whats wrong here? [php]

unknowngiver · 01-14-2007, 02:10 AM

hey
i am not getting an error..it just doesnt show anything when it should

PHP Code:

          function loadMedia($limit,$cid=-1)

????: NamePros.com http://www.namepros.com/programming/281056-whats-wrong-here-php.html
        {

        $media = array();

        if ($cid == -1) {

        $where = "WHERE `status` = 1"; }

        $q = "SELECT * FROM `media` $where order by `timestamp` desc LIMIT 0,$limit";

        $result = $this->query($q);

        for ($b = 0; $b <(mysql_num_rows($result));$b++)

        {

            $db = mysql_fetch_array($result);

????: NamePros.com http://www.namepros.com/showthread.php?t=281056
            $media[$b] = array($db['id'],$db['cat_id'],$db['by_id'],$db['file_name'],$db['dp'],$db['title'],

                        $db['description'],$db['views'],$db['timestamp']);

        }

        return $media or die ("cant");;

        }

and then :

PHP Code:

<?$media = $database->loadMedia(1); echo$media['0']['5'];?>

Shorty · 01-14-2007, 05:34 AM

It sounds like $cid != -1.

Either do "$cid = -1" before you call the function, or do it after $media = array();.

Matthew. · 01-14-2007, 05:44 AM

EDIT:
I didn't notice before.

Change:

PHP Code:

  return $media or die ("cant");;

To:

PHP Code:

  return $media; 

????: NamePros.com http://www.namepros.com/showthread.php?t=281056

That should fix it. With the 'or die' currently in the function, it will not be returning the array.

Possible Shorty but since he did not supply the second argument when using the function it should have been -1 by default. By it could be

Run this:

PHP Code:

  echo '<pre>';

print_r( $media )

echo '</pre>';

To see if the array is populated.

As a side note it does not matter in this case however you do not use '' for numerical array keys.

Dan · 01-14-2007, 10:00 AM

You should have had an error for an unexpected semi colon, too. When you are trying to find what's wrong, use error_reporting(E_ALL);

Matthew. · 01-14-2007, 10:07 AM

Originally Posted by Dan

You should have had an error for an unexpected semi colon, too. When you are trying to find what's wrong, use error_reporting(E_ALL);

The extra semi colon wasn't the problem. (although could well have been in other cases)

I tried a little experiment since i didn't know 100% if it was the problem but:

PHP Code:

  function test($test) 

{

    $string = 'hi';

    

    if($test == true)

    {

        return $string or die('cant');

    }

    else

    {

        return $string;

    }

}

PHP Code:

  echo test(false);

// will output "hi"

 
echo test(true);

// will output "1"

Unsure as to why but adding the or die seems to return 1.

ahtum · 01-14-2007, 10:18 AM

Here i rewrote a little your code.

Hope it works

If it's not working then you might have problems with your the mysql class that you are using.
????: NamePros.com http://www.namepros.com/showthread.php?t=281056

PHP Code:

  function loadMedia ( $limit , $cid = '-1' ){

    

    $media = array();

    

    if ($cid == '-1'){

        $where = "WHERE `status` = 1";

????: NamePros.com http://www.namepros.com/showthread.php?t=281056
    }else{

        $where = "";

    }

    

    $q = "SELECT * FROM `media` $where order by `timestamp` desc LIMIT 0,$limit";

    $result = $this->query($q);

    

    $db = mysql_fetch_array($result);

    

    for ($b = 0; $b <(mysql_num_rows($result));$b++){

            

        $media[$b] = array    (    $db['id'],$db['cat_id'],$db['by_id'],$db['file_name'],$db['dp'],

                                $db['title'],$db['description'],$db['views'],$db['timestamp']

                            );

    }

    

    return $media;

}

Dan · 01-14-2007, 10:20 AM

If you change

if($test == true)
to
if($test === true)

it will run just return $string;

I can't figure out why it makes it return 1..
"void exit ( [string status] )"

Ahtum, you need to have it get the row inside of the for loop to get all of the rows. What you did would only get one.

Matthew. · 01-14-2007, 10:23 AM

Originally Posted by Dan

If you change

if($test == true)
to
if($test === true)

it will run just return $string;

No it wont?

Dan · 01-14-2007, 10:25 AM

Originally Posted by Matthew.

No it wont?

... not if you use true... He's not using true as the input, so your test doesn't necessarily do anything except for showing that it's weird how it returns 1.

Matthew. · 01-14-2007, 10:30 AM

Originally Posted by Dan

... not if you use true... He's not using true as the input, so your test doesn't necessarily do anything except for showing that it's weird how it returns 1.

Which was the aim of my test

The conditional is irrelevant, it was just to make it easier to swap between the 2 (I'm bored gimme a break.)
????: NamePros.com http://www.namepros.com/showthread.php?t=281056

PHP Code:

  function test($string) 

{

    return $string or die("cant");

}

I'm determined to find out why it returns 1...

Amnezia · 01-14-2007, 12:21 PM

Originally Posted by Matthew.

Which was the aim of my test

The conditional is irrelevant, it was just to make it easier to swap between the 2 (I'm bored gimme a break.)
????: NamePros.com http://www.namepros.com/showthread.php?t=281056

PHP Code:

  function test($string) 

{

    return $string or die("cant");

}

I'm determined to find out why it returns 1...

return is a langauge construct not a function therefore it does not return an exit code.

in other words

return $somestring or die("some message") is a pointless line of code

Matthew. · 01-14-2007, 12:41 PM

Originally Posted by Amnezia

return is a langauge construct not a function therefore it does not return an exit code.
????: NamePros.com http://www.namepros.com/showthread.php?t=281056

in other words

return $somestring or die("some message") is a pointless line of code

That i know (which is why i said remove it from the original code)

My point/query was why it returns 1

It is not important, merely something i was wondering

01-14-2007, 02:10 AM	THREAD STARTER #1 (permalink)
unknowngiver Senior Member Join Date: May 2005 Location: Ontario Canada Posts: 3,088	whats wrong here? [php] hey i am not getting an error..it just doesnt show anything when it should PHP Code: function loadMedia($limit,$cid=-1) ????: NamePros.com http://www.namepros.com/programming/281056-whats-wrong-here-php.html { $media = array(); if ($cid == -1) { $where = "WHERE `status` = 1"; } $q = "SELECT * FROM `media` $where order by `timestamp` desc LIMIT 0,$limit"; $result = $this->query($q); for ($b = 0; $b <(mysql_num_rows($result));$b++) { $db = mysql_fetch_array($result); ????: NamePros.com http://www.namepros.com/showthread.php?t=281056 $media[$b] = array($db['id'],$db['cat_id'],$db['by_id'],$db['file_name'],$db['dp'],$db['title'], $db['description'],$db['views'],$db['timestamp']); } return $media or die ("cant");; } and then : PHP Code: `<?$media = $database->loadMedia(1); echo$media['0']['5'];?>` __________________ Simplifying Engineering \|\| $$ Currency Converter $$ Julsh - Community/Dating Website \|\| Free Image hosting Fresh Proxies \|\| Random Proxy \|\| Troll Board

01-14-2007, 05:44 AM	#3 (permalink)
Matthew. Senior Member Join Date: Dec 2006 Location: England Posts: 1,565	EDIT: I didn't notice before. Change: PHP Code: `return $media or die ("cant");;` To: PHP Code: `return $media; ????: NamePros.com http://www.namepros.com/showthread.php?t=281056` That should fix it. With the 'or die' currently in the function, it will not be returning the array. Possible Shorty but since he did not supply the second argument when using the function it should have been -1 by default. By it could be Run this: PHP Code: `echo '<pre>'; print_r( $media ) echo '</pre>';` To see if the array is populated. As a side note it does not matter in this case however you do not use '' for numerical array keys. __________________ University Grade Calculator juegosjuegos Your Hotel in Austria Matratzen Last edited by Matthew.; 01-14-2007 at 05:51 AM.

01-14-2007, 10:18 AM	#6 (permalink)
ahtum Senior Member Join Date: May 2006 Posts: 1,327	Here i rewrote a little your code. Hope it works If it's not working then you might have problems with your the mysql class that you are using. ????: NamePros.com http://www.namepros.com/showthread.php?t=281056 PHP Code: function loadMedia ( $limit , $cid = '-1' ){ $media = array(); if ($cid == '-1'){ $where = "WHERE `status` = 1"; ????: NamePros.com http://www.namepros.com/showthread.php?t=281056 }else{ $where = ""; } $q = "SELECT * FROM `media` $where order by `timestamp` desc LIMIT 0,$limit"; $result = $this->query($q); $db = mysql_fetch_array($result); for ($b = 0; $b <(mysql_num_rows($result));$b++){ $media[$b] = array ( $db['id'],$db['cat_id'],$db['by_id'],$db['file_name'],$db['dp'], $db['title'],$db['description'],$db['views'],$db['timestamp'] ); } return $media; }

01-14-2007, 05:34 AM	#2 (permalink)
Shorty Senior Member Join Date: Sep 2005 Location: England Posts: 1,034	It sounds like $cid != -1. Either do "$cid = -1" before you call the function, or do it after $media = array();.

01-14-2007, 10:00 AM	#4 (permalink)
Dan Buy my domains. Join Date: Feb 2006 Posts: 2,792	You should have had an error for an unexpected semi colon, too. When you are trying to find what's wrong, use error_reporting(E_ALL);

01-14-2007, 10:20 AM	#7 (permalink)
Dan Buy my domains. Join Date: Feb 2006 Posts: 2,792	If you change if($test == true) to if($test === true) it will run just return $string; I can't figure out why it makes it return 1.. "void exit ( [string status] )" Ahtum, you need to have it get the row inside of the for loop to get all of the rows. What you did would only get one.

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