[dizpers]

Hi! I create new module. I have a main file (in jamroom root dir: /myModule.php ). I use this file as router. The code is like this:

Code

<?php require('include/jamroom-include.inc.php'); require("modules/myModule/include/myModule.model.php"); $GLOBALS['JR_SCRIPT_NAME'] = 'myModule.php'; $mMyModule = new MyModule_Model(); $_POST = getPostVars(); $_user = sessionVerify(); $language = getLanguage($_user['user_language']); // make routing switch ($_POST['action']) {     case 'show':         if (checkType($_POST['band_id'], 'number_nz')){             $mMyModule::showBand($_POST['band_id']);         }elseif($_POST['band_id'] == 'all'){             $mMyModule::showAll();         }else{             //error         }         exit;         break;     default:         echo 'Not enough actual parameters!';         exit;         break; } ?>

And in myModule.model.php I have the code like this:

Code

<?php defined('IN_JAMROOM') or exit(); class myModule_Model {     public function __construct(){             }     public function showAll(){         $req = "SELECT album_name                 FROM {$jamroom_db['jrSongInfo']}";         $res = dbQuery($req, 'album_name');         htmlShowTemplate('myModule', 'myModule.tpl', $res);     }     public function showBand($band_id){         $req = "SELECT album_name                 FROM {$jamroom_db['jrSongInfo']}                 WHERE band_id = $band_id";         $res = dbQuery($req, 'album_name');         htmlShowTemplate('myModule', 'myModule.tpl', $res);     } } ?>

For debug this code, I use print_r($req) in methods. And I see that

Code

{$jamroom_db['jrSongInfo']}

don't insert the table name... but in /modules/myModule/include.php I have:

Code

$jamroom_db['jrSongInfo'] = $config['db_prefix'] .'song_info';

Why I can't use {$jamroom_db['jrSongInfo']} to retrieve the name of the table?

[Michael]

do you use an IDE? I recommend PhpStorm, its awesome.

Code

public function showBand($band_id){         $req = "SELECT album_name                 FROM {$jamroom_db['jrSongInfo']}                 WHERE band_id = $band_id";         $res = dbQuery($req, 'album_name');         htmlShowTemplate('myModule', 'myModule.tpl', $res);     }

If you had that file open in PhpStorm the error checking would put a red line under $jamroom_db and say "this variable is undefined."

you need to add global to your function and call the var you want to use in.

Code

public function showBand($band_id){ global $jamroom_db;         $req = "SELECT album_name                 FROM {$jamroom_db['jrSongInfo']}                 WHERE band_id = $band_id";         $res = dbQuery($req, 'album_name');         htmlShowTemplate('myModule', 'myModule.tpl', $res);     }

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Michael Ussher
Jamroom Network Team Member: http://www.jamroom.net
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[dizpers]

Thx, I will try this now:)

PS
Hm, I use phpStorm (and yes - its really awesome!) and it's no red underlining $jamroom_db ...

[Michael]

This is what it looks like for me:



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Michael Ussher
Jamroom Network Team Member: http://www.jamroom.net
Priority Support: http://www.jamroom.net/Support_Center

[dizpers]

Hm... I added

Code

global $jamroom_db;

and it's dont solve the problem

any suggestions?

upd:

I do like this:

Code

<?php defined('IN_JAMROOM') or exit(); class AlbumsList_Model {     private $jamroom_db;     public function __construct(){         $this->jamroom_db = $GLOBALS['GLOBALS']['jamroom_db'];     }     public function showAll(){         print_r($jamroom_db);         $req = "SELECT song_album                 FROM {$jamroom_db['jrSongInfo']}";         $temp = dbQuery($req, 'album_name');         foreach ($temp as $key => $value){             $res['ALBUMS'][$key] = $value['song_album'];         }         htmlShowTemplate('jrAlbumsList', 'jrAlbumsList.tpl', $res);     }     public function showBand($band_id){         print_r($this->jamroom_db);         $req = "SELECT song_album                 FROM {$jamroom_db['jrSongInfo']}                 WHERE band_id = $band_id";         print_r($req);         $temp = dbQuery($req, 'NUMERIC');         foreach ($temp as $key => $value){             $res['ALBUMS'][$key] = $value['song_album'];         }         htmlShowTemplate('jrAlbumsList', 'jrAlbumsList.tpl', $res);     } } ?>

And print_r() don't print me my defined (in include.php of my module) table names. My include.php:

Code

<?php defined('IN_JAMROOM') or exit(); $jrAlbumsList = "Albums List"; // Our version //default stuff goes here //... // Define custom database tables // HERE I define table names $jamroom_db['jrAlbumsList'] = $config['db_prefix'] .'jrAlbumsList'; $jamroom_db['jrSongInfo'] = $config['db_prefix'] .'song_info'; ?>

[Michael]

Try this:

Code

<?php defined('IN_JAMROOM') or exit(); class AlbumsList_Model {     public function showAll(){         global $jamroom_db;         print_r($jamroom_db);         $req = "SELECT song_album                 FROM {$jamroom_db['jrSongInfo']}";         $temp = dbQuery($req, 'album_name');         foreach ($temp as $key => $value){             $res['ALBUMS'][$key] = $value['song_album'];         }         htmlShowTemplate('jrAlbumsList', 'jrAlbumsList.tpl', $res);     }     public function showBand($band_id){         global $jamroom_db;         print_r($jamroom_db);         $req = "SELECT song_album                 FROM {$jamroom_db['jrSongInfo']}                 WHERE band_id = $band_id";         print_r($req);         $temp = dbQuery($req, 'NUMERIC');         foreach ($temp as $key => $value){             $res['ALBUMS'][$key] = $value['song_album'];         }         htmlShowTemplate('jrAlbumsList', 'jrAlbumsList.tpl', $res);     } }

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Michael Ussher
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Priority Support: http://www.jamroom.net/Support_Center

[dizpers]

Doesn't work.

PS
In $GLOBALS array I can't see my vars (assigned in include.php of my module) - I think - it's my main problem here.

[Michael]

why are you looking in $GLOBALS? what do you expect to find there?

Code

$somevalue = "hello world"; first(); second(); function first(){     print "the first value is: ". $somevalue; } function second(){     global $somevalue;     print "<br>the second value is: ". $somevalue; }

will print:

Code

the first value is: the second value is: Hello World

Because $somevalue was defined outside of the functions it is inaccessible INSIDE the functions. so we need to put that global line in the function in order to make it accessible from within the function.

do this:
print_r(get_defined_vars());

and see if any variables you want are anywhere in there, then you can get the name of the var you want.


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Michael Ussher
Jamroom Network Team Member: http://www.jamroom.net
Priority Support: http://www.jamroom.net/Support_Center

[dizpers]

Nevertheless defining $jamroom_db as global variable works fine. You are right, thank you! But one problem does not disappear: table names which I define in include.php of my module are not in $jamroom_db now.

Code

<?php defined('IN_JAMROOM') or exit(); $jrAlbumsList = "Albums List"; // Our version //default stuff goes here //... // Define custom database tables // HERE I define table names $jamroom_db['jrAlbumsList'] = $config['db_prefix'] .'jrAlbumsList'; $jamroom_db['jrSongInfo'] = $config['db_prefix'] .'song_info'; ?>

So, I don't see 'jrSongInfo' and 'jrAlbumsList' in global $jamroom_db.

PS
I think, that in include.php of my module I should define $jamroom_db as global variable too.

[Michael]

your includes info looks correct to me.

If they are not in $jamroom_db the places i would check are:
* is the include.php in the correct location for your module
* is the module turned on in the admin panel
* are there any errors in the admin panel in the php error log
* are there any errors in the admin panel in the activity log.


_________________
Michael Ussher
Jamroom Network Team Member: http://www.jamroom.net
Priority Support: http://www.jamroom.net/Support_Center

[jamesd116]

it may be in your schema if something in there is incorrect so if you check your php log file it should tell u where a mistake may be made if thats where the problem exists at.


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