysqli_query($this->connection, $query) or $this->choke("problem connecting to DB");
// Remove this
//while($row=mysqli_fetch_assoc($result)) {
// $resource[]=$row;
//}
And later, pass $result instead of $resource into your resource_get() method since it is $result that is the resource object.
Or, you might just skip the resource_get() call entirely and return $resource directly since it already contains the result array.