Need to output MYSQL as array(1 => 'January', 'feb', 'March');

russellharrower · 07-17-2009, 12:46 PM

Hi I need to be able to print the output of the mysql request ($quarry) into a array(1 => 'January', 'feb', 'March');

But instead of January, feb and March it needs to have the title output.

If you can help that would be great.

Thanks

Code:

$query = "SELECT id, title, content FROM push_content WHERE type = 0 ORDER by date DESC LIMIT 1";


if ($result = $mysqli->query($query)) {

    /* fetch associative array */
    while ($obj = mysqli_fetch_object($result)) {
        printf ("%s (%s)\n", $obj->title, $obj->content);
		echo"<br/>";
    }

    /* free result set */
    mysqli_free_result($result);

tony · 07-17-2009, 02:40 PM

why don't you use the fetch associative array static method instead of the mysqli_fetch_object?

russellharrower · 07-17-2009, 02:45 PM

because i need the ability to call each title by them self.
if you can do an example on how i would do that using that code, that would be helpful.

xenon · 07-17-2009, 05:15 PM

If you go object oriented style, keep it that way:

PHP Code:


			
$query = "SELECT id, title, content FROM push_content WHERE type = 0 ORDER by date DESC LIMIT 1";


if ($result = $mysqli->query($query)) {

    /* fetch associative array */
    while ($obj = $result->fetch_object()) {
        printf ("%s (%s)\n", $obj->title, $obj->content);
        echo"<br/>";
    }
}

What exactly are you trying to do, though? You may want to give some examples, as well, so we can figure out how to help you. I, for one, did not understand how using an object instead of an array help you.

russellharrower · 07-18-2009, 02:50 PM

Quote:

Originally Posted by xenon

If you go object oriented style, keep it that way:

PHP Code:


			
$query = "SELECT id, title, content FROM push_content WHERE type = 0 ORDER by date DESC LIMIT 1";





if ($result = $mysqli->query($query)) {



    /* fetch associative array */

    while ($obj = $result->fetch_object()) {

        printf ("%s (%s)\n", $obj->title, $obj->content);

        echo"<br/>";

    }

}

What exactly are you trying to do, though? You may want to give some examples, as well, so we can figure out how to help you. I, for one, did not understand how using an object instead of an array help you.

Thanks xexon,
I want to be able to call each title like
$title1 = "this is title 1";
$title2 = "this is title2";
$title3 = "this is title3";
etc...

As I want to allow the template designer to choose how many stories to show on the front page in there design.

I also want to show the content this way.
I then want to come up with a way to have global plugins that will allow the users to choose if they want to show the posters name, date published etc...

I hope that helps.

Thanks

adamdecaf · 07-18-2009, 04:58 PM

Well, this would be easier.

PHP Code:


			
$query = "SELECT id, title, content FROM push_content WHERE type = 0 ORDER by date DESC LIMIT 1"; 
$data = Array("title", "content");
$n = 0;

if ($result = $mysqli->query($query)) { 

    /* fetch associative array */ 
    while ($obj = $result->fetch_object()) { 
        $data['title'][$n] = $obj->title;
        $data['content'][$n] = $obj->content;
        $n++;
    } 
}

As for the option to display content I would store that into the DB and then only query for the ones that are allowed for the situation.

russellharrower · 07-18-2009, 05:02 PM

Quote:

Originally Posted by adamdecaf

Well, this would be easier.

PHP Code:


			
$query = "SELECT id, title, content FROM push_content WHERE type = 0 ORDER by date DESC LIMIT 1"; 

$data = Array("title", "content");

$n = 0;



if ($result = $mysqli->query($query)) { 



    /* fetch associative array */ 

    while ($obj = $result->fetch_object()) { 

        $data['title'][$n] = $obj->title;

        $data['content'][$n] = $obj->content;

        $n++;

    } 

}

As for the option to display content I would store that into the DB and then only query for the ones that are allowed for the situation.

Hi Just wondering what if i need to show more then 1 say like 10

adamdecaf · 07-18-2009, 05:07 PM

Quote:

Originally Posted by russellharrower

Hi Just wondering what if i need to show more then 1 say like 10

PHP Code:


			
for ($n = 0; $n <= 9; $n++) {

   echo "<p><strong>" . $data['title'][$n] . "</strong><br />";

   echo $data['content'][$n] . "</p>";

}

Something like that...

russellharrower · 07-18-2009, 05:11 PM

But i need to give each one a $title->1, $title->2 etc that code will only print the data

adamdecaf · 07-18-2009, 05:59 PM

Quote:

Originally Posted by russellharrower

But i need to give each one a $title->1, $title->2 etc that code will only print the data

Why will

PHP Code:


			
$data['title'][$n]

not work?

russellharrower · 07-19-2009, 06:20 AM

have a look what that does http://www.designvisa.com/beta

adamdecaf · 07-19-2009, 06:41 PM

Quote:

Originally Posted by russellharrower

have a look what that does http://www.designvisa.com/beta

The page is empty.

Village Idiot · 07-19-2009, 08:45 PM

Quote:

Originally Posted by russellharrower

have a look what that does http://www.designvisa.com/beta

I'm not sure objects are required for that.

russellharrower · 07-20-2009, 05:51 AM

http://www.designvisa.com/beta/index2.php sorry about that