OOP PHP MySQL return multiple rows and variable

Question

I am new w/ OPP and big pardon if my question maybe too simple :) Table category, navigation, etc contains multiple rows (category : samsung, apple, etc; and navigation : about us, terms, etc) and both stand as Menu in all pages (home, product,etc)

My old php code and work good is below

    <div id="categories">
    <ul>
        <?
        $mydbcategories = new myDBC();
        $resultcategories = $mydbcategories->runQuery("SELECT * FROM `category`");
        while ($rowcategories = $mydbcategories->runFetchArray($resultcategories)) {
            echo '<li><a href="'.ROOT_URL.$rowcategories[url].'">'.$rowcategories[title].'</a></li>';
        }
        ?>
    </ul>
    </div>
    <div id="navigation">

    <ul>
        <?
        $mydbnavigation = new myDBC();
        $resultnavigation = $mydbnavigation->runQuery("SELECT * FROM `navigation`");
        while ($rownavigation = $mydbnavigation->runFetchArray($resultnavigation)) { echo '<li><a href="'.ROOT_URL.$rownavigation [url].'">'.$rownavigation [title].'</a></li>';
        }
        ?>
    </ul>
    </div>

I would like to implement OOP PHP and create class then store in classes.php

   <?
   class Menu{
   var $title;
   var $url; 
   function setMenu($db){
   $mydbMenu= new myDBC();
   $resultmenu = $mydbMenu->runQuery("SELECT * FROM `$db`");
   $resultmenurows = mysqli_num_rows($resultmenu);
   while ($rowmenu = $mydbMenu->runFetchArray($resultmenu)){
        $this->title = $rowmenu[title];
        $this->url = $rowmenu[url];
    }
  }
  function getTitle() { return $this->title;}
  function getUrl() { return $this->url;}
  }
  ?>

Then i'm edit my old code with new one below;

    <div id="categories">
    <ul>
    <?
   $catmenu = new Menu();
   while ($catmenu ->setMenu('category')) { 
       echo '<li><a href="'.ROOT_URL.$catmenu->getUrl().'">'.$catmenu->getTitle().'</a></li>';
        }
        ?>
    </ul>
    </div>

    <div id="navigation">
    <ul>
        <?
        $navmenu = new Menu();
        while ($navmenu ->setMenu('category')) {
  echo '<li><a href="'.ROOT_URL.$navmenu ->getUrl().'">'.$navmenu ->getTitle().'</a></li>';
        }
        ?>
    </ul>
 </div>

I tested and error maybe because there are multiple rows (from table) in the setMenu func. How can i return this multiple rows ? should i use array ? Please help me to solve this and any reply really very appreciate

I tested and error - What is the error?

Daniel W.
– Daniel W.

2014-02-25 08:51:32 +00:00
Commented Feb 25, 2014 at 8:51 — Daniel W.
– Daniel W., Commented Feb 25, 2014 at 8:51

Daniel W. · Accepted Answer · 2014-02-25 08:56:32Z

2

You are coding PHP4 OOP style, this is very outdated. Don't use var, use public, protected, private.
$this->title = $rowmenu[title] in here, title is used as a constant (no quotes), proper: $this->title = $rowmenu['title'], same with $rowcategories[title]
"SELECT * FROM $db" is this correct? Or do you mean SELECT * FROM menu WHERE xxx='" . $db . "', do you catch errors if the lookup fails?

You should also look at PHP design patterns and code style to improve!

answered Feb 25, 2014 at 8:56

Daniel W.

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Comments

Hassan · Accepted Answer · 2014-02-25 08:59:19Z

-1

Try following PHP code

<?

class Menu {

var $title;
var $url;

    function setMenu($db) {
        $mydbMenu = new myDBC();
        $resultmenu = $mydbMenu->runQuery("SELECT * FROM `$db`");
        $resultmenurows = mysqli_num_rows($resultmenu);
        $this->title = array();
        $this->url = array();
        while ($rowmenu = $mydbMenu->runFetchArray($resultmenu)) {
            $this->title[] = $rowmenu['title'];
            $this->url[] = $rowmenu['url'];
        }
    }

    function getTitle($ind) {
        return $this->title[$ind];
    }

    function getUrl($ind) {
        return $this->url[$ind];
    }

}
?>

And HTML

<div id="categories">
    <ul>
        <?
        $catmenu = new Menu();
        $catmenu->setMenu('category');
        $i = 0;
        while ($catmenu->getTitle($i)) {
            echo '<li><a href="' . ROOT_URL . $catmenu->getUrl($i) . '">' . $catmenu->getTitle($i) . '</a></li>';
            $i++;
        }
        ?>
    </ul>
</div>

<div id="navigation">
    <ul>
        <?
        $navmenu = new Menu();
        $navmenu->setMenu('navigation');
        while ($navmenu->getTitle($i)) {
            echo '<li><a href="' . ROOT_URL . $navmenu->getUrl($i) . '">' . $navmenu->getTitle($i) . '</a></li>';
            $i++;
        }
        ?>
    </ul>
</div>

answered Feb 25, 2014 at 8:59

Hassan

7427 silver badges13 bronze badges

3 Comments

user3349980 Over a year ago

Thank you for answered, it's resolved now.once again thank you very much :)

Hassan Over a year ago

If this answer helps you, then please up vote or accept this answer as correct answer

Hassan Over a year ago

I think you need to do it again. Your first mark it as accepted and than you again click so it marked as not accepted :)

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