Home »
Imported messages »
comp.lang.php »
determine a mysqli_result object
| Re: determine a mysqli_result object [message #181269 is a reply to message #181268] |
Sat, 27 April 2013 23:18  |
Scott Johnson
Messages: 196
Registered: January 2012
Karma:
|
Senior Member |
|
|
On 4/27/2013 4:11 PM, Scott Johnson wrote:
> I have a class that I am converting from mysql to mysqli in PHP 5.2.x
>
> The problem I ran into is this.
>
> Previously I could check the mysql_query if it return a resource by using:
>
> is_resource($resource)
>
> and could parse the resource using mysql_fetch_assoc or similar
>
> and if it was an insert, update, etc.. it would not return a resource
> and I would just return a true and carry on.
>
> Now with mysqli it return a mysql_result object.
>
> How do I check for this so I know whether to parse or return
>
> Thanks
>
> Here is a nutshell of what was using for mysql
>
> // Already connected to mysqli
>
> public function query($query){
> $resource = $this->connection->query($query);
>
> if($resource) {
>
> if(is_resource($resource)) { // parse
>
> $data = array();
>
> $idx = 0;
>
> while($result = $this->connection->fetch_assoc()) {
>
> $data[$idx] = $result;
> // Object assembly
>
> }
> } else { // return
>
> // update, insert, etc..
> return true;
>
> }
> } else {
> // error
> }
> }
>
> Scotty
Crap figured it out. Was binging for the wrong answer.
if($result_obj) {
if($result_obj instanceof mysqli_result) {
echo 'good';
$data = array();
$idx = 0;
while($result = $result_obj->fetch_assoc()) {
$data[$idx] = $result;
$idx++;
}
print_r($data);
} else {
return true;
}
}
Scotty
|
|
|
|
Current Time: Sun Nov 24 20:00:41 GMT 2024
Total time taken to generate the page: 0.05425 seconds
Calendar
Search
Help
Members
Register
Login
Home
]